ways to factor… there a few different ways to approaching factoring an expression. however, the...
TRANSCRIPT
Ways to factorhellip
There a few different ways to approaching factoring an
expression However the first thing you should always
look for is the Greatest Common Factor (GCF)
How to find the GCF video
Factoring using GCF video
Why GCF firstYou look for the GCF first because it will help you
factor quadratics using the second method by making the numbers smaller
The second method of factor involves undoing the distributive propertyhellipI call it unFOILing
There are several videos posted on this method but herersquos an examplehellip
Factor x2 + 6x + 8
Look at the last numberIf the sign in a positive the signs in the parenthesis will be the same
x2 + 6x + 8Here the 8 is positive
Look at the sign on the middle numberWe know the signs will be the same because 8 is positive We look a the middle number and its also positive So both signs in the parenthesis will be positive
(x + )(x + )
Find factors of the last number that when you mulitply them you get that last number but when you combine them you get the middle numberSo were looking for factors of 8 that we multiply them we get an 8 but when we add them we get a 64 and 2
(x + 4)(x + 2)
Check it with FOILYou never get a factoring problem wrong You can always check it by multiplying
(x + 4)(x + 2) = x2 + 4x +2x +8
It works
Factor x2 - 3x - 54
Special Case The Difference of two Perfect Squares
The difference of two perfect squares is very easy tofactor but everyone always forgets about themTheyre in the form (ax)2 - c where a and c are perfect squares
Theres no visible b-valueso b = 0 You factor them by taking the square root of a and the square root of c and placing them in parenthesis that have opposite signs
Whenever you have a binomial that is subtraction always check to
see itrsquos this special case It usually does NOT have a GCF
Heres an examplehellip
ExampleFactor 4x2 ndash 9
Set up parenthesis with opposite signs
( + )( - )
Find the square root of a and place then answer in the front sections of the parenthesissqrt(4x2) = 2x
( 2x + )( 2x - )
Find the square root of c and place them at the end of the parenthesissqrt(9) = 3
( 2x + 3 )( 2x - 3 )
Difference of Two Perfect Squares Video
Practice Factoring1 x2 + 4x ndash 5 2 x2 - 3x + 2 3 x2 - 6x ndash 7 4 x2 + 4x + 4
Solutions1 x2 + 4x ndash 5 = (x+5)(x-1)2 x2 - 3x + 2 = (x-1)(x-2)3 x2 - 6x ndash 7 = (x-7)(x+1)4 x2 + 4x + 4 = (x +2)(x+2)
Practice Differen
ce of Two
Squares
Factoring with
Algebra Tiles
Practice
Common
Factors
Practice
Factor a = 1
What if the leading coefficient isnrsquot a 1Factor 3x2 + 11x - 4
Set up two pairs of parenthesis ( )( )Look over the equation
( + )( - )
Look at the a-valueUnfortunately the a-value is not a one so we need to list factors in a chart Were looking for the pair of factors that when I find the difference of the productswill yield the b-value
Factors of A Factors of C
1 3 22 and 14 12 - 32 = -4 NO
13 - 14= -1 NO 11 - 34= -12 YES
Enter in values (x - 4)(3x + 1)Check with FOILIts possible that you have the right numbers but in the wrong spots so you have to check
(x - 4)(3x + 1)= 3x^2 -12x + x - 4 = 3x^2 -11x -4
Factoring when ane 1
Terms in a quadratic expression may have some common factors before you break them down into
linear factors
Remember the greatest common factor GCF is thegreatest number that is a factor of all terms in the
expression
When a ne 1 weshould always check to see if the quadratic expression
has a greatest common factor
Factor 2x2 -22x +36Step 1
a ne 1 so we should check to see if the quadratic expression has a greatest common factor
It has a GCF of 22x2 -22x +36 = 2(x2 -11x +18)
Step 2Once we factor out the GCF the quadratic expression now
has a value of a =1 and we can use the process we just went through in the
previous examples
x2 -11x +18 = (x -2)(x-9)
Therefore 2x2 -22x +36 is = 2 (x -2)(x-9)
Ane 1 and NO GCF2x2 + 13x ndash 7
Step 1 a ne 1 so we should check to see if the quadratic expression has a greatest common factor
It does not have a GCFThis type of trinomial is much more difficult to
factor than the previous Instead of factoring the c value alone one has to also factor the a value
Our factors of a become coefficients of our x-terms and the factors of c will go right where they did in the previous examples
2x2 + 13x ndash 7Step 1 Find the product ac
ac= -14Step 2 Find two factors of ac that add to give
b 1048633 1 and -14 = -13 1048633 -1 and 14 = 13 This is our winner 1048633 2 and -7 = -5 1048633 -2 and 7 = 5
Step 3 Split the middle term into two terms using the numbers found in step above
2x2 -1x + 14x ndash 7
Step 4 Factor out the common binomial using the box method
2x2 -1x + 14x ndash 7
Quadratic Term
Factor 1
Factor 2 Constant Term
2x2 -1x
14x -7
Find the GCF for each column and row
Numbers in RED represent the GCF of each row and column
2x -1
x 2x2 -1x
7 14x -7
The factors are (x + 7)(2x - 1)
Practice Factoring1 2x2 11x + 5 2 3x2 - 5x - 2 3 7x2 - 16x + 4 4 3x2 + 12x + 12
Solutions1 2x2 +11x + 5 = (2x + 1)(x + 5)2 3x2 - 5x - 2 = (3x + 1)(x - 2)3 7x2 - 16x + 4 = (7x - 2)(x - 2)4 3x2 + 12x + 12 = 3(x + 2)(x + 2)
Special Products
Factoring Strategies
Prime Factors
RememberThis wonrsquot work for all quadratic trinomials because not all quadratic trinomials can be
factored into products of binomials with integer coefficients
We call these prime (Prime Numbers are 3 5 7 11 13 etc)Expressions such as x2 + 2x - 7 cannot be factored at all and is therefore known as a
prime polynomial
Practicing Factoring when a ne1
Please watch the demonstration below on factoring when a ne 1 There will be
interactive examples provided to help when a ne 1
MORE FACTORING
Upon completion of the video and demonstration please complete Mastery
Assignment Part 2
Gizmo Factoring
ax2 + bx + c
More InstructionPractice
Application
Problems
Practice All
Other Cases
Why GCF firstYou look for the GCF first because it will help you
factor quadratics using the second method by making the numbers smaller
The second method of factor involves undoing the distributive propertyhellipI call it unFOILing
There are several videos posted on this method but herersquos an examplehellip
Factor x2 + 6x + 8
Look at the last numberIf the sign in a positive the signs in the parenthesis will be the same
x2 + 6x + 8Here the 8 is positive
Look at the sign on the middle numberWe know the signs will be the same because 8 is positive We look a the middle number and its also positive So both signs in the parenthesis will be positive
(x + )(x + )
Find factors of the last number that when you mulitply them you get that last number but when you combine them you get the middle numberSo were looking for factors of 8 that we multiply them we get an 8 but when we add them we get a 64 and 2
(x + 4)(x + 2)
Check it with FOILYou never get a factoring problem wrong You can always check it by multiplying
(x + 4)(x + 2) = x2 + 4x +2x +8
It works
Factor x2 - 3x - 54
Special Case The Difference of two Perfect Squares
The difference of two perfect squares is very easy tofactor but everyone always forgets about themTheyre in the form (ax)2 - c where a and c are perfect squares
Theres no visible b-valueso b = 0 You factor them by taking the square root of a and the square root of c and placing them in parenthesis that have opposite signs
Whenever you have a binomial that is subtraction always check to
see itrsquos this special case It usually does NOT have a GCF
Heres an examplehellip
ExampleFactor 4x2 ndash 9
Set up parenthesis with opposite signs
( + )( - )
Find the square root of a and place then answer in the front sections of the parenthesissqrt(4x2) = 2x
( 2x + )( 2x - )
Find the square root of c and place them at the end of the parenthesissqrt(9) = 3
( 2x + 3 )( 2x - 3 )
Difference of Two Perfect Squares Video
Practice Factoring1 x2 + 4x ndash 5 2 x2 - 3x + 2 3 x2 - 6x ndash 7 4 x2 + 4x + 4
Solutions1 x2 + 4x ndash 5 = (x+5)(x-1)2 x2 - 3x + 2 = (x-1)(x-2)3 x2 - 6x ndash 7 = (x-7)(x+1)4 x2 + 4x + 4 = (x +2)(x+2)
Practice Differen
ce of Two
Squares
Factoring with
Algebra Tiles
Practice
Common
Factors
Practice
Factor a = 1
What if the leading coefficient isnrsquot a 1Factor 3x2 + 11x - 4
Set up two pairs of parenthesis ( )( )Look over the equation
( + )( - )
Look at the a-valueUnfortunately the a-value is not a one so we need to list factors in a chart Were looking for the pair of factors that when I find the difference of the productswill yield the b-value
Factors of A Factors of C
1 3 22 and 14 12 - 32 = -4 NO
13 - 14= -1 NO 11 - 34= -12 YES
Enter in values (x - 4)(3x + 1)Check with FOILIts possible that you have the right numbers but in the wrong spots so you have to check
(x - 4)(3x + 1)= 3x^2 -12x + x - 4 = 3x^2 -11x -4
Factoring when ane 1
Terms in a quadratic expression may have some common factors before you break them down into
linear factors
Remember the greatest common factor GCF is thegreatest number that is a factor of all terms in the
expression
When a ne 1 weshould always check to see if the quadratic expression
has a greatest common factor
Factor 2x2 -22x +36Step 1
a ne 1 so we should check to see if the quadratic expression has a greatest common factor
It has a GCF of 22x2 -22x +36 = 2(x2 -11x +18)
Step 2Once we factor out the GCF the quadratic expression now
has a value of a =1 and we can use the process we just went through in the
previous examples
x2 -11x +18 = (x -2)(x-9)
Therefore 2x2 -22x +36 is = 2 (x -2)(x-9)
Ane 1 and NO GCF2x2 + 13x ndash 7
Step 1 a ne 1 so we should check to see if the quadratic expression has a greatest common factor
It does not have a GCFThis type of trinomial is much more difficult to
factor than the previous Instead of factoring the c value alone one has to also factor the a value
Our factors of a become coefficients of our x-terms and the factors of c will go right where they did in the previous examples
2x2 + 13x ndash 7Step 1 Find the product ac
ac= -14Step 2 Find two factors of ac that add to give
b 1048633 1 and -14 = -13 1048633 -1 and 14 = 13 This is our winner 1048633 2 and -7 = -5 1048633 -2 and 7 = 5
Step 3 Split the middle term into two terms using the numbers found in step above
2x2 -1x + 14x ndash 7
Step 4 Factor out the common binomial using the box method
2x2 -1x + 14x ndash 7
Quadratic Term
Factor 1
Factor 2 Constant Term
2x2 -1x
14x -7
Find the GCF for each column and row
Numbers in RED represent the GCF of each row and column
2x -1
x 2x2 -1x
7 14x -7
The factors are (x + 7)(2x - 1)
Practice Factoring1 2x2 11x + 5 2 3x2 - 5x - 2 3 7x2 - 16x + 4 4 3x2 + 12x + 12
Solutions1 2x2 +11x + 5 = (2x + 1)(x + 5)2 3x2 - 5x - 2 = (3x + 1)(x - 2)3 7x2 - 16x + 4 = (7x - 2)(x - 2)4 3x2 + 12x + 12 = 3(x + 2)(x + 2)
Special Products
Factoring Strategies
Prime Factors
RememberThis wonrsquot work for all quadratic trinomials because not all quadratic trinomials can be
factored into products of binomials with integer coefficients
We call these prime (Prime Numbers are 3 5 7 11 13 etc)Expressions such as x2 + 2x - 7 cannot be factored at all and is therefore known as a
prime polynomial
Practicing Factoring when a ne1
Please watch the demonstration below on factoring when a ne 1 There will be
interactive examples provided to help when a ne 1
MORE FACTORING
Upon completion of the video and demonstration please complete Mastery
Assignment Part 2
Gizmo Factoring
ax2 + bx + c
More InstructionPractice
Application
Problems
Practice All
Other Cases
Factor x2 + 6x + 8
Look at the last numberIf the sign in a positive the signs in the parenthesis will be the same
x2 + 6x + 8Here the 8 is positive
Look at the sign on the middle numberWe know the signs will be the same because 8 is positive We look a the middle number and its also positive So both signs in the parenthesis will be positive
(x + )(x + )
Find factors of the last number that when you mulitply them you get that last number but when you combine them you get the middle numberSo were looking for factors of 8 that we multiply them we get an 8 but when we add them we get a 64 and 2
(x + 4)(x + 2)
Check it with FOILYou never get a factoring problem wrong You can always check it by multiplying
(x + 4)(x + 2) = x2 + 4x +2x +8
It works
Factor x2 - 3x - 54
Special Case The Difference of two Perfect Squares
The difference of two perfect squares is very easy tofactor but everyone always forgets about themTheyre in the form (ax)2 - c where a and c are perfect squares
Theres no visible b-valueso b = 0 You factor them by taking the square root of a and the square root of c and placing them in parenthesis that have opposite signs
Whenever you have a binomial that is subtraction always check to
see itrsquos this special case It usually does NOT have a GCF
Heres an examplehellip
ExampleFactor 4x2 ndash 9
Set up parenthesis with opposite signs
( + )( - )
Find the square root of a and place then answer in the front sections of the parenthesissqrt(4x2) = 2x
( 2x + )( 2x - )
Find the square root of c and place them at the end of the parenthesissqrt(9) = 3
( 2x + 3 )( 2x - 3 )
Difference of Two Perfect Squares Video
Practice Factoring1 x2 + 4x ndash 5 2 x2 - 3x + 2 3 x2 - 6x ndash 7 4 x2 + 4x + 4
Solutions1 x2 + 4x ndash 5 = (x+5)(x-1)2 x2 - 3x + 2 = (x-1)(x-2)3 x2 - 6x ndash 7 = (x-7)(x+1)4 x2 + 4x + 4 = (x +2)(x+2)
Practice Differen
ce of Two
Squares
Factoring with
Algebra Tiles
Practice
Common
Factors
Practice
Factor a = 1
What if the leading coefficient isnrsquot a 1Factor 3x2 + 11x - 4
Set up two pairs of parenthesis ( )( )Look over the equation
( + )( - )
Look at the a-valueUnfortunately the a-value is not a one so we need to list factors in a chart Were looking for the pair of factors that when I find the difference of the productswill yield the b-value
Factors of A Factors of C
1 3 22 and 14 12 - 32 = -4 NO
13 - 14= -1 NO 11 - 34= -12 YES
Enter in values (x - 4)(3x + 1)Check with FOILIts possible that you have the right numbers but in the wrong spots so you have to check
(x - 4)(3x + 1)= 3x^2 -12x + x - 4 = 3x^2 -11x -4
Factoring when ane 1
Terms in a quadratic expression may have some common factors before you break them down into
linear factors
Remember the greatest common factor GCF is thegreatest number that is a factor of all terms in the
expression
When a ne 1 weshould always check to see if the quadratic expression
has a greatest common factor
Factor 2x2 -22x +36Step 1
a ne 1 so we should check to see if the quadratic expression has a greatest common factor
It has a GCF of 22x2 -22x +36 = 2(x2 -11x +18)
Step 2Once we factor out the GCF the quadratic expression now
has a value of a =1 and we can use the process we just went through in the
previous examples
x2 -11x +18 = (x -2)(x-9)
Therefore 2x2 -22x +36 is = 2 (x -2)(x-9)
Ane 1 and NO GCF2x2 + 13x ndash 7
Step 1 a ne 1 so we should check to see if the quadratic expression has a greatest common factor
It does not have a GCFThis type of trinomial is much more difficult to
factor than the previous Instead of factoring the c value alone one has to also factor the a value
Our factors of a become coefficients of our x-terms and the factors of c will go right where they did in the previous examples
2x2 + 13x ndash 7Step 1 Find the product ac
ac= -14Step 2 Find two factors of ac that add to give
b 1048633 1 and -14 = -13 1048633 -1 and 14 = 13 This is our winner 1048633 2 and -7 = -5 1048633 -2 and 7 = 5
Step 3 Split the middle term into two terms using the numbers found in step above
2x2 -1x + 14x ndash 7
Step 4 Factor out the common binomial using the box method
2x2 -1x + 14x ndash 7
Quadratic Term
Factor 1
Factor 2 Constant Term
2x2 -1x
14x -7
Find the GCF for each column and row
Numbers in RED represent the GCF of each row and column
2x -1
x 2x2 -1x
7 14x -7
The factors are (x + 7)(2x - 1)
Practice Factoring1 2x2 11x + 5 2 3x2 - 5x - 2 3 7x2 - 16x + 4 4 3x2 + 12x + 12
Solutions1 2x2 +11x + 5 = (2x + 1)(x + 5)2 3x2 - 5x - 2 = (3x + 1)(x - 2)3 7x2 - 16x + 4 = (7x - 2)(x - 2)4 3x2 + 12x + 12 = 3(x + 2)(x + 2)
Special Products
Factoring Strategies
Prime Factors
RememberThis wonrsquot work for all quadratic trinomials because not all quadratic trinomials can be
factored into products of binomials with integer coefficients
We call these prime (Prime Numbers are 3 5 7 11 13 etc)Expressions such as x2 + 2x - 7 cannot be factored at all and is therefore known as a
prime polynomial
Practicing Factoring when a ne1
Please watch the demonstration below on factoring when a ne 1 There will be
interactive examples provided to help when a ne 1
MORE FACTORING
Upon completion of the video and demonstration please complete Mastery
Assignment Part 2
Gizmo Factoring
ax2 + bx + c
More InstructionPractice
Application
Problems
Practice All
Other Cases
Factor x2 - 3x - 54
Special Case The Difference of two Perfect Squares
The difference of two perfect squares is very easy tofactor but everyone always forgets about themTheyre in the form (ax)2 - c where a and c are perfect squares
Theres no visible b-valueso b = 0 You factor them by taking the square root of a and the square root of c and placing them in parenthesis that have opposite signs
Whenever you have a binomial that is subtraction always check to
see itrsquos this special case It usually does NOT have a GCF
Heres an examplehellip
ExampleFactor 4x2 ndash 9
Set up parenthesis with opposite signs
( + )( - )
Find the square root of a and place then answer in the front sections of the parenthesissqrt(4x2) = 2x
( 2x + )( 2x - )
Find the square root of c and place them at the end of the parenthesissqrt(9) = 3
( 2x + 3 )( 2x - 3 )
Difference of Two Perfect Squares Video
Practice Factoring1 x2 + 4x ndash 5 2 x2 - 3x + 2 3 x2 - 6x ndash 7 4 x2 + 4x + 4
Solutions1 x2 + 4x ndash 5 = (x+5)(x-1)2 x2 - 3x + 2 = (x-1)(x-2)3 x2 - 6x ndash 7 = (x-7)(x+1)4 x2 + 4x + 4 = (x +2)(x+2)
Practice Differen
ce of Two
Squares
Factoring with
Algebra Tiles
Practice
Common
Factors
Practice
Factor a = 1
What if the leading coefficient isnrsquot a 1Factor 3x2 + 11x - 4
Set up two pairs of parenthesis ( )( )Look over the equation
( + )( - )
Look at the a-valueUnfortunately the a-value is not a one so we need to list factors in a chart Were looking for the pair of factors that when I find the difference of the productswill yield the b-value
Factors of A Factors of C
1 3 22 and 14 12 - 32 = -4 NO
13 - 14= -1 NO 11 - 34= -12 YES
Enter in values (x - 4)(3x + 1)Check with FOILIts possible that you have the right numbers but in the wrong spots so you have to check
(x - 4)(3x + 1)= 3x^2 -12x + x - 4 = 3x^2 -11x -4
Factoring when ane 1
Terms in a quadratic expression may have some common factors before you break them down into
linear factors
Remember the greatest common factor GCF is thegreatest number that is a factor of all terms in the
expression
When a ne 1 weshould always check to see if the quadratic expression
has a greatest common factor
Factor 2x2 -22x +36Step 1
a ne 1 so we should check to see if the quadratic expression has a greatest common factor
It has a GCF of 22x2 -22x +36 = 2(x2 -11x +18)
Step 2Once we factor out the GCF the quadratic expression now
has a value of a =1 and we can use the process we just went through in the
previous examples
x2 -11x +18 = (x -2)(x-9)
Therefore 2x2 -22x +36 is = 2 (x -2)(x-9)
Ane 1 and NO GCF2x2 + 13x ndash 7
Step 1 a ne 1 so we should check to see if the quadratic expression has a greatest common factor
It does not have a GCFThis type of trinomial is much more difficult to
factor than the previous Instead of factoring the c value alone one has to also factor the a value
Our factors of a become coefficients of our x-terms and the factors of c will go right where they did in the previous examples
2x2 + 13x ndash 7Step 1 Find the product ac
ac= -14Step 2 Find two factors of ac that add to give
b 1048633 1 and -14 = -13 1048633 -1 and 14 = 13 This is our winner 1048633 2 and -7 = -5 1048633 -2 and 7 = 5
Step 3 Split the middle term into two terms using the numbers found in step above
2x2 -1x + 14x ndash 7
Step 4 Factor out the common binomial using the box method
2x2 -1x + 14x ndash 7
Quadratic Term
Factor 1
Factor 2 Constant Term
2x2 -1x
14x -7
Find the GCF for each column and row
Numbers in RED represent the GCF of each row and column
2x -1
x 2x2 -1x
7 14x -7
The factors are (x + 7)(2x - 1)
Practice Factoring1 2x2 11x + 5 2 3x2 - 5x - 2 3 7x2 - 16x + 4 4 3x2 + 12x + 12
Solutions1 2x2 +11x + 5 = (2x + 1)(x + 5)2 3x2 - 5x - 2 = (3x + 1)(x - 2)3 7x2 - 16x + 4 = (7x - 2)(x - 2)4 3x2 + 12x + 12 = 3(x + 2)(x + 2)
Special Products
Factoring Strategies
Prime Factors
RememberThis wonrsquot work for all quadratic trinomials because not all quadratic trinomials can be
factored into products of binomials with integer coefficients
We call these prime (Prime Numbers are 3 5 7 11 13 etc)Expressions such as x2 + 2x - 7 cannot be factored at all and is therefore known as a
prime polynomial
Practicing Factoring when a ne1
Please watch the demonstration below on factoring when a ne 1 There will be
interactive examples provided to help when a ne 1
MORE FACTORING
Upon completion of the video and demonstration please complete Mastery
Assignment Part 2
Gizmo Factoring
ax2 + bx + c
More InstructionPractice
Application
Problems
Practice All
Other Cases
Special Case The Difference of two Perfect Squares
The difference of two perfect squares is very easy tofactor but everyone always forgets about themTheyre in the form (ax)2 - c where a and c are perfect squares
Theres no visible b-valueso b = 0 You factor them by taking the square root of a and the square root of c and placing them in parenthesis that have opposite signs
Whenever you have a binomial that is subtraction always check to
see itrsquos this special case It usually does NOT have a GCF
Heres an examplehellip
ExampleFactor 4x2 ndash 9
Set up parenthesis with opposite signs
( + )( - )
Find the square root of a and place then answer in the front sections of the parenthesissqrt(4x2) = 2x
( 2x + )( 2x - )
Find the square root of c and place them at the end of the parenthesissqrt(9) = 3
( 2x + 3 )( 2x - 3 )
Difference of Two Perfect Squares Video
Practice Factoring1 x2 + 4x ndash 5 2 x2 - 3x + 2 3 x2 - 6x ndash 7 4 x2 + 4x + 4
Solutions1 x2 + 4x ndash 5 = (x+5)(x-1)2 x2 - 3x + 2 = (x-1)(x-2)3 x2 - 6x ndash 7 = (x-7)(x+1)4 x2 + 4x + 4 = (x +2)(x+2)
Practice Differen
ce of Two
Squares
Factoring with
Algebra Tiles
Practice
Common
Factors
Practice
Factor a = 1
What if the leading coefficient isnrsquot a 1Factor 3x2 + 11x - 4
Set up two pairs of parenthesis ( )( )Look over the equation
( + )( - )
Look at the a-valueUnfortunately the a-value is not a one so we need to list factors in a chart Were looking for the pair of factors that when I find the difference of the productswill yield the b-value
Factors of A Factors of C
1 3 22 and 14 12 - 32 = -4 NO
13 - 14= -1 NO 11 - 34= -12 YES
Enter in values (x - 4)(3x + 1)Check with FOILIts possible that you have the right numbers but in the wrong spots so you have to check
(x - 4)(3x + 1)= 3x^2 -12x + x - 4 = 3x^2 -11x -4
Factoring when ane 1
Terms in a quadratic expression may have some common factors before you break them down into
linear factors
Remember the greatest common factor GCF is thegreatest number that is a factor of all terms in the
expression
When a ne 1 weshould always check to see if the quadratic expression
has a greatest common factor
Factor 2x2 -22x +36Step 1
a ne 1 so we should check to see if the quadratic expression has a greatest common factor
It has a GCF of 22x2 -22x +36 = 2(x2 -11x +18)
Step 2Once we factor out the GCF the quadratic expression now
has a value of a =1 and we can use the process we just went through in the
previous examples
x2 -11x +18 = (x -2)(x-9)
Therefore 2x2 -22x +36 is = 2 (x -2)(x-9)
Ane 1 and NO GCF2x2 + 13x ndash 7
Step 1 a ne 1 so we should check to see if the quadratic expression has a greatest common factor
It does not have a GCFThis type of trinomial is much more difficult to
factor than the previous Instead of factoring the c value alone one has to also factor the a value
Our factors of a become coefficients of our x-terms and the factors of c will go right where they did in the previous examples
2x2 + 13x ndash 7Step 1 Find the product ac
ac= -14Step 2 Find two factors of ac that add to give
b 1048633 1 and -14 = -13 1048633 -1 and 14 = 13 This is our winner 1048633 2 and -7 = -5 1048633 -2 and 7 = 5
Step 3 Split the middle term into two terms using the numbers found in step above
2x2 -1x + 14x ndash 7
Step 4 Factor out the common binomial using the box method
2x2 -1x + 14x ndash 7
Quadratic Term
Factor 1
Factor 2 Constant Term
2x2 -1x
14x -7
Find the GCF for each column and row
Numbers in RED represent the GCF of each row and column
2x -1
x 2x2 -1x
7 14x -7
The factors are (x + 7)(2x - 1)
Practice Factoring1 2x2 11x + 5 2 3x2 - 5x - 2 3 7x2 - 16x + 4 4 3x2 + 12x + 12
Solutions1 2x2 +11x + 5 = (2x + 1)(x + 5)2 3x2 - 5x - 2 = (3x + 1)(x - 2)3 7x2 - 16x + 4 = (7x - 2)(x - 2)4 3x2 + 12x + 12 = 3(x + 2)(x + 2)
Special Products
Factoring Strategies
Prime Factors
RememberThis wonrsquot work for all quadratic trinomials because not all quadratic trinomials can be
factored into products of binomials with integer coefficients
We call these prime (Prime Numbers are 3 5 7 11 13 etc)Expressions such as x2 + 2x - 7 cannot be factored at all and is therefore known as a
prime polynomial
Practicing Factoring when a ne1
Please watch the demonstration below on factoring when a ne 1 There will be
interactive examples provided to help when a ne 1
MORE FACTORING
Upon completion of the video and demonstration please complete Mastery
Assignment Part 2
Gizmo Factoring
ax2 + bx + c
More InstructionPractice
Application
Problems
Practice All
Other Cases
ExampleFactor 4x2 ndash 9
Set up parenthesis with opposite signs
( + )( - )
Find the square root of a and place then answer in the front sections of the parenthesissqrt(4x2) = 2x
( 2x + )( 2x - )
Find the square root of c and place them at the end of the parenthesissqrt(9) = 3
( 2x + 3 )( 2x - 3 )
Difference of Two Perfect Squares Video
Practice Factoring1 x2 + 4x ndash 5 2 x2 - 3x + 2 3 x2 - 6x ndash 7 4 x2 + 4x + 4
Solutions1 x2 + 4x ndash 5 = (x+5)(x-1)2 x2 - 3x + 2 = (x-1)(x-2)3 x2 - 6x ndash 7 = (x-7)(x+1)4 x2 + 4x + 4 = (x +2)(x+2)
Practice Differen
ce of Two
Squares
Factoring with
Algebra Tiles
Practice
Common
Factors
Practice
Factor a = 1
What if the leading coefficient isnrsquot a 1Factor 3x2 + 11x - 4
Set up two pairs of parenthesis ( )( )Look over the equation
( + )( - )
Look at the a-valueUnfortunately the a-value is not a one so we need to list factors in a chart Were looking for the pair of factors that when I find the difference of the productswill yield the b-value
Factors of A Factors of C
1 3 22 and 14 12 - 32 = -4 NO
13 - 14= -1 NO 11 - 34= -12 YES
Enter in values (x - 4)(3x + 1)Check with FOILIts possible that you have the right numbers but in the wrong spots so you have to check
(x - 4)(3x + 1)= 3x^2 -12x + x - 4 = 3x^2 -11x -4
Factoring when ane 1
Terms in a quadratic expression may have some common factors before you break them down into
linear factors
Remember the greatest common factor GCF is thegreatest number that is a factor of all terms in the
expression
When a ne 1 weshould always check to see if the quadratic expression
has a greatest common factor
Factor 2x2 -22x +36Step 1
a ne 1 so we should check to see if the quadratic expression has a greatest common factor
It has a GCF of 22x2 -22x +36 = 2(x2 -11x +18)
Step 2Once we factor out the GCF the quadratic expression now
has a value of a =1 and we can use the process we just went through in the
previous examples
x2 -11x +18 = (x -2)(x-9)
Therefore 2x2 -22x +36 is = 2 (x -2)(x-9)
Ane 1 and NO GCF2x2 + 13x ndash 7
Step 1 a ne 1 so we should check to see if the quadratic expression has a greatest common factor
It does not have a GCFThis type of trinomial is much more difficult to
factor than the previous Instead of factoring the c value alone one has to also factor the a value
Our factors of a become coefficients of our x-terms and the factors of c will go right where they did in the previous examples
2x2 + 13x ndash 7Step 1 Find the product ac
ac= -14Step 2 Find two factors of ac that add to give
b 1048633 1 and -14 = -13 1048633 -1 and 14 = 13 This is our winner 1048633 2 and -7 = -5 1048633 -2 and 7 = 5
Step 3 Split the middle term into two terms using the numbers found in step above
2x2 -1x + 14x ndash 7
Step 4 Factor out the common binomial using the box method
2x2 -1x + 14x ndash 7
Quadratic Term
Factor 1
Factor 2 Constant Term
2x2 -1x
14x -7
Find the GCF for each column and row
Numbers in RED represent the GCF of each row and column
2x -1
x 2x2 -1x
7 14x -7
The factors are (x + 7)(2x - 1)
Practice Factoring1 2x2 11x + 5 2 3x2 - 5x - 2 3 7x2 - 16x + 4 4 3x2 + 12x + 12
Solutions1 2x2 +11x + 5 = (2x + 1)(x + 5)2 3x2 - 5x - 2 = (3x + 1)(x - 2)3 7x2 - 16x + 4 = (7x - 2)(x - 2)4 3x2 + 12x + 12 = 3(x + 2)(x + 2)
Special Products
Factoring Strategies
Prime Factors
RememberThis wonrsquot work for all quadratic trinomials because not all quadratic trinomials can be
factored into products of binomials with integer coefficients
We call these prime (Prime Numbers are 3 5 7 11 13 etc)Expressions such as x2 + 2x - 7 cannot be factored at all and is therefore known as a
prime polynomial
Practicing Factoring when a ne1
Please watch the demonstration below on factoring when a ne 1 There will be
interactive examples provided to help when a ne 1
MORE FACTORING
Upon completion of the video and demonstration please complete Mastery
Assignment Part 2
Gizmo Factoring
ax2 + bx + c
More InstructionPractice
Application
Problems
Practice All
Other Cases
Practice Factoring1 x2 + 4x ndash 5 2 x2 - 3x + 2 3 x2 - 6x ndash 7 4 x2 + 4x + 4
Solutions1 x2 + 4x ndash 5 = (x+5)(x-1)2 x2 - 3x + 2 = (x-1)(x-2)3 x2 - 6x ndash 7 = (x-7)(x+1)4 x2 + 4x + 4 = (x +2)(x+2)
Practice Differen
ce of Two
Squares
Factoring with
Algebra Tiles
Practice
Common
Factors
Practice
Factor a = 1
What if the leading coefficient isnrsquot a 1Factor 3x2 + 11x - 4
Set up two pairs of parenthesis ( )( )Look over the equation
( + )( - )
Look at the a-valueUnfortunately the a-value is not a one so we need to list factors in a chart Were looking for the pair of factors that when I find the difference of the productswill yield the b-value
Factors of A Factors of C
1 3 22 and 14 12 - 32 = -4 NO
13 - 14= -1 NO 11 - 34= -12 YES
Enter in values (x - 4)(3x + 1)Check with FOILIts possible that you have the right numbers but in the wrong spots so you have to check
(x - 4)(3x + 1)= 3x^2 -12x + x - 4 = 3x^2 -11x -4
Factoring when ane 1
Terms in a quadratic expression may have some common factors before you break them down into
linear factors
Remember the greatest common factor GCF is thegreatest number that is a factor of all terms in the
expression
When a ne 1 weshould always check to see if the quadratic expression
has a greatest common factor
Factor 2x2 -22x +36Step 1
a ne 1 so we should check to see if the quadratic expression has a greatest common factor
It has a GCF of 22x2 -22x +36 = 2(x2 -11x +18)
Step 2Once we factor out the GCF the quadratic expression now
has a value of a =1 and we can use the process we just went through in the
previous examples
x2 -11x +18 = (x -2)(x-9)
Therefore 2x2 -22x +36 is = 2 (x -2)(x-9)
Ane 1 and NO GCF2x2 + 13x ndash 7
Step 1 a ne 1 so we should check to see if the quadratic expression has a greatest common factor
It does not have a GCFThis type of trinomial is much more difficult to
factor than the previous Instead of factoring the c value alone one has to also factor the a value
Our factors of a become coefficients of our x-terms and the factors of c will go right where they did in the previous examples
2x2 + 13x ndash 7Step 1 Find the product ac
ac= -14Step 2 Find two factors of ac that add to give
b 1048633 1 and -14 = -13 1048633 -1 and 14 = 13 This is our winner 1048633 2 and -7 = -5 1048633 -2 and 7 = 5
Step 3 Split the middle term into two terms using the numbers found in step above
2x2 -1x + 14x ndash 7
Step 4 Factor out the common binomial using the box method
2x2 -1x + 14x ndash 7
Quadratic Term
Factor 1
Factor 2 Constant Term
2x2 -1x
14x -7
Find the GCF for each column and row
Numbers in RED represent the GCF of each row and column
2x -1
x 2x2 -1x
7 14x -7
The factors are (x + 7)(2x - 1)
Practice Factoring1 2x2 11x + 5 2 3x2 - 5x - 2 3 7x2 - 16x + 4 4 3x2 + 12x + 12
Solutions1 2x2 +11x + 5 = (2x + 1)(x + 5)2 3x2 - 5x - 2 = (3x + 1)(x - 2)3 7x2 - 16x + 4 = (7x - 2)(x - 2)4 3x2 + 12x + 12 = 3(x + 2)(x + 2)
Special Products
Factoring Strategies
Prime Factors
RememberThis wonrsquot work for all quadratic trinomials because not all quadratic trinomials can be
factored into products of binomials with integer coefficients
We call these prime (Prime Numbers are 3 5 7 11 13 etc)Expressions such as x2 + 2x - 7 cannot be factored at all and is therefore known as a
prime polynomial
Practicing Factoring when a ne1
Please watch the demonstration below on factoring when a ne 1 There will be
interactive examples provided to help when a ne 1
MORE FACTORING
Upon completion of the video and demonstration please complete Mastery
Assignment Part 2
Gizmo Factoring
ax2 + bx + c
More InstructionPractice
Application
Problems
Practice All
Other Cases
Solutions1 x2 + 4x ndash 5 = (x+5)(x-1)2 x2 - 3x + 2 = (x-1)(x-2)3 x2 - 6x ndash 7 = (x-7)(x+1)4 x2 + 4x + 4 = (x +2)(x+2)
Practice Differen
ce of Two
Squares
Factoring with
Algebra Tiles
Practice
Common
Factors
Practice
Factor a = 1
What if the leading coefficient isnrsquot a 1Factor 3x2 + 11x - 4
Set up two pairs of parenthesis ( )( )Look over the equation
( + )( - )
Look at the a-valueUnfortunately the a-value is not a one so we need to list factors in a chart Were looking for the pair of factors that when I find the difference of the productswill yield the b-value
Factors of A Factors of C
1 3 22 and 14 12 - 32 = -4 NO
13 - 14= -1 NO 11 - 34= -12 YES
Enter in values (x - 4)(3x + 1)Check with FOILIts possible that you have the right numbers but in the wrong spots so you have to check
(x - 4)(3x + 1)= 3x^2 -12x + x - 4 = 3x^2 -11x -4
Factoring when ane 1
Terms in a quadratic expression may have some common factors before you break them down into
linear factors
Remember the greatest common factor GCF is thegreatest number that is a factor of all terms in the
expression
When a ne 1 weshould always check to see if the quadratic expression
has a greatest common factor
Factor 2x2 -22x +36Step 1
a ne 1 so we should check to see if the quadratic expression has a greatest common factor
It has a GCF of 22x2 -22x +36 = 2(x2 -11x +18)
Step 2Once we factor out the GCF the quadratic expression now
has a value of a =1 and we can use the process we just went through in the
previous examples
x2 -11x +18 = (x -2)(x-9)
Therefore 2x2 -22x +36 is = 2 (x -2)(x-9)
Ane 1 and NO GCF2x2 + 13x ndash 7
Step 1 a ne 1 so we should check to see if the quadratic expression has a greatest common factor
It does not have a GCFThis type of trinomial is much more difficult to
factor than the previous Instead of factoring the c value alone one has to also factor the a value
Our factors of a become coefficients of our x-terms and the factors of c will go right where they did in the previous examples
2x2 + 13x ndash 7Step 1 Find the product ac
ac= -14Step 2 Find two factors of ac that add to give
b 1048633 1 and -14 = -13 1048633 -1 and 14 = 13 This is our winner 1048633 2 and -7 = -5 1048633 -2 and 7 = 5
Step 3 Split the middle term into two terms using the numbers found in step above
2x2 -1x + 14x ndash 7
Step 4 Factor out the common binomial using the box method
2x2 -1x + 14x ndash 7
Quadratic Term
Factor 1
Factor 2 Constant Term
2x2 -1x
14x -7
Find the GCF for each column and row
Numbers in RED represent the GCF of each row and column
2x -1
x 2x2 -1x
7 14x -7
The factors are (x + 7)(2x - 1)
Practice Factoring1 2x2 11x + 5 2 3x2 - 5x - 2 3 7x2 - 16x + 4 4 3x2 + 12x + 12
Solutions1 2x2 +11x + 5 = (2x + 1)(x + 5)2 3x2 - 5x - 2 = (3x + 1)(x - 2)3 7x2 - 16x + 4 = (7x - 2)(x - 2)4 3x2 + 12x + 12 = 3(x + 2)(x + 2)
Special Products
Factoring Strategies
Prime Factors
RememberThis wonrsquot work for all quadratic trinomials because not all quadratic trinomials can be
factored into products of binomials with integer coefficients
We call these prime (Prime Numbers are 3 5 7 11 13 etc)Expressions such as x2 + 2x - 7 cannot be factored at all and is therefore known as a
prime polynomial
Practicing Factoring when a ne1
Please watch the demonstration below on factoring when a ne 1 There will be
interactive examples provided to help when a ne 1
MORE FACTORING
Upon completion of the video and demonstration please complete Mastery
Assignment Part 2
Gizmo Factoring
ax2 + bx + c
More InstructionPractice
Application
Problems
Practice All
Other Cases
Practice Differen
ce of Two
Squares
Factoring with
Algebra Tiles
Practice
Common
Factors
Practice
Factor a = 1
What if the leading coefficient isnrsquot a 1Factor 3x2 + 11x - 4
Set up two pairs of parenthesis ( )( )Look over the equation
( + )( - )
Look at the a-valueUnfortunately the a-value is not a one so we need to list factors in a chart Were looking for the pair of factors that when I find the difference of the productswill yield the b-value
Factors of A Factors of C
1 3 22 and 14 12 - 32 = -4 NO
13 - 14= -1 NO 11 - 34= -12 YES
Enter in values (x - 4)(3x + 1)Check with FOILIts possible that you have the right numbers but in the wrong spots so you have to check
(x - 4)(3x + 1)= 3x^2 -12x + x - 4 = 3x^2 -11x -4
Factoring when ane 1
Terms in a quadratic expression may have some common factors before you break them down into
linear factors
Remember the greatest common factor GCF is thegreatest number that is a factor of all terms in the
expression
When a ne 1 weshould always check to see if the quadratic expression
has a greatest common factor
Factor 2x2 -22x +36Step 1
a ne 1 so we should check to see if the quadratic expression has a greatest common factor
It has a GCF of 22x2 -22x +36 = 2(x2 -11x +18)
Step 2Once we factor out the GCF the quadratic expression now
has a value of a =1 and we can use the process we just went through in the
previous examples
x2 -11x +18 = (x -2)(x-9)
Therefore 2x2 -22x +36 is = 2 (x -2)(x-9)
Ane 1 and NO GCF2x2 + 13x ndash 7
Step 1 a ne 1 so we should check to see if the quadratic expression has a greatest common factor
It does not have a GCFThis type of trinomial is much more difficult to
factor than the previous Instead of factoring the c value alone one has to also factor the a value
Our factors of a become coefficients of our x-terms and the factors of c will go right where they did in the previous examples
2x2 + 13x ndash 7Step 1 Find the product ac
ac= -14Step 2 Find two factors of ac that add to give
b 1048633 1 and -14 = -13 1048633 -1 and 14 = 13 This is our winner 1048633 2 and -7 = -5 1048633 -2 and 7 = 5
Step 3 Split the middle term into two terms using the numbers found in step above
2x2 -1x + 14x ndash 7
Step 4 Factor out the common binomial using the box method
2x2 -1x + 14x ndash 7
Quadratic Term
Factor 1
Factor 2 Constant Term
2x2 -1x
14x -7
Find the GCF for each column and row
Numbers in RED represent the GCF of each row and column
2x -1
x 2x2 -1x
7 14x -7
The factors are (x + 7)(2x - 1)
Practice Factoring1 2x2 11x + 5 2 3x2 - 5x - 2 3 7x2 - 16x + 4 4 3x2 + 12x + 12
Solutions1 2x2 +11x + 5 = (2x + 1)(x + 5)2 3x2 - 5x - 2 = (3x + 1)(x - 2)3 7x2 - 16x + 4 = (7x - 2)(x - 2)4 3x2 + 12x + 12 = 3(x + 2)(x + 2)
Special Products
Factoring Strategies
Prime Factors
RememberThis wonrsquot work for all quadratic trinomials because not all quadratic trinomials can be
factored into products of binomials with integer coefficients
We call these prime (Prime Numbers are 3 5 7 11 13 etc)Expressions such as x2 + 2x - 7 cannot be factored at all and is therefore known as a
prime polynomial
Practicing Factoring when a ne1
Please watch the demonstration below on factoring when a ne 1 There will be
interactive examples provided to help when a ne 1
MORE FACTORING
Upon completion of the video and demonstration please complete Mastery
Assignment Part 2
Gizmo Factoring
ax2 + bx + c
More InstructionPractice
Application
Problems
Practice All
Other Cases
What if the leading coefficient isnrsquot a 1Factor 3x2 + 11x - 4
Set up two pairs of parenthesis ( )( )Look over the equation
( + )( - )
Look at the a-valueUnfortunately the a-value is not a one so we need to list factors in a chart Were looking for the pair of factors that when I find the difference of the productswill yield the b-value
Factors of A Factors of C
1 3 22 and 14 12 - 32 = -4 NO
13 - 14= -1 NO 11 - 34= -12 YES
Enter in values (x - 4)(3x + 1)Check with FOILIts possible that you have the right numbers but in the wrong spots so you have to check
(x - 4)(3x + 1)= 3x^2 -12x + x - 4 = 3x^2 -11x -4
Factoring when ane 1
Terms in a quadratic expression may have some common factors before you break them down into
linear factors
Remember the greatest common factor GCF is thegreatest number that is a factor of all terms in the
expression
When a ne 1 weshould always check to see if the quadratic expression
has a greatest common factor
Factor 2x2 -22x +36Step 1
a ne 1 so we should check to see if the quadratic expression has a greatest common factor
It has a GCF of 22x2 -22x +36 = 2(x2 -11x +18)
Step 2Once we factor out the GCF the quadratic expression now
has a value of a =1 and we can use the process we just went through in the
previous examples
x2 -11x +18 = (x -2)(x-9)
Therefore 2x2 -22x +36 is = 2 (x -2)(x-9)
Ane 1 and NO GCF2x2 + 13x ndash 7
Step 1 a ne 1 so we should check to see if the quadratic expression has a greatest common factor
It does not have a GCFThis type of trinomial is much more difficult to
factor than the previous Instead of factoring the c value alone one has to also factor the a value
Our factors of a become coefficients of our x-terms and the factors of c will go right where they did in the previous examples
2x2 + 13x ndash 7Step 1 Find the product ac
ac= -14Step 2 Find two factors of ac that add to give
b 1048633 1 and -14 = -13 1048633 -1 and 14 = 13 This is our winner 1048633 2 and -7 = -5 1048633 -2 and 7 = 5
Step 3 Split the middle term into two terms using the numbers found in step above
2x2 -1x + 14x ndash 7
Step 4 Factor out the common binomial using the box method
2x2 -1x + 14x ndash 7
Quadratic Term
Factor 1
Factor 2 Constant Term
2x2 -1x
14x -7
Find the GCF for each column and row
Numbers in RED represent the GCF of each row and column
2x -1
x 2x2 -1x
7 14x -7
The factors are (x + 7)(2x - 1)
Practice Factoring1 2x2 11x + 5 2 3x2 - 5x - 2 3 7x2 - 16x + 4 4 3x2 + 12x + 12
Solutions1 2x2 +11x + 5 = (2x + 1)(x + 5)2 3x2 - 5x - 2 = (3x + 1)(x - 2)3 7x2 - 16x + 4 = (7x - 2)(x - 2)4 3x2 + 12x + 12 = 3(x + 2)(x + 2)
Special Products
Factoring Strategies
Prime Factors
RememberThis wonrsquot work for all quadratic trinomials because not all quadratic trinomials can be
factored into products of binomials with integer coefficients
We call these prime (Prime Numbers are 3 5 7 11 13 etc)Expressions such as x2 + 2x - 7 cannot be factored at all and is therefore known as a
prime polynomial
Practicing Factoring when a ne1
Please watch the demonstration below on factoring when a ne 1 There will be
interactive examples provided to help when a ne 1
MORE FACTORING
Upon completion of the video and demonstration please complete Mastery
Assignment Part 2
Gizmo Factoring
ax2 + bx + c
More InstructionPractice
Application
Problems
Practice All
Other Cases
Factoring when ane 1
Terms in a quadratic expression may have some common factors before you break them down into
linear factors
Remember the greatest common factor GCF is thegreatest number that is a factor of all terms in the
expression
When a ne 1 weshould always check to see if the quadratic expression
has a greatest common factor
Factor 2x2 -22x +36Step 1
a ne 1 so we should check to see if the quadratic expression has a greatest common factor
It has a GCF of 22x2 -22x +36 = 2(x2 -11x +18)
Step 2Once we factor out the GCF the quadratic expression now
has a value of a =1 and we can use the process we just went through in the
previous examples
x2 -11x +18 = (x -2)(x-9)
Therefore 2x2 -22x +36 is = 2 (x -2)(x-9)
Ane 1 and NO GCF2x2 + 13x ndash 7
Step 1 a ne 1 so we should check to see if the quadratic expression has a greatest common factor
It does not have a GCFThis type of trinomial is much more difficult to
factor than the previous Instead of factoring the c value alone one has to also factor the a value
Our factors of a become coefficients of our x-terms and the factors of c will go right where they did in the previous examples
2x2 + 13x ndash 7Step 1 Find the product ac
ac= -14Step 2 Find two factors of ac that add to give
b 1048633 1 and -14 = -13 1048633 -1 and 14 = 13 This is our winner 1048633 2 and -7 = -5 1048633 -2 and 7 = 5
Step 3 Split the middle term into two terms using the numbers found in step above
2x2 -1x + 14x ndash 7
Step 4 Factor out the common binomial using the box method
2x2 -1x + 14x ndash 7
Quadratic Term
Factor 1
Factor 2 Constant Term
2x2 -1x
14x -7
Find the GCF for each column and row
Numbers in RED represent the GCF of each row and column
2x -1
x 2x2 -1x
7 14x -7
The factors are (x + 7)(2x - 1)
Practice Factoring1 2x2 11x + 5 2 3x2 - 5x - 2 3 7x2 - 16x + 4 4 3x2 + 12x + 12
Solutions1 2x2 +11x + 5 = (2x + 1)(x + 5)2 3x2 - 5x - 2 = (3x + 1)(x - 2)3 7x2 - 16x + 4 = (7x - 2)(x - 2)4 3x2 + 12x + 12 = 3(x + 2)(x + 2)
Special Products
Factoring Strategies
Prime Factors
RememberThis wonrsquot work for all quadratic trinomials because not all quadratic trinomials can be
factored into products of binomials with integer coefficients
We call these prime (Prime Numbers are 3 5 7 11 13 etc)Expressions such as x2 + 2x - 7 cannot be factored at all and is therefore known as a
prime polynomial
Practicing Factoring when a ne1
Please watch the demonstration below on factoring when a ne 1 There will be
interactive examples provided to help when a ne 1
MORE FACTORING
Upon completion of the video and demonstration please complete Mastery
Assignment Part 2
Gizmo Factoring
ax2 + bx + c
More InstructionPractice
Application
Problems
Practice All
Other Cases
Factor 2x2 -22x +36Step 1
a ne 1 so we should check to see if the quadratic expression has a greatest common factor
It has a GCF of 22x2 -22x +36 = 2(x2 -11x +18)
Step 2Once we factor out the GCF the quadratic expression now
has a value of a =1 and we can use the process we just went through in the
previous examples
x2 -11x +18 = (x -2)(x-9)
Therefore 2x2 -22x +36 is = 2 (x -2)(x-9)
Ane 1 and NO GCF2x2 + 13x ndash 7
Step 1 a ne 1 so we should check to see if the quadratic expression has a greatest common factor
It does not have a GCFThis type of trinomial is much more difficult to
factor than the previous Instead of factoring the c value alone one has to also factor the a value
Our factors of a become coefficients of our x-terms and the factors of c will go right where they did in the previous examples
2x2 + 13x ndash 7Step 1 Find the product ac
ac= -14Step 2 Find two factors of ac that add to give
b 1048633 1 and -14 = -13 1048633 -1 and 14 = 13 This is our winner 1048633 2 and -7 = -5 1048633 -2 and 7 = 5
Step 3 Split the middle term into two terms using the numbers found in step above
2x2 -1x + 14x ndash 7
Step 4 Factor out the common binomial using the box method
2x2 -1x + 14x ndash 7
Quadratic Term
Factor 1
Factor 2 Constant Term
2x2 -1x
14x -7
Find the GCF for each column and row
Numbers in RED represent the GCF of each row and column
2x -1
x 2x2 -1x
7 14x -7
The factors are (x + 7)(2x - 1)
Practice Factoring1 2x2 11x + 5 2 3x2 - 5x - 2 3 7x2 - 16x + 4 4 3x2 + 12x + 12
Solutions1 2x2 +11x + 5 = (2x + 1)(x + 5)2 3x2 - 5x - 2 = (3x + 1)(x - 2)3 7x2 - 16x + 4 = (7x - 2)(x - 2)4 3x2 + 12x + 12 = 3(x + 2)(x + 2)
Special Products
Factoring Strategies
Prime Factors
RememberThis wonrsquot work for all quadratic trinomials because not all quadratic trinomials can be
factored into products of binomials with integer coefficients
We call these prime (Prime Numbers are 3 5 7 11 13 etc)Expressions such as x2 + 2x - 7 cannot be factored at all and is therefore known as a
prime polynomial
Practicing Factoring when a ne1
Please watch the demonstration below on factoring when a ne 1 There will be
interactive examples provided to help when a ne 1
MORE FACTORING
Upon completion of the video and demonstration please complete Mastery
Assignment Part 2
Gizmo Factoring
ax2 + bx + c
More InstructionPractice
Application
Problems
Practice All
Other Cases
Ane 1 and NO GCF2x2 + 13x ndash 7
Step 1 a ne 1 so we should check to see if the quadratic expression has a greatest common factor
It does not have a GCFThis type of trinomial is much more difficult to
factor than the previous Instead of factoring the c value alone one has to also factor the a value
Our factors of a become coefficients of our x-terms and the factors of c will go right where they did in the previous examples
2x2 + 13x ndash 7Step 1 Find the product ac
ac= -14Step 2 Find two factors of ac that add to give
b 1048633 1 and -14 = -13 1048633 -1 and 14 = 13 This is our winner 1048633 2 and -7 = -5 1048633 -2 and 7 = 5
Step 3 Split the middle term into two terms using the numbers found in step above
2x2 -1x + 14x ndash 7
Step 4 Factor out the common binomial using the box method
2x2 -1x + 14x ndash 7
Quadratic Term
Factor 1
Factor 2 Constant Term
2x2 -1x
14x -7
Find the GCF for each column and row
Numbers in RED represent the GCF of each row and column
2x -1
x 2x2 -1x
7 14x -7
The factors are (x + 7)(2x - 1)
Practice Factoring1 2x2 11x + 5 2 3x2 - 5x - 2 3 7x2 - 16x + 4 4 3x2 + 12x + 12
Solutions1 2x2 +11x + 5 = (2x + 1)(x + 5)2 3x2 - 5x - 2 = (3x + 1)(x - 2)3 7x2 - 16x + 4 = (7x - 2)(x - 2)4 3x2 + 12x + 12 = 3(x + 2)(x + 2)
Special Products
Factoring Strategies
Prime Factors
RememberThis wonrsquot work for all quadratic trinomials because not all quadratic trinomials can be
factored into products of binomials with integer coefficients
We call these prime (Prime Numbers are 3 5 7 11 13 etc)Expressions such as x2 + 2x - 7 cannot be factored at all and is therefore known as a
prime polynomial
Practicing Factoring when a ne1
Please watch the demonstration below on factoring when a ne 1 There will be
interactive examples provided to help when a ne 1
MORE FACTORING
Upon completion of the video and demonstration please complete Mastery
Assignment Part 2
Gizmo Factoring
ax2 + bx + c
More InstructionPractice
Application
Problems
Practice All
Other Cases
2x2 + 13x ndash 7Step 1 Find the product ac
ac= -14Step 2 Find two factors of ac that add to give
b 1048633 1 and -14 = -13 1048633 -1 and 14 = 13 This is our winner 1048633 2 and -7 = -5 1048633 -2 and 7 = 5
Step 3 Split the middle term into two terms using the numbers found in step above
2x2 -1x + 14x ndash 7
Step 4 Factor out the common binomial using the box method
2x2 -1x + 14x ndash 7
Quadratic Term
Factor 1
Factor 2 Constant Term
2x2 -1x
14x -7
Find the GCF for each column and row
Numbers in RED represent the GCF of each row and column
2x -1
x 2x2 -1x
7 14x -7
The factors are (x + 7)(2x - 1)
Practice Factoring1 2x2 11x + 5 2 3x2 - 5x - 2 3 7x2 - 16x + 4 4 3x2 + 12x + 12
Solutions1 2x2 +11x + 5 = (2x + 1)(x + 5)2 3x2 - 5x - 2 = (3x + 1)(x - 2)3 7x2 - 16x + 4 = (7x - 2)(x - 2)4 3x2 + 12x + 12 = 3(x + 2)(x + 2)
Special Products
Factoring Strategies
Prime Factors
RememberThis wonrsquot work for all quadratic trinomials because not all quadratic trinomials can be
factored into products of binomials with integer coefficients
We call these prime (Prime Numbers are 3 5 7 11 13 etc)Expressions such as x2 + 2x - 7 cannot be factored at all and is therefore known as a
prime polynomial
Practicing Factoring when a ne1
Please watch the demonstration below on factoring when a ne 1 There will be
interactive examples provided to help when a ne 1
MORE FACTORING
Upon completion of the video and demonstration please complete Mastery
Assignment Part 2
Gizmo Factoring
ax2 + bx + c
More InstructionPractice
Application
Problems
Practice All
Other Cases
Step 4 Factor out the common binomial using the box method
2x2 -1x + 14x ndash 7
Quadratic Term
Factor 1
Factor 2 Constant Term
2x2 -1x
14x -7
Find the GCF for each column and row
Numbers in RED represent the GCF of each row and column
2x -1
x 2x2 -1x
7 14x -7
The factors are (x + 7)(2x - 1)
Practice Factoring1 2x2 11x + 5 2 3x2 - 5x - 2 3 7x2 - 16x + 4 4 3x2 + 12x + 12
Solutions1 2x2 +11x + 5 = (2x + 1)(x + 5)2 3x2 - 5x - 2 = (3x + 1)(x - 2)3 7x2 - 16x + 4 = (7x - 2)(x - 2)4 3x2 + 12x + 12 = 3(x + 2)(x + 2)
Special Products
Factoring Strategies
Prime Factors
RememberThis wonrsquot work for all quadratic trinomials because not all quadratic trinomials can be
factored into products of binomials with integer coefficients
We call these prime (Prime Numbers are 3 5 7 11 13 etc)Expressions such as x2 + 2x - 7 cannot be factored at all and is therefore known as a
prime polynomial
Practicing Factoring when a ne1
Please watch the demonstration below on factoring when a ne 1 There will be
interactive examples provided to help when a ne 1
MORE FACTORING
Upon completion of the video and demonstration please complete Mastery
Assignment Part 2
Gizmo Factoring
ax2 + bx + c
More InstructionPractice
Application
Problems
Practice All
Other Cases
Numbers in RED represent the GCF of each row and column
2x -1
x 2x2 -1x
7 14x -7
The factors are (x + 7)(2x - 1)
Practice Factoring1 2x2 11x + 5 2 3x2 - 5x - 2 3 7x2 - 16x + 4 4 3x2 + 12x + 12
Solutions1 2x2 +11x + 5 = (2x + 1)(x + 5)2 3x2 - 5x - 2 = (3x + 1)(x - 2)3 7x2 - 16x + 4 = (7x - 2)(x - 2)4 3x2 + 12x + 12 = 3(x + 2)(x + 2)
Special Products
Factoring Strategies
Prime Factors
RememberThis wonrsquot work for all quadratic trinomials because not all quadratic trinomials can be
factored into products of binomials with integer coefficients
We call these prime (Prime Numbers are 3 5 7 11 13 etc)Expressions such as x2 + 2x - 7 cannot be factored at all and is therefore known as a
prime polynomial
Practicing Factoring when a ne1
Please watch the demonstration below on factoring when a ne 1 There will be
interactive examples provided to help when a ne 1
MORE FACTORING
Upon completion of the video and demonstration please complete Mastery
Assignment Part 2
Gizmo Factoring
ax2 + bx + c
More InstructionPractice
Application
Problems
Practice All
Other Cases
Practice Factoring1 2x2 11x + 5 2 3x2 - 5x - 2 3 7x2 - 16x + 4 4 3x2 + 12x + 12
Solutions1 2x2 +11x + 5 = (2x + 1)(x + 5)2 3x2 - 5x - 2 = (3x + 1)(x - 2)3 7x2 - 16x + 4 = (7x - 2)(x - 2)4 3x2 + 12x + 12 = 3(x + 2)(x + 2)
Special Products
Factoring Strategies
Prime Factors
RememberThis wonrsquot work for all quadratic trinomials because not all quadratic trinomials can be
factored into products of binomials with integer coefficients
We call these prime (Prime Numbers are 3 5 7 11 13 etc)Expressions such as x2 + 2x - 7 cannot be factored at all and is therefore known as a
prime polynomial
Practicing Factoring when a ne1
Please watch the demonstration below on factoring when a ne 1 There will be
interactive examples provided to help when a ne 1
MORE FACTORING
Upon completion of the video and demonstration please complete Mastery
Assignment Part 2
Gizmo Factoring
ax2 + bx + c
More InstructionPractice
Application
Problems
Practice All
Other Cases
Solutions1 2x2 +11x + 5 = (2x + 1)(x + 5)2 3x2 - 5x - 2 = (3x + 1)(x - 2)3 7x2 - 16x + 4 = (7x - 2)(x - 2)4 3x2 + 12x + 12 = 3(x + 2)(x + 2)
Special Products
Factoring Strategies
Prime Factors
RememberThis wonrsquot work for all quadratic trinomials because not all quadratic trinomials can be
factored into products of binomials with integer coefficients
We call these prime (Prime Numbers are 3 5 7 11 13 etc)Expressions such as x2 + 2x - 7 cannot be factored at all and is therefore known as a
prime polynomial
Practicing Factoring when a ne1
Please watch the demonstration below on factoring when a ne 1 There will be
interactive examples provided to help when a ne 1
MORE FACTORING
Upon completion of the video and demonstration please complete Mastery
Assignment Part 2
Gizmo Factoring
ax2 + bx + c
More InstructionPractice
Application
Problems
Practice All
Other Cases
Special Products
Factoring Strategies
Prime Factors
RememberThis wonrsquot work for all quadratic trinomials because not all quadratic trinomials can be
factored into products of binomials with integer coefficients
We call these prime (Prime Numbers are 3 5 7 11 13 etc)Expressions such as x2 + 2x - 7 cannot be factored at all and is therefore known as a
prime polynomial
Practicing Factoring when a ne1
Please watch the demonstration below on factoring when a ne 1 There will be
interactive examples provided to help when a ne 1
MORE FACTORING
Upon completion of the video and demonstration please complete Mastery
Assignment Part 2
Gizmo Factoring
ax2 + bx + c
More InstructionPractice
Application
Problems
Practice All
Other Cases
Factoring Strategies
Prime Factors
RememberThis wonrsquot work for all quadratic trinomials because not all quadratic trinomials can be
factored into products of binomials with integer coefficients
We call these prime (Prime Numbers are 3 5 7 11 13 etc)Expressions such as x2 + 2x - 7 cannot be factored at all and is therefore known as a
prime polynomial
Practicing Factoring when a ne1
Please watch the demonstration below on factoring when a ne 1 There will be
interactive examples provided to help when a ne 1
MORE FACTORING
Upon completion of the video and demonstration please complete Mastery
Assignment Part 2
Gizmo Factoring
ax2 + bx + c
More InstructionPractice
Application
Problems
Practice All
Other Cases
Prime Factors
RememberThis wonrsquot work for all quadratic trinomials because not all quadratic trinomials can be
factored into products of binomials with integer coefficients
We call these prime (Prime Numbers are 3 5 7 11 13 etc)Expressions such as x2 + 2x - 7 cannot be factored at all and is therefore known as a
prime polynomial
Practicing Factoring when a ne1
Please watch the demonstration below on factoring when a ne 1 There will be
interactive examples provided to help when a ne 1
MORE FACTORING
Upon completion of the video and demonstration please complete Mastery
Assignment Part 2
Gizmo Factoring
ax2 + bx + c
More InstructionPractice
Application
Problems
Practice All
Other Cases
Practicing Factoring when a ne1
Please watch the demonstration below on factoring when a ne 1 There will be
interactive examples provided to help when a ne 1
MORE FACTORING
Upon completion of the video and demonstration please complete Mastery
Assignment Part 2
Gizmo Factoring
ax2 + bx + c
More InstructionPractice
Application
Problems
Practice All
Other Cases
Gizmo Factoring
ax2 + bx + c
More InstructionPractice
Application
Problems
Practice All
Other Cases