factoring polynomials. copyright © by houghton mifflin company, inc. all rights reserved. 2...

Factoring Polynomials

Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 2

The simplest method of factoring a polynomial is to factor out the greatest common factor (GCF) of each term.

Example: Factor 18x3 + 60x.

GCF = 6x18x3 + 60x = 6x (3x2) + 6x (10)

18x3 = 2 · 3 · 3 · x · x · x

Apply the distributive law

to factor the polynomial.

6x (3x2 + 10) = 6x (3x2) + 6x (10) = 18x3 + 60x

Check the answer by multiplication.

Factor each term.

Find the GCF.

60x = 2 · 2 · 3 · 5 · x

= 6x (3x2 + 10)

= (2 · 3 · x) · 3 · x · x

= (2 · 3 · x) · 2 · 5


Example: Factor 4x2 – 12x + 20.

GCF = 4.

4(x2 – 3x + 5) = 4x2 – 12x + 20Check the answer.

A common binomial factor can be factored out of certain expressions.

Example: Factor the expression 5(x + 1) – y(x + 1).

5(x + 1) – y(x + 1) = (x + 1) (5 – y)

(x + 1) (5 – y) = 5(x + 1) – y(x + 1)Check.

= 4(x2 – 3x + 5)


A difference of two squares can be factored using the formula

Example: Factor x2 – 9y2.

= (x)2 – (3y)2

= (x + 3y)(x – 3y)

Write terms as perfect squares.

Use the formula.

The same method can be used to factor any expression which can be written as a difference of squares.

Example: Factor (x + 1)2 – 25y 4.

= (x + 1)2 – (5y2)2

= [(x + 1) + (5y2)][(x + 1) – (5y2)]

= (x + 1 + 5y2)(x + 1 – 5y2)

a2 – b2 = (a + b)(a – b).

x2 – 9y2

(x + 1)2 – 25y 4


2. Factor 2a2 + 3bc – 2ab – 3ac.

Some polynomials can be factored by grouping terms to produce a common binomial factor.

= 2a2 – 2ab + 3bc – 3ac

= y (2x + 3) – 2(2x + 3)

= (2a2 – 2ab) + (3bc – 3ac)

= 2a(a – b) + 3c(b – a)

= (2xy + 3y) – (4x + 6) Group terms.

Examples: 1. Factor 2xy + 3y – 4x – 6.

Factor each pair of terms.

= (2x + 3) ( y – 2) Factor out the common binomial.

Rearrange terms.

Group terms.

Factor.

= 2a(a – b) – 3c(a – b) b – a = – (a – b).

= (a – b) (2a – 3c) Factor.

Examples: Factor

2xy + 3y – 4x – 6

2a2 + 3bc – 2ab – 3ac

Notice the sign!


Factoring these trinomials is based on reversing the distributive property process.

To factor a simple trinomial of the form x2 + bx + c, express the trinomial as the product of two binomials. For example,

x2 + 10x + 24 = (x + 4)(x + 6).

Example: Factor x2 + 3x + 2. = (x + a)(x + b)

Express the trinomial as a product of two binomials with leading term x and unknown constant terms a and b.

= x2 Multiply the binomials.

= x2 + (b + a) x + ba Since ab = 2 and a + b = 3, it follows that a = 1 and b = 2.

= x2 + (1 + 2) x + 1 · 2

Therefore, x2 + 3x + 2 = (x + 1)(x + 2).

+ bx + ax + ba

x2 + 3x + 2

(Product-sum method)


Example: Factor x2 – 8x + 15.

= (x + a)(x + b)

(x – 3)(x – 5) = x2 – 5x – 3x + 15

x2 – 8x + 15 = (x – 3)(x – 5).

Therefore a + b = -8

Check:

= x2 + (a + b)x + ab

It follows that both a and b are negative.

= x2 – 8x + 15.

SumNegative Factors of 15

-3, - 5 - 8

- 1, - 15 -15

and ab = 15.

x2 – 8x + 15


Example: Factor x2 + 13x + 36.

= (x + a)(x + b)

Check: (x + 4)(x + 9)

Therefore a and b are:

x2 + 13x + 36

= x2 + 9x + 4x + 36 = x2 + 13x + 36.

= (x + 4)(x + 9)

= x2 + (a + b) x + ab

SumPositive Factors of 36

1, 36 37

153, 12

4, 9 13

6, 6 12

2, 18 20

x2 + 13x + 36

two positive factors of 36

whose sum is 13.

36 13


Example: Factor 4x3 – 40x2 + 100x.

A polynomial is factored completely when it is written as a product of factors that can not be factored further.

The GCF is 4x.

= 4x(x2 – 10x + 25) Use distributive property to factor out the GCF.

= 4x(x – 5)(x – 5) Factor the trinomial.

4x(x – 5)(x – 5) = 4x(x2 – 5x – 5x + 25)

= 4x(x2 – 10x + 25)

= 4x3 – 40x2 + 100x

4x3 – 40x2 + 100x

Check:


Factoring complex trinomials of the form ax2 + bx + c, (a 1) can be done by decomposition or cross-check method.

Example: Factor 3x2 + 8x + 4.

1. Find the product of

first and last terms

3 4 =

12Decomposition Method

2. We need to find factors of 12whose sum is 8

1, 122, 63, 4

3. Rewrite the middle

term decomposed

into the two numbers

3x2 + 2x + 6x + 4

= x(3x + 2) + 2(3x + 2)

= (3x2 + 2x) + (6x + 4)4. Factor by grouping

in pairs = (3x + 2) (x + 2)

3x2 + 8x + 4 = (3x + 2) (x + 2)


Example: Factor 4x2 + 8x – 5.

4x2 + 8x – 5 = (2x –1)(2x – 5)

4 5 =

20 We need to find factors of 20

whose difference is 81, 202, 104, 5

Rewrite the middle term

decomposed into the

two numbers

4x2 – 2x + 10x – 5

= 2x(2x – 1) + 5(2x – 1)

= (4x2 – 2x) + (10x – 5)

= (2x – 1) (2x + 5)

Factor by grouping

in pairs

Solve by Factoring

• To solve a quadratic by factoring you first get the equation equal to zero and then factor.

• Once factored you set each expression equal to zero and solve for the variable.


Solving Quadratic Equations by

Completing the Square

Perfect Square Trinomials

Examples x2 + 6x + 9 x2 - 10x + 25 x2 + 12x + 36

Creating a Perfect Square Trinomial

In the following perfect square trinomial, the constant term is missing. X2 + 14x + ____

Find the constant term by squaring half the coefficient of the linear term.

(14/2)2

X2 + 14x + 49

Perfect Square Trinomials

Create perfect square trinomials.

x2 + 20x + ___ x2 - 4x + ___ x2 + 5x + ___

100

4

25/4

Solving Quadratic Equations by Completing

the SquareSolve the following

equation by completing the square:

Step 1: Move quadratic term, and linear term to left side of the equation

2 8 20 0x x

2 8 20x x

Solving Quadratic Equations by Completing

the SquareStep 2: Find the term

that completes the square on the left side of the equation. Add that term to both sides.

2 8 =20 + x x 21

( ) 4 then square it, 4 162

8

2 8 2016 16x x

Solving Quadratic Equations by Completing the Square

Step 3: Factor

the perfect square trinomial on the left side of the equation. Simplify the right side of the equation.

2 8 2016 16x x

2

( 4)( 4) 36

( 4) 36

x x

x


Step 4: Take the square root of each side

2( 4) 36x

( 4) 6x


Step 5: Set up the two possibilities and solve

4 6

4 6 an

d 4 6

10 and 2 x=

x

x x

x

Completing the Square-Example #2

Solve the following equation by completing the square:

Step 1: Move quadratic term, and linear term to left side of the equation, the constant to the right side of the equation.

22 7 12 0x x

22 7 12x x


2

2

2

2

2

1. 2 63 0

2. 8 84 0

3. 5 24 0

4. 7 13 0

5. 3 5 6 0

x x

x x

x x

x x

x x

Try the following examples. Do your work on your paper and then check your answers.

1. 9,7

2.(6, 14)

3. 3,8

7 34.

2

5 475.

6

i

i

The Quadratic Formula.

a

acbbx

2

42

What Does The Formula Do ?

The Quadratic formula allows you to find the roots of a quadratic

equation (if they exist) even if the quadratic equation does not

factorise.

The formula states that for a quadratic equation of the form :

ax2 + bx + c = 0

The roots of the quadratic equation are given by :

a

acbbx

2

42

Example 1

Use the quadratic formula to solve the equation :

x 2 + 5x + 6= 0

Solution:

x 2 + 5x + 6= 0

a = 1 b = 5 c = 6

a

acbbx

2

42

12

)614(55 2

x

2

)24(255 x

2

15 x

2

15

2

15

xorx

x = - 2 or x = - 3

These are the roots of the equation.

Example 2


8x 2 + 2x - 3= 0

Solution:

8x 2 + 2x - 3= 0

a = 8 b = 2 c = -3

a

acbbx

2

42

82

)384(22 2

x

16

)96(42 x

16

1002 x

16

102

16

102

xorx

x = ½ or x = - ¾


Example 3


8x 2 - 22x + 15= 0

Solution:

8x 2 - 22x + 15= 0

a = 8 b = -22 c = 15

a

acbbx

2

42

82

)1584()22()22( 2

x

16

)480(484(22 x

16

422 x

16

222

16

222

xorx

x = 3/2 or x = 5/4



8x 2 - 22x + 15= 0

Solution:

8x 2 - 22x + 15= 0

a = 8 b = -22 c = 15

x = 3/2 or x = 5/4


Example 4

Use the quadratic formula to solve for x to 2 d.p :

2x 2 +3x - 7= 0

Solution:

2x 2 + 3x – 7 = 0

a = 2 b = 3 c = - 7

a

acbbx

2

42

22

)724(33 2

x

4

)56(93 x

4

653 x

4

0622.83

4

0622.83

xorx

x = 1.27 or x = - 2.77


factoring polynomials. copyright © by houghton mifflin company, inc. all rights reserved. 2...

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