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Unit 5 Notes FACTORING POLYNOMIALS

Factorization

How many pairs of monomials can you find whose product is

416x ?

The process they went through, finding pairs of monomials with a

product of

416x is called FACTORING.

Each pair of monomials is called a FACTORIZATION.

FACTOR MONOMIAL

Example 1

Find the factorization of 15𝑥3

Factors are 1,-1,3,-3,5,-5,15,-15,x, 𝑥2, 𝑥3.

Some factorizations are

(15x)( 𝑥2) (5x)( 3𝑥2

) (3)( 5𝑥3)

(1)( 15𝑥3) (3x)( 5𝑥2

) (5)( 3𝑥3)

Since (-1)(-1) = 1, we could have

(-15𝑥3)(-1) (-3)( −5𝑥3

) (−𝑥3)(-15)

There are still other ways to factor 15𝑥3.

Try This. Find 3 factorizations of each monomial.

a. 8𝑥4

b. 6𝑚5

c. 12𝑎3𝑏2

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FACTORING TERMS WITH A COMMON FACTOR

Factor out is explained as the opposite of the Distributive Property.

Always make sure to use the greatest possible coefficient and the variable

to the greatest power.

DISTRIBUTE (Multiply) FACTOR (Divide)

2 2

2

3 ( 2) 3 ( ) 3 (2)

................. 3 6

a b a b a

ab a

25 15 5 5 3

................ 5 ( 3)

x x x x x

x x

Examples

1. 3𝑥2 + 3 = 3(𝑥2) + 3(1)

= 3(𝑥2 + 1)

2. 16𝑎2𝑏2 + 20𝑎2 = 4𝑎2(4𝑏2) + 4𝑎2(5)

= 4𝑎2(4𝑏2 + 5)

3. 15𝑥5 − 12𝑥4 + 27𝑥3 − 3𝑥2

= 3𝑥2(5𝑥3) − 3𝑥2(4𝑥2) + 3𝑥2(9𝑥) − 3𝑥2(1)

= 3𝑥2(5𝑥3 − 4𝑥2 + 9𝑥 − 1)

Try This

d. 𝑥2 + 3𝑥

e. 3𝑥6 − 5𝑥3 + 2𝑥2

f. 9𝑥4 − 15𝑥3 + 3𝑥2

g. 12𝑚4𝑛4 + 3𝑚3𝑛2 + 6𝑚2𝑛 ≈

3

Factoring Difference of Two Squares

DIFFERENCE OF TWO SQUARES

For a binomial to be the difference of 2 squares; two conditions must be

met:

There must be two terms, both perfect squares (

2 4 816, ,9 ,25y x x )

There must be a minus sign between the two terms (difference)

Examples

1. Is 9𝑥2 − 36 a difference of two squares?

The first term is a square. 9𝑥2 = (3𝑥)2

The second term is a square. 36 = 62

There is a minus sign between them. Thus we have a difference of

two squares.

2. Is −4𝑥2 + 16 a difference of two squares?

−4𝑥2 + 16 = 16 − 4𝑥2

16 = 42 𝑎𝑛𝑑 −4𝑥2 = (2𝑥)2

Since there is a minus sign between 16 and −4𝑥2 , we have a

difference of two squares.

Try This. State whether each expression is a difference of two squares.

a. 𝑥2 − 25

b. 𝑥2 − 24

c. −36 − 𝑥2

d. 4𝑥2 − 15

e. −49 + 16𝑎2

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FACTORING THE DIFFERENCE OF TWO SQUARES

2 2( )( )a b a b a b …if we look carefully, the product IS the

difference of two squares; so writing the equation in reverse is factoring

the difference of two squares…

2 2( ) ( )( )a b a b a b

Examples

3. 4𝑥2 − 25 = (2𝑥)2 − (5)2 = (2𝑥 + 5)(2𝑥 − 5)

=(𝐴)2 − (𝐵)2 = (𝐴 + 𝐵)(𝐴 − 𝐵)

4. 𝑚6 − 16𝑛2 = (𝑚3)2 − (4𝑛)2 = (𝑚3 + 4𝑛)(𝑚3 − 4𝑛)

5. 36𝑥2 − 25𝑦6 = (6𝑥)2 − (5𝑦3)2 = (6𝑥 − 5𝑦3)(6𝑥 + 5𝑦3)

6. 9𝑎8𝑏4 − 49 = (3𝑎4𝑏2 + 7)(3𝑎4𝑏2 − 7)

Try This. Factor

f. 4𝑦2 − 49

g. 16𝑥2 − 25𝑦2

h. 𝑎8𝑏4 − 4

i. 25𝑎10 − 36𝑏8

If the terms of the binomial have a common factor, first factor out the

common factor. Then continue factoring.

4 6 4 2

4 2 2

4

49 9 (49 9 )

.................. (7) (3 )

................. (7 3 )(7 3 )

x x x x

x x

x x x

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Example

9. 18𝑎2𝑏2 − 50𝑎6 = 2𝑎2(9𝑏2 − 25𝑎4)

= 2𝑎2[(3𝑏)2 − (5𝑎 )2]

= 2𝑎2(3𝑏 − 5𝑎2)(3𝑏 + 5𝑎2)

Try This. Factor

j. 32𝑦6 − 8𝑦2

k. 5 − 20𝑦6

l. 𝑎3𝑏 − 4𝑎𝑏3

m. 64𝑥6𝑦4 − 25𝑥4𝑦8

FACTORING COMPLETELY

After you factor the difference of 2 squares, you can sometimes continue

to factor. Factoring Completely means factor until factoring is no longer

possible (other than for a common factor of 1)

Example. Factor

10. 1 − 16𝑥12 = (1)2 − (4𝑥6)2

= (1 − 4𝑥6)(1 + 4𝑥6)

= [12 − (2𝑥3)2](1 + 4𝑥6)

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= (1 + 2𝑥3)(1 − 2𝑥3)(1 + 4𝑥6)

Reminder, that one factoring does NOT mean they are finished. Assume

there will always be another factoring available and that way never stop

after the first step.

Try This. Factor

h. 81𝑥4 − 1

i. 16𝑚4 − 𝑛8

TRINOMIAL SQUARES.

RECOGNIZE A TRINOMIAL SQUARE

2( 3)x 2( 3)x

2 6 9x x 2 6 9x x

Use the following to help identify a trinomial square:

Two of the terms must be squares (

2 2A andB )

There must be NO minus sign before either of the two squares

If we multiply A and B and double the result we get the third term,

either 2AB or the additive inverse, -2AB

Examples

1. Is 𝑥2 + 6𝑥 + 9 a trinomial square?

A. 𝑥2 = (𝑥)2 𝑎𝑛𝑑 9 = (3)2

B. There is no minus sign before 𝑥2 and 9.

C. If we multiply x and 3 and double the results, we get the third

term, 2 ∙ 3 ∙ 𝑥, 𝑜𝑟 6𝑥

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Thus 𝑥2 + 6𝑥 + 9 is the square of the binomial (x + 3).

2. Is 𝑥2 + 6𝑥 + 11 a trinomial square?

The answer is no because only one term is a square.

3. Is 16𝑎2 − 56𝑎𝑏 + 49𝑏2 a trinomial square?

A. 16𝑎2 = (4𝑎)2 𝑎𝑛𝑑 49𝑏2 = (7𝑏)2

B. There is no minus sign before 16𝑎2and 49𝑏2

.

C. If you multiply 4a and 7b and double the result, we get the

additive inverse of the third term, 2 ∙ 4𝑎 ∙ 7𝑏 = 56𝑎𝑏

Thus 16𝑎2 − 56𝑎𝑏 + 49𝑏2 is the square of (4a - 7b)

Try This. Which of the following are trinomial squares?

a. 𝑥2 + 8𝑥 + 16

b. 𝑥2 − 12𝑥 + 4

c. 4𝑥2 + 20𝑥 + 25

d. 9𝑥2 − 14𝑥 + 16

e. 16𝑥2 + 40𝑥𝑦 + 25𝑦2

FACTORING TRINOMIAL SQUARES

2 2 22 ( )A AB B A B Notice the addition sign between the first 2

terms gives an addition sign in the factorization.

2 2 22 ( )A AB B A B Again notice, the subtraction sign between the

first two terms, gives a subtraction sign in the factorization.

***Remember to ALWAYS factor out a common factor first if possible***

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Examples. Factor

4. 𝑥2 + 6𝑥 + 9 = 𝑥2 + 2 ∙ 𝑥 ∙ 3 + 32 = (𝑥 + 3)2

5. 𝑥2 − 14𝑥 + 49 = 𝑥2 − 2 ∙ 𝑥 ∙ 7 + 72 = (𝑥 − 7)2

6. 16𝑎2 − 40𝑎𝑏 + 25𝑏2 = (4𝑎 − 5𝑏)2

7. 27𝑚2 + 72𝑚𝑛 + 48𝑛2 = 3(9𝑚2 + 24𝑚𝑛 + 16𝑛2)

=3(3𝑚 + 4𝑛)2

Try This. Factor

f. 𝑥2 + 2𝑥 + 1

g. 𝑥2 − 2𝑥 + 1

h. 25𝑥2 − 70𝑥 + 49

i. 48𝑚2 + 120𝑚𝑛 + 75𝑛2

FACTORING

2x bx c , where c > 0 (Constant term is

Positive)

Explain to students that in the polynomial

2x bx c , c is called

the constant term or just the constant.

If the constant of the polynomial is NOT a perfect square, the trinomial

cannot be factored into the square of a binomial like

2 8 16x x 2( 4)x ,is for example.

It may be possible to be factored as the product of two DIFFERENT

binomials.

Example 1

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Factor 𝑥2 + 5𝑥 + 6.

The first term of each factor is x.

(x+_)(x+_)

Look for the two numbers whose product is 6 and whose sum is 5.

Product of 6 Sum

1,6 7

2,3 5

𝑥2 + 5𝑥 + 6 = (𝑥 + 2)(𝑥 + 3)

Try This. Factor

a. 𝑥2 + 7𝑥 + 12

b. 𝑥2 + 13𝑥 + 36

In Example 2 on we now have to worry about the second term being

negative. Using FOIL in reverse tells us that out factors will be negative

(negative times a negative is a positive) and the sum will be a negative

(negative plus negative is negative)

Example 2

Factor 𝑥2 − 8𝑥 + 12

Since the coefficient of the middle term is negative, we need two negative

numbers whose products are 12 and whose sum is -8.

Product of 12 Sum

-1,-12 -13

-2,-6 -8

-3,-4 -7

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𝑥2 − 8𝑥 + 12 = (𝑥 − 2)(𝑥 − 6)

Try This. Factor

c. 𝑥2 − 8𝑥 + 15

d. 𝑥2 − 9𝑥 + 20

e. 𝑥2 − 7𝑥 + 12

Example 3 takes what we have done and added the twist of having two

variables.

Factor 𝑎2 + 7𝑎𝑏 + 10𝑏2

Since 𝑎2 is the product of a times a, and 𝑏2

is the product of b times b, we

are looking for binomials of the form.

(a + _b)(a + _b)

Find the two numbers whose sum is 7 and whose product is 10.

Product of 10 Sum

1,10 11

2,5 7

𝑎2 + 7𝑎𝑏 + 10𝑏2 = (𝑎 + 2𝑏)(𝑎 + 5𝑏)

Try This . Factor

f. 𝑚2 + 8𝑚𝑛 + 15𝑛2

g. 𝑎2 + 5𝑎𝑏 + 6𝑏2

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h. 𝑝2 + 6𝑝𝑞 + 8𝑞2

CONSTANT TERM IS NEGATIVE

2x bx c , where c < 0

Because the constant term is negative, or less than zero, the middle term

MAY be positive or negative.

2

2

( 5)( 2) 2 5 10

.......................... 3 10

x x x x x

x x

2

2

( 5)( 2) 5 2 10

............................ 3 10

x x x x x

x x

In both cases the constant term is less than zero, or negative, so the

second term will determine how we factor.

Example 4

Factor 𝑥2 − 8𝑥 − 20.

Find two numbers whose sum is -8 and whose product is -20.

Product of -20 Sum

-1,20 19

1,-20 -19

-2,10 8

2,-10 -8

4,-5 -1

-4,5 1

𝑥2 − 8𝑥 − 20 = (𝑥 + 2)(𝑥 − 10)

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Example 5 demonstrates 2 variables.

Factor 𝑎2 − 𝑎𝑏 − 6𝑏2

We are looking for binomials of the form (a_b)(a_b). Find two numbers

whose sum is 1 and whose product is -6.

Product of -6 Sum of 1

1,-6 -5

-1,6 5

2,-3 -1

-2,3 1

𝑎2 − 𝑎𝑏 − 6𝑏2 = (𝑎 + 2𝑏)(𝑎 − 3𝑏)

Try This. Factor.

i. 𝑥2 + 4𝑥 − 12

j. 𝑥2 − 13𝑥 + 12

k. 𝑎2 + 5𝑎𝑏 − 14𝑏2

l. 𝑥2 − 𝑥𝑦 − 30𝑦2

FACTORING

2ax bx c

What is (2x+5)(3x+4) ?

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Students should work this out using FOIL or the Box method;

26 23 20x x

Factoring Trinomials

To factor 𝑎𝑥2 + 𝑏𝑥 + 𝑐, we look for binomials

(_x + _)(_x + _) where products of numbers in the blanks are as follows.

1. The numbers in the first blanks of each binomial have a product of

a.

2. The numbers in the last blanks of each binomial have a product of

c.

3. The outside product and the inside product have a sum of b.

Examples

1. Factor 3𝑥2 + 5𝑥 + 2.

First look for a factor common to all terms. There is none. Next look for

two numbers whose product is 3.

1,3 -1,-3

Now look for numbers whose product is 2.

1,2 -1,-2

Since the last term of the trinomial is positive, the signs of the second

terms must be the same. Here are some possible factorizations.

(x + 1)(3x + 2) (x + 2)(3x + 1)

(x – 1)(3x – 2) (x - 2)(3x – 1)

When we multiply, the first term will be 3𝑥2 and the last term will be 2 in

each case. Only the first multiplication gives a middle term of 5x.

3𝑥2 + 5𝑥 + 2 = (𝑥 + 1)(3𝑥 + 2)

2. Factor 2𝑥2 + 5𝑥 − 12.

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First term: Find two numbers whose product is 2.

Last term: Find two numbers with product of -12.

(2x + 3)(x – 4) (2x – 2)(x + 6) (2x – 1)(x + 12)

(2x – 3)(x + 4) (2x + 2)(x – 6) (2x - 12)(x + 1)

The outside product plus the inside product must equal 5x.

2𝑥2 + 5𝑥 − 12 = (2𝑥 − 3)(𝑥 + 4)

Try This

a. 6𝑥2 + 7𝑥 + 2

b. 8𝑥2 + 10𝑥 − 3

c. 6𝑥2 + 41𝑥 − 7

3. Factor 8𝑚2 + 8𝑚 − 6.

8𝑚2 + 8𝑚 − 6 = 2(4𝑚2 + 4𝑚 − 3)

1st

term: Find two numbers whose product is 4.

2nd

term: Find two numbers whose product is -3.

(4m + 3)(m - 1) (4m – 3)(m + 1) (2m + 3)(2m – 1)

(4m – 1)(m + 3) (4m + 1)(m - 3) (2m – 3)(2m + 1)

The outside product plus the inside product must be equal to 4m.

8𝑚2 + 8𝑚 − 6 = 2(4𝑚2 + 4𝑚 − 3) = 2(2𝑚 + 3)(2𝑚 − 1)

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Try This

4. 3𝑥2 − 21𝑥 + 36

5. 9𝑎2 − 15𝑎 − 6

6. 4𝑎2 + 2𝑎 − 6

7. 6𝑚2 + 15𝑚𝑛 − 9𝑛2

FACTORING BY GROUPING

Can I use the distributive property to factor this polynomial? If so, show

me. 4 3 23 12 6 9x x x x

3 23 ( 4 2 3)x x x x The common factor 3x was removed from each

term within the polynomial.

Can I use the distributive property to factor this polynomial?

3 2 2 2x x x …..Unfortunately there is no one common factor

for all the terms, but maybe students may see the following:

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3 2x x has a common factor (

2x )(x+1)

2x + 2 has a common factor (2)(x + 1)

Now, if we put this all together we would get

2

2

( 1) 2( 1)

( 2)( 1)

x x x

x x

The (x+1) can be factored from both.

Because we factored in sections or groups set apart using parentheses,

this is called factoring by grouping.

Examples

1. 6𝑥3 − 9𝑥2 + 4𝑥 − 6 = (6𝑥3 − 9𝑥2) + (4𝑥 − 6)

= 3𝑥2(2𝑥 − 3) + 2(2𝑥 − 3)

= (2𝑥 − 3)(3𝑥2 + 2)

2. 𝑥3 + 𝑥2 + 𝑥 + 1 = (𝑥3 + 𝑥2) + (𝑥 + 1)

= 𝑥2(𝑥 + 1) + 1(𝑥 + 1)

= (𝑥 + 1)(𝑥2 + 1)

3. 𝑥3 + 2𝑥2 − 𝑥 − 2 = (𝑥3 + 2𝑥2) + (−𝑥 − 2)

= 𝑥2(𝑥 + 2) + 1(−𝑥 − 2)

= 𝑥2(𝑥 + 2) − 1(𝑥 + 2)

= (𝑥 + 2)(𝑥2 − 1)

= (𝑥 + 2)(𝑥 − 1)(𝑥 + 1)

4. 𝑥2𝑦2 + 𝑎𝑦2 + 𝑎𝑏 + 𝑏𝑥2 = 𝑦2(𝑥2 + 𝑎) + 𝑏(𝑥2 + 𝑎)

= (𝑥2 + 𝑎)(𝑦2 + 𝑏)

5. 𝑥3 + 𝑥2 + 2𝑥 − 2 = 𝑥2(𝑥 + 1) + 2(𝑥 − 1)

Cannot be factored.

Try This. Factor

a. 8𝑥3 + 2𝑥2 + 12𝑥 + 3

b. 4𝑥3 − 6𝑥2 − 6𝑥 + 9

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c. 𝑥3 + 𝑥2 − 𝑥 − 1

d. 3𝑎 − 6𝑏 + 5𝑎2 − 10𝑎𝑏

FACTORING POLYNOMIALS

A. ALWAYS look first for a common factor

B. Then look at the number of terms

2 terms: Is it the difference of 2 squares

3 terms: Is the trinomial a square of a binomial? If not, test the

Factors of the terms

4 terms: try factoring by grouping

C. ALWAYS factor completely

SOLVING EQUATIONS BY FACTORING

The Principle of Zero Products

The product of two or more factors is zero if any of the factors are

equal to zero.

If a product is zero, then one or more of the factors must be zero

For any rational number a and b, if ab=0, then a=0 or b=0 and if

a=0 or b=0, then ab=0

If we have an equation with zero on one side and a factorization on the

other, we can solve the equation by finding the values that make the

factors zero. This means there can be more than ONE right answer.

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(5 1)( 7) 0

5 1 0......... ........ 7 0

5 1.......................... 7

1

5

x x

x or x

x x

x

Check to see if these solutions

work by substituting, one at a time, both values for x and calculate

1 1(5( ) 1)( 7)

5 5

1( 1 1)( 7 )

5

10( 7 )

5

0

(5 7 1)(7 7)

(35 1)(0)

(36)(0)

0

FACTORING and SOLVING

Use the following steps to solve equations using the principle of zero

Get zero on one side of the equation by using the addition property

Factor the expression on the other side of the equation

Set each factor equal to zero

Solve each equation.

Check your solutions.

Example 1. Solve.

𝑥2 − 6𝑥 = 16

𝑥2 − 8𝑥 + 16 = 0

(𝑥 − 8)(𝑥 + 2) = 0

𝑥 − 8 = 0 𝑜𝑟 𝑥 + 2 = 0

𝑥 = 8 𝑜𝑟 𝑥 = −2

Check:

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𝑥2 − 6𝑥 = 16

(8)2 − 6(8) = 16

64 − 48 = 16

𝑥2 − 6𝑥 = 16

(−2)2 − 6(−2) = 16

4 + 12 = 16

Example 2

𝑥2 + 5𝑥 + 6 = 0

(𝑥 + 2)(𝑥 + 3) = 0

𝑥 + 2 = 0 𝑜𝑟 𝑥 + 3 = 0

𝑥 = −2 𝑜𝑟 𝑥 = −3

Check:

𝑥2 + 5𝑥 + 6 = 0

(−2)2 + 5(−2) + 6 = 0

4 − 10 + 6 = 0

𝑥2 + 5𝑥 + 6 = 0

(−3)2 + 5(−3) + 6 = 0

9 − 15 + 6 = 0

The solutions are -2 and -3.

Try This

a. 𝑥2 − 𝑥 − 6 = 0

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b. 𝑥2 − 𝑥 = 56

c. 𝑥2 − 3𝑥 = 28

The answers found to solving the equations are called the root of the

polynomial; the value of the variable that makes the polynomial equal to

zero.

Example 3 shows how the word “root” will be used.

Find the roots of 4𝑥2 − 25

4𝑥2 − 25 = 0

(2𝑥 − 5)(2𝑥 + 5) = 0

2𝑥 − 5 = 0 𝑜𝑟 2𝑥 + 5 = 0

2𝑥 = 5 𝑜𝑟 2𝑥 = −5

𝑥 =5

2 𝑜𝑟 𝑥 =

−5

2

Check:

4𝑥2 − 25 = 0

4(5

2)2 − 25 = 0

4(25

4) − 25 = 0

0 = 0

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4𝑥2 − 25 = 0

4(−5

2)2 − 25 = 0

4(25

4)2 − 25 = 0

0 = 0

Try This. Find the roots.

d. 𝑥2 + 6𝑥 + 9

e. 25𝑥2 − 16

SOLVE PROBLEMS BY WRITING & SOLVING EQUATIONS

Examples Translate to an equation and solve.

1. The product of one more than a number and one less than the

number is 8. Find the number.

Let x = the number.

One more than a number times one less than the number is 8.

(x + 1)(x – 1) = 8

𝑥2 − 1 = 8

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𝑥2 − 1 − 8 = 0

𝑥2 − 9 = 0

(x – 3)(x + 3) = 0

x – 3 = 0 or x + 3 = 0

x = 3 or x= -3

Check: Both 3 and -3 are the solutions.

2. The square of a number minus twice the number is 48. Find the

number.

Let x = the number.

𝑥2 − 2𝑥 = 48

𝑥2 − 2𝑥 − 48 = 0

(𝑥 − 8)(𝑥 + 6) = 0

𝑥 − 8 = 0 𝑜𝑟 𝑥 + 6 = 0

𝑥 = 8 𝑜𝑟 𝑥 = −6

Both 8 and -6 check. They are both solutions.

Try This. Translate to an equation and solve.

a. The product of seven less than a number and eight less than the

number is 0.

.

b. The product of one more than a number and one less than the

number is 24.

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c. The square of a number minus the number is 20.

d. One more than twice the square of a number is 73.

3. The area of the foresail on a 12-meter racing yacht is 93.75 square

meters. The sail’s height is 8.75 meters greater than its base. Find

its base and height.

Area = 1

2∙ 𝑏𝑎𝑠𝑒 ∙ ℎ𝑒𝑖𝑔ℎ𝑡

Let h = the sail’s height and h – 8.75 = the length of the sail’s base.

1

2(ℎ − 8.75)ℎ = 93.75

(ℎ − 8.75)ℎ = 187.5

(ℎ2 − 8.75ℎ) = 187.5

(ℎ2 − 8.75ℎ − 187.5) = 0

(ℎ − 18.75)(ℎ + 10) = 0

ℎ − 18.75 = 0 𝑜𝑟 ℎ + 10 = 0

ℎ = 18.75 𝑜𝑟 ℎ = −10

The solution of the equation are 18.75 and -10. The height of the sail

cannot have a negative value, so the height must be 18.75 meters. The

height of the base is then 8.75 meters shorter, or 10 meters.

24

4. The product of two consecutive integers is 156. Find the integers.

Let x represent the first integer, Then x+1 represent the second

integer.

𝑥 ∙ (𝑥 + 1) = 156

𝑥(𝑥 + 1) = 156

𝑥2 + 𝑥 = 156

𝑥2 + 𝑥 − 156 = 0

(𝑥 − 12)(𝑥 + 13) = 0

𝑥 − 12 = 0 𝑜𝑟 𝑥 + 13 = 0

𝑥 = 12 𝑜𝑟 𝑥 = −13

When x = 12, x+1=13 and 12(13)=156

When x = -13, x+1=-12 and -13(-12)=156

We have two pairs of solutions, 12 and 13 and -12 and -13. Both are

pairs of consecutive integers whose product is 156.

Try This

e. The width of a rectangular card is 2cm less than the length. The

area is 15 cm squared. Find the length and width?

f. The product of two consecutive integers is 462. Find the integers?