Math 8H

Algebra 1 Glencoe McGraw-Hill JoAnn Evans

8-4 (day 2)More Factoring Practice

+●

-24

266 -4-4

24x2y + 6xy – 9y

3y(2x - 1)(4x + 3) is the factored

form.

Check your solution!

3y(8x2 + 2x – 3)

Start every factoring problem by checking for a greatest common monomial factor.

+●

150

3166 2525

10y2 + 31y + 15

(2y + 5)(5y + 3) is the factored form.

Check your solution!

+●

-24

232424 -1-1

23x - 6 + 4x2 Put the polynomial in standard form before you begin to

factor.

(x + 6)(4x - 1) is the factored form.

4x2 + 23x – 6

Practice Problems

4 - 25n + 25n2 x4 - 10x2 + 25

+●

100

-25-5-5 +

●

25

-10

-5-5-20-20 -5-5

(5n - 1)(5n - 4) (x2 - 5)(x2 - 5)

25n2 - 25n + 4 Notice the power of the first term.

Solve Equations by Factoring (find the roots)

x2 = -8x

x2 + 8x = 0

x(x + 8)= 0

x = 0 or x + 8 = 0

0 or 8x =

Set the equation equal to 0.

Factor the polynomial completely.

Set each of the factors equal to 0.

Solve Equations by Factoring (find the roots)

9y2 = 12y - 4

9y2 – 12y + 4 = 0

+●

36

-12

-6-6 -6-6

(3y – 2)(3y - 2)= 0

3y - 2 = 02

y3

Set the equation equal to 0. Make sure it’s in standard form.

Factor the polynomial completely.

Set each of the factors equal to 0.

Find two consecutive positive even integers whose product is 80.

Let x = the 1st integer Let x + 2 = the 2nd integer1st integer 2nd integer = 80

x(x + 2) = 80x2 + 2x = 80

x2 + 2x – 80 = 0+●

-80

21010 -8-8

(x + 10) (x – 8) = 0

x = -10 or 8 Remember the problem asked for positive

consecutive even integers. Only one of the roots will fit the context of this problem.

The 1st integer is 8 and the 2nd integer is

10.

x + 10 = 0 or x – 8 = 0

The width of a rectangular poster is 7 in. less than its length. If the poster’s area is 78 in2,

find the dimensions.

Let x = length Let x - 7 = width

length width = areax(x - 7) = 78

x2 - 7x = 78

x2 - 7x – 78 = 0+●

-78

-766 -13-13

(x + 6) (x – 13) = 0

x = -6 or 13Only the positive root makes sense in this

context. Length can’t be negative. Discard

the negative root.Length is 13 and

width is 6.

x

x - 7

The sum of the squares of two negative consecutive integers is 145. Find the integers.

Let x = 1st integer

Let x + 1 = 2nd integer

square of 1st + square of 2nd = 145x2 + (x + 1)2 = 145

x2 + (x + 1)(x + 1) = 145x2 + x2 + x + x + 1 = 145

+●

-72

199 -8-8

2x2 + 2x + 1 = 145

2x2 + 2x – 144 = 0

2(x2 + x – 72) = 0

2(x + 9)(x – 8) = 0

x + 9 = 0 or x – 8 = 0

x = -9 or 8

Which answer doesn’t make sense in the

context of this problem?

The integers are -9 & -8.