1. Algebra 1 Glencoe McGraw-Hill JoAnn Evans
Math 8H
8-4 (day 2)
More Factoring Practice
Algebra Glencoe McGraw-Hill JoAnn Evans

2. 3y(2x - 1)(4x + 3) is the factored form.
24x2y + 6xy – 9y
Start every factoring problem by checking for a greatest common monomial factor.
3y(8x2 + 2x – 3)
-24
Check your solution! ● 6 -4 + 2
3y(2x - 1)(4x + 3) is the factored form.

3. (2y + 5)(5y + 3) is the factored form.
150 ●
Check your solution! 6 25 + 31
(2y + 5)(5y + 3) is the factored form.

4. (x + 6)(4x - 1) is the factored form.
Put the polynomial in standard form before you begin to factor.
4x2 + 23x – 6
-24 ● 24 -1 + 23
(x + 6)(4x - 1) is the factored form.

5. Notice the power of the first term.
Practice Problems
4 - 25n + 25n2
x4 - 10x2 + 25
25n2 - 25n + 4
Notice the power of the first term.
100 25 ● -5 ●
-20 -5 -5 + +
-25
-10
(5n - 1)(5n - 4)
(x2 - 5)(x2 - 5)

6. Solve Equations by Factoring (find the roots)
x2 = -8x
Set the equation equal to 0.
x2 + 8x = 0
Factor the polynomial completely.
x(x + 8)= 0
Set each of the factors equal to 0.
x = 0 or x + 8 = 0
x =

7. Solve Equations by Factoring (find the roots)
9y2 = 12y - 4 36
Set the equation equal to 0. Make sure it’s in standard form. ● -6 -6 +
-12
9y2 – 12y + 4 = 0
Factor the polynomial completely.
(3y – 2)(3y - 2)= 0
Set each of the factors equal to 0.
3y - 2 = 0

8. 1st integer  2nd integer = 80
Find two consecutive positive even integers whose product is 80.
Let x = the 1st integer
Let x + 2 = the 2nd integer
1st integer  2nd integer = 80
x(x + 2) = 80
x2 + 2x = 80
-80
x2 + 2x – 80 = 0 ● 10 -8 +
(x + 10) (x – 8) = 0 2
x + 10 = 0 or 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.

9. The width of a rectangular poster is 7 in. less than its length
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
x - 7
length  width = area x
x(x - 7) = 78
-78
x2 - 7x = 78 ● 6
-13 +
x2 - 7x – 78 = 0 -7
(x + 6) (x – 13) = 0
Only the positive root makes sense in this context. Length can’t be negative. Discard the negative root.
x = -6 or 13
Length is 13 and width is 6.

10. square of 1st + square of 2nd = 145
The sum of the squares of two negative consecutive integers is Find the integers.
Let x = 1st integer
Let x + 1 = 2nd integer
square of 1st + square of 2nd = 145
x2 + (x + 1)2 = 145
-72
x2 + (x + 1)(x + 1) = 145 ● 9 -8 +
x2 + x2 + x + x + 1 = 145 1
2x2 + 2x + 1 = 145
x = -9 or 8
2x2 + 2x – 144 = 0
Which answer doesn’t make sense in the context of this problem?
2(x2 + x – 72) = 0
2(x + 9)(x – 8) = 0
x + 9 = 0 or x – 8 = 0
The integers are -9 & -8.