Pearson Algebra 1 Common Core, 2011
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Pearson Algebra 1 Common Core, 2011 View details
6. Factoring ax²+ bx + c
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Exercise 31 Page 521

Start by identifying a, b, and c. What are the given values? What is the missing value?

Example Values: 13 and 20
Example Factored Expressions: (g+1)(9g+4) and (g+2)(9g+2)

Practice makes perfect

First, we will find two different values that complete the expression so that the trinomial can be factored into the product of two binomials. Then we will factor the resulting trinomials.

Finding Two Different Values

Let's start by identifying the given and missing coefficients in the quadratic expression. 9g^2+ g+4 ⇔ 9g^2+ b g+ 4

We have that a= 9 and c= 4. Next, we need to find the product of a and c. 9 * 4= 36 Now, we will find two ways of writing 36 as a product of two factors. Two possible values for b are the sum of those factors.

Written as a Product Factors b
36 = 9 * 4 9 and 4 9+4= 13
36=18 * 2 18 and 2 18 + 2= 20

Note that there are several possible missing values, these are only two options.

Factoring the First Trinomial

We will assume that b= 13 and factor the quadratic expression.
9g^2+ 13g+4
9g^2+9g+4g+4
Factor out 9g & 4
9g(g+1)+4g+4
9g(g+1)+4(g+1)
(g+1)(9g+4)

Factoring the Second Trinomial

Now, we will consider b= 20 and factor the trinomial.
9g^2+ 20g+4
9g^2+18g+2g+4
Factor out 9g & 2
9g(g+2)+2g+4
9g(g+2)+2(g+2)
(g+2)(9g+2)