McGraw Hill Integrated II, 2012
MH
McGraw Hill Integrated II, 2012 View details
Study Guide and Review
Continue to next subchapter

Exercise 11 Page 217

Is there a greatest common factor between all of the terms in the given expression? If so, factor that out first.

x=-3, x=4

Practice makes perfect

Let's start factoring by first identifying the greatest common factor. Then, we will rewrite the expression as a trinomial with a leading coefficient of one.

Factor Out the GCF

The greatest common factor (GCF) of an expression is a common factor of the terms in the expression. It is the common factor with the greatest coefficient and the greatest exponent. In this case, the GCF is 2.

2x^2-2x-24=0
2* x^2- 2* x- 2* 12
2(x^2-x-12)

The result of factoring out a GCF from the given expression is a trinomial with a leading coefficient of one. 2( x^2-x-12)

Let's temporarily only focus on this trinomial, and we will bring back the GCF after factoring.

Factor the Expression

To factor a trinomial with a leading coefficient of one, think of the process as multiplying two binomials in reverse. Let's start by taking a look at the constant term. x^2-x+(-12) In this case, we have -12. This is a negative number, so for the product of the constant terms in the factors to be negative, these constants must have the opposite sign (one positive and one negative.)

Factor Constants Product of Constants
-1 and 12 -12
1 and -12 -12
-2 and 6 -12
2 and -6 -12
-3 and 4 -12
3 and -4 -12

Next, let's consider the coefficient of the linear term. x^2+(-1)x+(-12) For this term, we need the sum of the factors that produced the constant term to equal the coefficient of the linear term, -1.

Factors Sum of Factors
-1 and 12 11
1 and -12 -11
-2 and 6 4
2 and -6 -4
-3 and 4 1
3 and -4 -1

We found the factors whose product is -12 and whose sum is -1. x^2+(-1)x+(-12) ⇔ (x+3)(x-4)

Wait! Before we finish, remember that we factored out a GCF from the original expression. To fully complete the factored expression, let's reintroduce that GCF now. 2(x+3)(x-4) Now we can use the Zero Product Property to find the solutions of the equation by setting our factors equal to 0.

2(x+3)(x-4)=0
lc2=0 * & (I) x+3=0 & (II) x-4=0 & (III)
lc2=0 * & (I) x=- 3 & (II) x-4=0 & (III)
lc2=0 * & (I) x=- 3 & (II) x=4 & (III)

The solutions to the equation are x=-3 and x=4.