McGraw Hill Integrated II, 2012
MH
McGraw Hill Integrated II, 2012 View details
8. Differences of Squares
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Exercise 65 Page 63

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

G

Practice makes perfect

We want to find the sum of the roots of the given equation. In order to do that, we will factor the equation and then use the Zero Product Property to solve it.

Factoring

Let's start by rewriting the equation in such way that all the terms are on the left side of the equality sign. x^2+3x=54 ⇕ x^2+3x-54=0 Now, to factor a trinomial with a leading coefficient of 1, think of the process as multiplying two binomials in reverse. Let's take a look at the constant term.

x^2+3x -54=0 In this case, we have -54. 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 - 54 - 54
-1 and 54 - 54
2 and - 27 - 54
-2 and 27 - 54
3 and - 18 - 54
-3 and 18 - 54
6 and - 9 - 54
-6 and 9 - 54

Next, let's consider the coefficient of the linear term. x^2+ 3x -54=0 For this term, we need the sum of the factors that produced the constant term to equal the coefficient of the linear term, 3.

Factors Sum of Factors
1 and - 54 - 53
-1 and 54 53
2 and - 27 - 25
-2 and 27 25
3 and - 18 - 15
-3 and 18 15
6 and - 9 - 3
-6 and 9 3

We found the factors whose product is - 54 and whose sum is 3. x^2+ 3x -54=0 ⇕ (x-6)(x+9)=0

Solving the Equation

Now, as we have already factored the equation, we can apply the Zero Product Property to solve it.
(x-6)(x+9)=0
lcx-6=0 & (I) x+9=0 & (II)
lx=6 x+9=0
lx=6 x=- 9
We found that the roots of the equation are 6 and - 9. Now, we can calculate the sum of the roots. 6+( - 9)=- 3 Therefore, our answer is G.