Pearson Algebra 1 Common Core, 2011
PA
Pearson Algebra 1 Common Core, 2011 View details
6. Trigonometric Ratios
Continue to next subchapter

Exercise 60 Page 651

Try to rewrite a trinomial as a product of two binomials.

(x+3)(x-5)

Practice makes perfect
To factor a trinomial with a leading coefficient of 1, think of the process as multiplying two binomials in reverse. Let's start by taking a look at the constant term. x^2-2x-15 In this case, we have - 15. This is a negative number, so for the product of the constant terms in the factors to be negative, these constants must have different signs - positive or negative.
Factor Constants Product of Constants
1 and - 15 - 15
-1 and 15 - 15
3 and - 5 - 15
-3 and 5 - 15

Next, let's consider the coefficient of the linear term. x^2- 2x-15 For this term, we need the sum of the factors that produced the constant term to equal the coefficient of the linear term, - 2.

Factors Sum of Factors
1 and - 15 - 14
-1 and 15 14
3 and - 5 - 2
-3 and 5 2
We found the factors whose product is - 2 and whose sum is - 15. x^2- 2x-15 ⇔ (x+3)(x-5)

Checking Our Answer

Check your answer âś“
We can check our answer by applying the Distributive Property and comparing the result with the given expression.
(x+3) (x - 5)
x (x-5) +3 (x - 5 )
x^2 - 5x +3 (x - 5)
x^2 - 5x +3x -15
x^2-2x-15
After applying the Distributive Property and simplifying, the result is the same as the given expression. Therefore, we can be sure our solution is correct!