Exercise 9 Page 217

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+13x+40 In this case, we have 40. This is a positive number, so for the product of the constant terms in the factors to be positive, these constants must have the same sign (both positive or both negative.) Let's also consider the coefficient of the linear term.

x^2+13x+40 For this term, we need the sum of the factors that produced the constant term to equal the coefficient of the linear term, 13. Since a sum of two negative numbers is never positive, we should only consider positive constants.

Constants	Product of Constants	Sum of Constants
1 and 40	40	41
2 and 20	40	22
4 and 10	40	14
5 and 8	40	13

Now we can write the products of the constants and variables. 5x and 8x We found the factor constants whose product is 40 and also cause the sum of the products of x and the factor constants to equal 13x. x^2+13x+40 ⇔ (x+5)(x+8) or x^2+13x+40 ⇔ (x+8)(x+5)

Checking Our Answer

Check your answer ✓

We can check our answer by applying the Distributive Property and comparing the result with the given expression.

(x+8)(x+5)

Distr

Distribute x+5

x(x+5)+8(x+5)

Distr

Distribute x

x^2+5x+8(x+5)

Distr

Distribute 8

x^2+5x+8x+40

AddTerms

Add terms

x^2+13x+40

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!