## Algebra 1: Common Core (15th Edition)

Published by Prentice Hall

# Chapter 8 - Polynomials and Factoring - 8-5 Factoring x2+bx+c - Practice and Problem-Solving Exercises - Page 515: 11

#### Answer

$x^2-7x+10=(x-5)(x-2)$

#### Work Step by Step

We are trying to fill in the blank in the equation $x^2-7x+10=(x-5)(x-\square)$. In order to do so, we will factor the trinomial on the left side of the equation, to determine the second factor. To factor a trinomial in the form $x^2+bx+c$, we must find two numbers whose product is $c$ and whose sum is $b$. We then insert these two numbers into the blanks of the factors $(x+\_)(x+\_)$. In the case of $x^2-7x+10$, we are looking for two numbers whose product is $10$ and whose sum is $-7$. The numbers $-2$ and $-5$ meet these criteria, because $$-5\times(-2)=10\;\text{and}\;-5+(-2)=-7$$When we insert these numbers into the blanks, we arrive at the factors $(x-5)(x-2)$. Inserting factored form back into the original equation, we have the complete equation, which is $x^2-7x+10=(x-5)(x-2)$.

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