# Chapter 8 - Polynomials and Factoring - 8-5 Factoring x2+bx+c - Practice and Problem-Solving Exercises - Page 515: 22

$n^2-5n-50=(n+5)(n-10)$

#### Work Step by Step

We are trying to fill in the blank in the equation $n^2-5n-50=(n+5)(n-\square)$. 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 $n^2+bn+c$, 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 $(n+\_)(n+\_)$. In the case of $n^2-5n-50$, we are looking for two numbers whose product is $-50$ and whose sum is $-5$. The numbers $5$ and $-10$ meet these criteria, because $$5\times(-10)=-50\;\text{and}\;5+(-10)=-5$$When we insert these numbers into the blanks, we arrive at the factors $(n+5)(n-10)$.

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