# Chapter 5 - Section 5.2 - Factoring Trinomials - 5.2 Exercises: 10

$(n-8)$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $n^2-14n+48 ,$ find two numbers, $m_1$ and $m_2,$ whose product is $c$ and whose sum is $b$ in the quadratic expression $x^2+bx+c.$ Then, express the factored form as $(x+m_1)(x+m_2).$ $\bf{\text{Solution Details:}}$ In the given expression, the value of $c$ is $48$ and the value of $b$ is $-14 .$ The possible pairs of integers whose product is $c$ are \begin{array}{l}\require{cancel} \{ 1,48 \}, \{ 2,24 \}, \{ 3,16 \}, \{ 4,12 \}, \{ 6,8 \}, \{ -1,-48 \}, \{ -2,-24 \}, \{ -3,-16 \}, \{ -4,-12 \}, \{ -6,-8 \} .\end{array} Among these pairs, the one that gives a sum of $b$ is $\{ -6,-8 \}.$ Hence, the factored form of the given expression is \begin{array}{l}\require{cancel} (n-6)(n-8) .\end{array} The missing factor in the given expression is $(n-8) .$

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