Answer
$(n-8)$
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)
.$