QuestionFind the value of $m$ such that $x^{3}-2 m x^{2}+16$ is divisible by $x+2$ .

Studdy Solution

STEP 1

Assumptions1. We are given a cubic polynomial $x^{3}- m x^{}+16$ . . We are told that this polynomial is divisible by the binomial $x+$ .
3. We are asked to find the value of $m$ that makes this divisibility true.

STEP 2

A polynomial $f(x)$ is divisible by a binomial $x-a$ if and only if $f(a) =0$ . Since we know that $x^{}-2 m x^{2}+16$ is divisible by $x+2$ , we can set $x = -2$ in the polynomial and equate it to zero.
$(-2)^{}-2 m (-2)^{2}+16 =0$

STEP 3

implify the equation.
$-8 -8m +16 =0$

STEP 4

Rearrange the equation to solve for $m$ .
$-8m =8 -16$

STEP 5

olve for $m$ .
$m = \frac{8 -16}{-8}$

STEP 6

Calculate the value of $m$ .
$m = \frac{8 -16}{-8} =1$ For $m =1$ , the polynomial $x^{3}-2 m x^{2}+16$ is divisible by $x+2$ .

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