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Polynomial function
g is defined as
g(x)=x^(3)-ax^(2)-17 x+12, where
a is a constant. If
x+4 is a factor of the polynomial, then what is the value of
a ?

Polynomial function $g$ is defined as $g(x)=x^{3}-a x^{2}-17 x+12$ , where $a$ is a constant. If $x+4$ is a factor of the polynomial, then what is the value of $a$ ?

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Find derivatives using the chain rule II

Full solution

Q. Polynomial function

g

is defined as

g(x)=x^{3}-a x^{2}-17 x+12

, where

a

is a constant. If

x+4

is a factor of the polynomial, then what is the value of

a

?

Apply Factor Theorem: Since $x+4$ is a factor of the polynomial $g(x)$ , we can use the Factor Theorem which states that if $x+4$ is a factor, then $g(-4)=0$ .
Substitute $x = -4$ : Substitute $x = -4$ into the polynomial $g(x)$ to find the value of $a$ . $\newline$ $g(-4) = (-4)^3 - a(-4)^2 - 17(-4) + 12$
Calculate $g(-4)$ : Calculate the value of $g(-4)$ .
$g(-4) = (-64) - a(16) + 68 + 12$
$g(-4) = -64 - 16a + 68 + 12$
$g(-4) = 4 - 16a$
Set $g(-4)$ equal to $0$ : Set $g(-4)$ equal to $0$ and solve for $a$ . $0 = 4 - 16a$
Add $16a$ to both sides: Add $16a$ to both sides of the equation to isolate the term with $a$ . $16a = 4$
Divide both sides: Divide both sides by $16$ to solve for $a$ . $\newline$ $a = \frac{4}{16}$ $\newline$ $a = \frac{1}{4}$

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