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Find the value of
c so that the polynomial
p(x) is divisible by
(x-4).

p(x)=cx^(3)-15 x-68

c=

Find the value of $c$ so that the polynomial $p(x)$ is divisible by $(x-4)$ . $\newline$ $p(x)=c x^{3}-15 x-68$ $\newline$ $c=$

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Composition of linear and quadratic functions: find a value

Full solution

Q. Find the value of

c

so that the polynomial

p(x)

is divisible by

(x-4)

.

\newline

p(x)=c x^{3}-15 x-68

\newline

c=

Remainder Theorem: To determine the value of $c$ that makes the polynomial $p(x)$ divisible by $(x-4)$ , we need to use the Remainder Theorem. According to the Remainder Theorem, if a polynomial $p(x)$ is divisible by $(x - a)$ , then $p(a) = 0$ . In this case, $a$ is $4$ , so we need to find $p(4)$ and set it equal to $0$ .
Substituting $x = 4$ : Let's substitute $x = 4$ into the polynomial $p(x) = cx^3 - 15x - 68$ and set it equal to $0$ . $\newline$ $p(4) = c(4)^3 - 15(4) - 68 = 0$
Calculating $p(4)$ : Now, we calculate the value of $p(4)$ with the given $x$ value. $\newline$ $p(4) = c(4)^3 - 15(4) - 68$ $\newline$ $p(4) = c(64) - 60 - 68$ $\newline$ $p(4) = 64c - 128$
Setting $p(4) = 0$ : Since $p(4)$ must be equal to $0$ for $p(x)$ to be divisible by $(x-4)$ , we set the equation $64c - 128 = 0$ and solve for $c$ . $64c - 128 = 0$
Isolating the term with $c$ : Add $128$ to both sides of the equation to isolate the term with $c$ . $\newline$ $64c = 128$
Solving for c: Divide both sides of the equation by $64$ to solve for $c$ . $\newline$ $c = \frac{128}{64}$
Finding the value of c: Perform the division to find the value of $c$ . $c = 2$

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