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$Find the value of c so that (x+1) is a factor of the polynomial p(x). {:[p(x)=5x^(4)+7x^(3)-2x^(2)-3x+c],[c=]:}$

Find the value of $c$ so that $(x+1)$ is a factor of the polynomial $p(x)$ . $\newline$ $\begin{array}{l} p(x)=5 x^{4}+7 x^{3}-2 x^{2}-3 x+c \\ c= \end{array}$

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Find derivatives of sine and cosine functions

Full solution

Q. Find the value of

c

so that

(x+1)

is a factor of the polynomial

p(x)

.

\newline

\begin{array}{l} p(x)=5 x^{4}+7 x^{3}-2 x^{2}-3 x+c \\ c= \end{array}

Apply Remainder Theorem: Since $(x+1)$ is a factor of $p(x)$ , we can use the Remainder Theorem which states that if $(x+1)$ is a factor, then $p(-1) = 0$ .
Substitute $x = -1$ : Plug $x = -1$ into the polynomial $p(x)$ to find the value of $c$ . $\newline$ $p(-1) = 5(-1)^{4} + 7(-1)^{3} - 2(-1)^{2} - 3(-1) + c$
Simplify the expression: Simplify the expression. $\newline$ $p(-1) = 5(1) + 7(-1) - 2(1) + 3 + c$
Combine like terms: Combine like terms. $p(-1) = 5 - 7 - 2 + 3 + c$
Further simplify: Further simplify to find the value of $c$ . $0 = 5 - 7 - 2 + 3 + c$
Add up the numbers: Add up the numbers. $0 = -1 + c$
Solve for c: Solve for $c$ . $c = 1$

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