What is the remainder when $(- 2 x^{4} - 6 x^{2} + 3 x + 1) \div (x + 1)$ $(-2x^4 - 6x^2 + 3x + 1) div ( x + 1)$ ?

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What is the remainder when $(- 2 x^{4} - 6 x^{2} + 3 x + 1) \div (x + 1)$ $(-2x^4 - 6x^2 + 3x + 1) div ( x + 1)$ ?

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Answer 1 · 2015-12-31T12:13:32

$- 10$ $-10$

Explanation:

From the remainder theorem theory, we may simply find the required remainder by evaluating $f (- 1)$ $f(-1)$ in $(f (x) = - 2 x^{4} - 6 x^{2} + 3 x + 1$ $(f(x)=-2x^4-6x^2+3x+1$ .

Doing so yields

$f (- 1) = - 2 {(- 1)}^{4} - 6 {(- 1)}^{2} + 3 (- 1) + 1$ $f(-1)=-2(-1)^4-6(-1)^2+3(-1)+1$

$= - 2 - 6 - 3 + 1$ $=-2-6-3+1$

$= - 10$ $=-10$ .

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What is the remainder when $(- 2 x^{4} - 6 x^{2} + 3 x + 1) \div (x + 1)$ $(-2x^4 - 6x^2 + 3x + 1) div ( x + 1)$ ?

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Explanation:

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