How do you find the remainder for $(x^{4} - x^{3} - 3 x^{2} + 4 x + 2) \div (x + 2)$ $(x^4-x^3-3x^2+4x+2)div(x+2)$ ?

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Answer 1 · 2017-11-16T12:38:00

The remainder is $= 6$ $=6$

Apply the remainder theorem

When we divide a polynomial $f (x)$ $f(x)$ by $(x - c)$ $(x-c)$

$f (x) = (x - c) q (x) + r$ $f(x)=(x-c)q(x)+r$

$f (c) = 0 + r$ $f(c)=0+r$

Here,

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

and $(x - c)$ $(x-c)$ is $(x - (- 2))$ $(x-(-2))$

Therefore,

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

$= 16 + 8 - 12 - 8 + 2$ $=16+8-12-8+2$

$= 6$ $=6$

The remainder is $= 6$ $=6$

How do you find the remainder for (x^4-x^3-3x^2+4x+2)div(x+2)(x4−x3−3x2+4x+2)÷(x+2)(x^4-x^3-3x^2+4x+2)div(x+2)?