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

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Answer 1 · 2017-11-16T12:38:00

The remainder is $=6$

Apply the remainder theorem

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

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

$f(c)=0+r$

Here,

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

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

Therefore,

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

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

$=6$

The remainder is $=6$

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