How do you find the remainder when $1.f(x)=x^3+2x^2-6x+8; x+4$ ?

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Answer 1 · 2018-03-22T01:38:45

The remainder is $0$

The remainder theorem tell us that if we divide a polynomial $f(x)$ by a linear factor $(x-a)$ then the remainder is $f(a)$

Thus the remainder we seek for:

$f(x) = x^3+2x^2-6x+8$

divided by $(x+4)$ is given by:

$f(-4) = (-4)^3+ 2(-4)^2-6(-4)+8$
$\ \ \ \ \ \ \ \ \ \ \ = -64+ 32+24+8$
$\ \ \ \ \ \ \ \ \ \ \ = 0$

And we can conclude that $x+4$ is a factor

How do you find the remainder when 1.f(x)=x^3+2x^2-6x+8; x+41.f(x)=x^3+2x^2-6x+8; x+4?