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

Mar 22, 2018

The remainder is $0$

#### Explanation:

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

Thus the remainder we seek for:

$f \left(x\right) = {x}^{3} + 2 {x}^{2} - 6 x + 8$

divided by $\left(x + 4\right)$ is given by:

$f \left(- 4\right) = {\left(- 4\right)}^{3} + 2 {\left(- 4\right)}^{2} - 6 \left(- 4\right) + 8$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = - 64 + 32 + 24 + 8$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = 0$

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