# How do you find the remainder when f(x)=x^4+8x^3+12x^2; x+1?

5

#### Explanation:

Simply perform long division. That is;

$\frac{{x}^{4} + 8 {x}^{3} + 12 {x}^{2} + 0 x + 0}{x + 1}$

Dec 31, 2015

The remainder is $f \left(- 1\right) = 5$

#### Explanation:

The remainder of dividing $f \left(x\right)$ by a linear factor of the form $\left(x - a\right)$ is $f \left(a\right)$.

In our case $x + 1 = x - \left(- 1\right)$, so evaluate $f \left(- 1\right)$ to find the remainder:

$f \left(- 1\right) = {\left(- 1\right)}^{4} + 8 {\left(- 1\right)}^{3} + 12 {\left(- 1\right)}^{2} = 1 - 8 + 12 = 5$