# How do you find all zeros with multiplicities of f(x)=x^3+4x^2-11x+6?

Mar 4, 2017

The zeros of $f \left(x\right)$ are $1$ with multiplicity $2$ and $- 6$ with multiplicity $1$.

#### Explanation:

Given:

$f \left(x\right) = {x}^{3} + 4 {x}^{2} - 11 x + 6$

First note that the sum of the coefficients is $0$. That is:

$1 + 4 - 11 + 6 = 0$

Hence $x = 1$ is a zero and $\left(x - 1\right)$ a factor:

${x}^{3} + 4 {x}^{2} - 11 x + 6 = \left(x - 1\right) \left({x}^{2} + 5 x - 6\right)$

The sum of the coefficients of the remaining quadratic is also $0$, so $x = 1$ is a zero again and $\left(x - 1\right)$ a factor:

${x}^{2} + 5 x - 6 = \left(x - 1\right) \left(x + 6\right)$

So the remaining zero is $x = - 6$.

So the zeros of $f \left(x\right)$ are $1$ with multiplicity $2$ and $- 6$ with multiplicity $1$.