# How do you find all zeros with multiplicities of f(x)=x^3-2x^2-5x+6?

Aug 22, 2017

The zeros of $f \left(x\right)$ are $1 , 3 , - 2$, each with multiplicity $1$.

#### Explanation:

Given:

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

Note that the sum of the coefficients is $0$. That is:

$1 - 2 - 5 + 6 = 0$

Hence we find $f \left(1\right) = 0$ and $\left(x - 1\right)$ is a factor:

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

To factor the remaining quadratic find a pair of factors of $6$ which differ by $1$. The pair $3 , 2$ works, so we find:

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

So the zeros of $f \left(x\right)$ are $1 , 3 , - 2$, each with multiplicity $1$.