# How do you find all the zeros of f(x)= x^3 - 8x^2 +19x - 12 with its multiplicities?

Aug 11, 2016

$f \left(x\right)$ has zeros $1$, $3$ and $4$ all with multiplicity $1$.

#### Explanation:

$f \left(x\right) = {x}^{3} - 8 {x}^{2} + 19 x - 12$

First note that the sum of the coefficients is zero. That is:

$1 - 8 + 19 - 12 = 0$

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

${x}^{3} - 8 {x}^{2} + 19 x - 12$

$= \left(x - 1\right) \left({x}^{2} - 7 x + 12\right)$

To factor the remaining quadratic, note that $3 + 4 = 7$ and $3 \times 4 = 12$.

So we find:

${x}^{2} - 7 x + 12 = \left(x - 3\right) \left(x - 4\right)$

with associated zeros $x = 3$ and $x = 4$.