# How do you find the rest of the zeros given one of the zero c=1 and the function f(x)=x^3-6x^2+11x-6?

Aug 22, 2016

Other zeros are $2$ and $3$.

#### Explanation:

$f \left(x\right) = {x}^{3} - 6 {x}^{2} + 11 x - 6$ has one of the zeros as $1$, $\left(x - 1\right)$ is factor of $f \left(x\right)$.

Dividing $f \left(x\right) = {x}^{3} - 6 {x}^{2} + 11 x - 6$ by $\left(x - 1\right)$, we get ${x}^{2} \left(x - 1\right) - 5 x \left(x - 1\right) + 6 \left(x - 1\right)$ or

$\left(x - 1\right) \left({x}^{2} - 5 x + 6\right)$, which can be further factorized as

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

= $\left(x - 1\right) \left(x \left(x - 3\right) - 2 \left(x - 3\right)\right)$

= $\left(x - 1\right) \left(x - 2\right) \left(x - 3\right)$

Hence, other zeros are $2$ and $3$.
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