# How do you factor x^3-12x^2+41x-42?

Apr 3, 2018

${x}^{3} - 12 {x}^{2} + 41 x - 42 = \left(x - 2\right) \left(x - 3\right) \left(x - 7\right)$

#### Explanation:

Given:

$f \left(x\right) = {x}^{3} - 12 {x}^{2} + 41 x - 42$

By the rational roots theorem, any rational zeros of $f \left(x\right)$ are expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $- 42$ and $q$ a divisor of the coefficient $1$ of the leading term.

So the only possible rational zeros are:

$\pm 1 , \pm 2 , \pm 3 , \pm 6 , \pm 7 , \pm 14 , \pm 21 , \pm 42$

In addition, note that the pattern of signs of the coefficients of $f \left(x\right)$ is $+ - + -$. With $3$ changes of signs, Descartes' Rule of Signs tells us that $f \left(x\right)$ has $3$ or $1$ positive real zeros. Also, the signs of the coefficients of $f \left(- x\right)$ are in the pattern $- - - -$. With no changes of sign, Descartes' Rule of Signs tells us that $f \left(x\right)$ has no negative real zeros.

So the only possible rational zeros are:

$1 , 2 , 3 , 6 , 7 , 14 , 21 , 42$

We find:

$f \left(2\right) = {\left(\textcolor{b l u e}{2}\right)}^{3} - 12 {\left(\textcolor{b l u e}{2}\right)}^{2} + 41 \left(\textcolor{b l u e}{2}\right) - 42$

$\textcolor{w h i t e}{f \left(2\right)} = 8 - 48 + 82 - 42 = 0$

So $x = 2$ is a zero and $\left(x - 2\right)$ a factor:

${x}^{3} - 12 {x}^{2} + 41 x - 42 = \left(x - 2\right) \left({x}^{2} - 10 x + 21\right)$

To factor the remaining quadratic note that $3 + 7 = 10$ and $3 \cdot 7 = 21$. So:

${x}^{2} - 10 x + 21 = \left(x - 3\right) \left(x - 7\right)$

Putting it all together:

${x}^{3} - 12 {x}^{2} + 41 x - 42 = \left(x - 2\right) \left(x - 3\right) \left(x - 7\right)$