How do you factor x^4-2x^3-13x^2+38x-24?

Jul 18, 2015

Use sum of coefficients short cut, then rational roots theorem to find:

${x}^{4} - 2 {x}^{3} - 13 {x}^{2} + 38 x - 24$

$= \left(x - 1\right) \left(x - 2\right) \left(x - 3\right) \left(x + 4\right)$

Explanation:

Let $f \left(x\right) = {x}^{4} - 2 {x}^{3} - 13 {x}^{2} + 38 x - 24$

First notice that the sum of the coefficients is $0$, so $f \left(1\right) = 0$ and $\left(x - 1\right)$ is a factor of $f \left(x\right)$.

${x}^{4} - 2 {x}^{3} - 13 {x}^{2} + 38 x - 24$

$= \left({x}^{4} - {x}^{3}\right) - \left({x}^{3} - {x}^{2}\right) - \left(14 {x}^{2} - 14 x\right) + \left(24 x - 24\right)$

$= {x}^{3} \left(x - 1\right) - {x}^{2} \left(x - 1\right) - 14 x \left(x - 1\right) + 24 \left(x - 1\right)$

$= \left({x}^{3} - {x}^{2} - 14 x + 24\right) \left(x - 1\right)$

Let $g \left(x\right) = {x}^{3} - {x}^{2} - 14 x + 24$

By the rational roots theorem, any rational root of $g \left(x\right) = 0$ must be a factor of $24$ and hence one of:

$\pm 1$, $\pm 2$, $\pm 3$, $\pm 4$, $\pm 6$, $\pm 12$, $\pm 24$

We soon find $g \left(2\right) = 8 - 4 - 28 + 24 = 0$,

so $\left(x - 2\right)$ is a factor of $g \left(x\right)$

${x}^{3} - {x}^{2} - 14 x + 24$

$= \left({x}^{3} - 2 {x}^{2}\right) + \left({x}^{2} - 2 x\right) - \left(12 x - 24\right)$

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

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

Then ${x}^{2} + x - 12 = \left(x + 4\right) \left(x - 3\right)$ by finding two numbers whose product is $12$ and whose difference is $1$

Putting it all together:

${x}^{4} - 2 {x}^{3} - 13 {x}^{2} + 38 x - 24$

$= \left(x - 1\right) \left(x - 2\right) \left(x - 3\right) \left(x + 4\right)$