# How do you factor x^3-x^2-4x+5?

May 20, 2016

Factoring by grouping does not work with this polynomial, but read on...

#### Explanation:

Factoring by grouping does not immediately help with this cubic polynomial. If it were ${x}^{3} - {x}^{2} - 4 x + 4$ then it would be much easier:

${x}^{3} - {x}^{2} - 4 x + 4$

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

$= {x}^{2} \left(x - 1\right) - 4 \left(x - 1\right)$

$= \left({x}^{2} - 4\right) \left(x - 1\right)$

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

Was there a typo in the question?

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Can we factor $f \left(x\right) = {x}^{3} - {x}^{2} - 4 x + 5$ ?

The only possible rational zeros are $\pm 1$, $\pm 5$, neither of which work.

This cubic turns out to have one Real zero and two Complex zeros.

It is possible, if a little lengthy, to find them using Cardano's method, which will give you:

Real zero:

${x}_{1} = \frac{1}{3} \left(1 - \sqrt{\frac{97 - 3 \sqrt{69}}{2}} - \sqrt{\frac{97 + 3 \sqrt{69}}{2}}\right)$

Complex zeros:

${x}_{2} = \frac{1}{3} \left(1 - \omega \sqrt{\frac{97 - 3 \sqrt{69}}{2}} - {\omega}^{2} \sqrt{\frac{97 + 3 \sqrt{69}}{2}}\right)$

${x}_{3} = \frac{1}{3} \left(1 - {\omega}^{2} \sqrt{\frac{97 - 3 \sqrt{69}}{2}} - \omega \sqrt{\frac{97 + 3 \sqrt{69}}{2}}\right)$

where $\omega = - \frac{1}{2} + \frac{\sqrt{3}}{2} i$ is the primitive Complex cube root of $1$.

Then:

${x}^{3} - {x}^{2} - 4 x + 5 = \left(x - {x}_{1}\right) \left(x - {x}_{2}\right) \left(x - {x}_{3}\right)$