How do you factor completely y=2x^3 - 11x^2 + 14x - 3?

Apr 24, 2016

$2 {x}^{3} - 11 {x}^{2} + 14 x - 3 = \left(2 x - 3\right) \left(x - 2 - \sqrt{3}\right) \left(x - 2 + \sqrt{3}\right)$

Explanation:

$f \left(x\right) = 2 {x}^{3} - 11 {x}^{2} + 14 x - 3$

By the rational root theorem, any zeros of $f \left(x\right)$ must be expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $- 3$ and $q$ a divisor of the coefficient $2$ of the leading term.

That means that the only possible rational zeros are:

$\pm \frac{1}{2}$, $\pm 1$, $\pm \frac{3}{2}$, $\pm 3$

Trying each in turn, we eventually find:

$f \left(\frac{3}{2}\right) = \frac{27}{4} - \frac{99}{4} + 21 - 3 = \frac{27 - 99 + 84 - 12}{4} = 0$

So $x = \frac{3}{2}$ is a zero and $\left(2 x - 3\right)$ a factor:

$2 {x}^{3} - 11 {x}^{2} + 14 x - 3 = \left(2 x - 3\right) \left({x}^{2} - 4 x + 1\right)$

The remaining quadratic factor can be factored by completing the square and using the difference of squares identity:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

with $a = \left(x - 2\right)$ and $b = \sqrt{3}$ as follows:

${x}^{2} - 4 x + 1 = {x}^{2} - 4 x + 4 - 3$

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

$= \left(\left(x - 2\right) - \sqrt{3}\right) \left(\left(x - 2\right) + \sqrt{3}\right)$

$= \left(x - 2 - \sqrt{3}\right) \left(x - 2 + \sqrt{3}\right)$

Putting it all together:

$2 {x}^{3} - 11 {x}^{2} + 14 x - 3 = \left(2 x - 3\right) \left(x - 2 - \sqrt{3}\right) \left(x - 2 + \sqrt{3}\right)$