# How do you factor x^5 -4x^4 + 8x^3 - 14x^2 +15x -6 ?

Nov 12, 2015

Find factor $\left(x - 1\right)$ by examining the sum of the coefficients, divide and repeat, then factor by grouping to find:

${x}^{5} - 4 {x}^{4} + 8 {x}^{3} - 14 {x}^{2} - 14 {x}^{2} + 15 x - 6$

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

#### Explanation:

$f \left(x\right) = {x}^{5} - 4 {x}^{4} + 8 {x}^{3} - 14 {x}^{2} - 14 {x}^{2} + 15 x - 6$

First notice that $f \left(1\right) = 0$, since the sum of the coefficients is $0$.

So $\left(x - 1\right)$ is a factor of $f \left(x\right)$.

$f \left(x\right) = \left(x - 1\right) \left({x}^{4} - 3 {x}^{3} + 5 {x}^{2} - 9 x + 6\right)$

Again, the remaining quartic factor is divisible by $\left(x - 1\right)$ since its coefficients also add up to $0$.

${x}^{4} - 3 {x}^{3} + 5 {x}^{2} - 9 x + 6 = \left(x - 1\right) \left({x}^{3} - 2 {x}^{2} + 3 x - 6\right)$

The remaining cubic factor is factorizable by grouping:

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

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

The remaining quadratic factor has no linear factors with Real coefficients since ${x}^{2} + 3 \ge 3 > 0$ for all $x \in \mathbb{R}$

Putting this together we get:

${x}^{5} - 4 {x}^{4} + 8 {x}^{3} - 14 {x}^{2} - 14 {x}^{2} + 15 x - 6 = \left(x - 1\right) \left(x - 1\right) \left(x - 2\right) \left({x}^{2} + 3\right)$

If you allow Complex coefficients then you can factor further using:

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

So:

${x}^{5} - 4 {x}^{4} + 8 {x}^{3} - 14 {x}^{2} - 14 {x}^{2} + 15 x - 6$

$= \left(x - 1\right) \left(x - 1\right) \left(x - 2\right) \left(x - \sqrt{3} i\right) \left(x + \sqrt{3} i\right)$