How do you factor given that f(6)=0 and f(x)=5x^3-27x^2-17x-6?

Nov 11, 2016

We know that $\left(x - 6\right)$ is a factor of $f \left(x\right)$ because of the given root.

Using synthetic division, I divided f(x) by (x-6) to get the other factor.

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

Nov 11, 2016

The only real factor of $f \left(x\right)$ are $\left(5 {x}^{2} + 3 x + 1\right) \left(x - 6\right)$. However, complex factors are $5 \left(x - 6\right) \left(x + \frac{3 + i \sqrt{11}}{10}\right) \left(x + \frac{3 - i \sqrt{11}}{10}\right)$

Explanation:

As $f \left(x\right) = 5 {x}^{3} - 27 x - 17 x - 6$ and $f \left(6\right) = 0$, $\left(x - 6\right)$ is a factor of $f \left(x\right)$.

Hence $f \left(x\right) = 5 {x}^{3} - 27 {x}^{2} - 17 x - 6$

$\textcolor{w h i t e}{X X X X X x} = 5 {x}^{2} \left(x - 6\right) + 3 x \left(x - 6\right) + 1 \left(x - 6\right)$

$\textcolor{w h i t e}{X X X X X x} = \left(5 {x}^{2} + 3 x + 1\right) \left(x - 6\right)$

As determinant (given by ${b}^{2} - 4 a c$) in $5 {x}^{2} + 3 x + 1$ is ${3}^{2} - 4 \times 5 \times 1 = - 11 < 0$, $5 {x}^{2} + 3 x + 1$ cannot be factorized further with real roots.

Hence only factor are $\left(5 {x}^{2} + 3 x + 1\right) \left(x - 6\right)$.

However for $5 {x}^{2} + 3 x + 1 = 0$, $x = \frac{- 3 \pm \sqrt{- 11}}{10}$

Hence, we can have complex factors

$f \left(x\right) = 5 {x}^{3} - 27 x - 17 x - 6$

= $5 \left(x - 6\right) \left(x + \frac{3 + i \sqrt{11}}{10}\right) \left(x + \frac{3 - i \sqrt{11}}{10}\right)$