How do you factor given that f(10)=0 and f(x)=x^3-12x^2+12x+80?

Dec 19, 2016

$f \left(x\right) = {x}^{3} - 12 {x}^{2} + 12 x + 80 = \left(x - 10\right) \left(x + 2\right) \left(x - 4\right)$

Explanation:

As $f \left(x\right) = {x}^{3} - 12 {x}^{2} + 12 x + 80$ and $f \left(10\right) = 0$

$\left(x - 10\right)$ is a factor of $f \left(x\right) = {x}^{3} - 12 {x}^{2} + 12 x + 80$

Now dividing $f \left(x\right) = {x}^{3} - 12 {x}^{2} + 12 x + 80$ by (x-10)#, we can get a quadratic polynomial which can be factorized further by splitting the middle term.

$f \left(x\right) = {x}^{3} - 12 {x}^{2} + 12 x + 80$

= ${x}^{2} \left(x - 10\right) - 2 x \left(x - 10\right) - 8 \left(x - 10\right)$

= $\left(x - 10\right) \left({x}^{2} - 2 x - 8\right)$

= $\left(x - 10\right) \left({x}^{2} - 4 x + 2 x - 8\right)$

= $\left(x - 10\right) \left(x \left(x - 4\right) + 2 \left(x - 4\right)\right)$

= $\left(x - 10\right) \left(x + 2\right) \left(x - 4\right)$