# How do you factor completely x^3 - 8x^2 + 5x + 50 = 0?

Nov 13, 2015

$\left(x + 2\right) {\left(x - 5\right)}^{2}$.

#### Explanation:

We know that if a number solves such an equation, it must divide its last coefficient, i.e. $50$. So, we can try with some of its divisors: if $f \left(x\right) = {x}^{3} - 8 {x}^{2} + 5 x + 50$, then:

• $f \left(1\right) = 48 \ne 0$;
• $f \left(- 1\right) = 36 \ne 0$
• $f \left(2\right) = 36 \ne 0$
• $f \left(- 2\right) = 0$.

So, $- 2$ is a solution, which means that $f \left(x\right)$ can be divided by $\left(x + 2\right)$. Do the long division, and you have that

${x}^{3} - 8 {x}^{2} + 5 x + 50 = \left(x + 2\right) \left({x}^{2} - 10 x + 25\right)$

Now we must factor the second parenthesis, but you can see that it is the square of $\left(x - 5\right)$, and so $f \left(x\right)$ is completely factored.