# How do you factor completely 2x^2+20=x^2+9x?

Dec 28, 2017

GIven: $2 {x}^{2} + 20 = {x}^{2} + 9 x$

Combine terms so that the quadratic is equal to 0:

${x}^{2} - 9 x + 20 = 0$

This will factor into $\left(x - {r}_{1}\right) \left(x - {r}_{2}\right) = 0$, if we can find numbers such that ${r}_{1} {r}_{2} = 20$ and $- \left({r}_{1} + {r}_{2}\right) = - 9$.

4 and 5 will do it $\left(4\right) \left(5\right) = 20$ and $- \left(4 + 5\right) = - 9$:

$\left(x - 4\right) \left(x - 5\right) = 0$

Dec 28, 2017

Set it equal to zero, then find factors of $c$ that add to $b$.

#### Explanation:

Set equal to zero:

${x}^{2} - 9 x + 20 = 0$

Looking at the discriminate: (${b}^{2} - 4 a c$)

$81 - 4 \cdot 1 \cdot 20 = 81 - 80 = 1$

Since $1$ is a perfect square, we know it factors.

Factors of $20$ that add to $- 9$ are $- 5$ and $- 4$.

$\left(x - 5\right) \left(x - 4\right) = {x}^{2} - 9 x + 20$