# How do you factor y= x^2 - 10x - 9 ?

Dec 3, 2015

Prime; or, $y = \left(x - 5 - \sqrt{34}\right) \left(x - 5 + \sqrt{34}\right)$

#### Explanation:

We need to look for numbers whose SUM is $- 10$ and whose PRODUCT is $- 9$.

The pairs we could use to have a product of $9$ are $\left(- 1 , 9\right) , \left(1 , - 9\right) ,$ and $\left(3 , - 3\right)$. Unfortunately, none of these pairs' sums is $- 10$.

This means the expression is PRIME, if we limit ourselves to integers.

However, if we throw irrationals into the mix, we can use the quadratic formula or complete the square to find this quadratic's irrational roots.

$x = \frac{10 \pm \sqrt{{10}^{2} - 4 \left(- 9\right) \left(1\right)}}{2 \left(1\right)} = \frac{10 \pm \sqrt{136}}{2} = \frac{10 \pm 2 \sqrt{34}}{2}$
$= 5 \pm \sqrt{34}$

Thus, the roots of this are $5 + \sqrt{34}$ and $5 - \sqrt{34}$.

Normally, if a root were, say, $7$, you could figure out how that would look in a factored quadratic by saying that $x = 7$ and seeing when that holds true and is equal to $0$, since that is where the roots of an equation occur.

You could rewrite $x = 7$ as $x - 7 = 0$, and $\left(x - 7\right)$ would be one of the factored terms in the expression.

We can do the same for the irrational monstrosity we have:

$x = 5 + \sqrt{34}$

$x - 5 - \sqrt{34} = 0$

$x = 5 - \sqrt{34}$

$x - 5 + \sqrt{34} = 0$

Thus, factored, we get $y = \left(x - 5 - \sqrt{34}\right) \left(x - 5 + \sqrt{34}\right)$