# How do you solve x^2-6x-16=0 by factoring?

Feb 4, 2015

Factoring a polynomial $p \left(x\right)$ means to find its roots ${x}_{i}$ and divide the polynomial by the term $\left(x - {x}_{i}\right)$, writing $p \left(x\right)$ as $\left(x - {x}_{i}\right) q \left(x\right)$, and then iterating the process searching the roots of $q \left(x\right)$ (which of course are roots of $p \left(x\right)$, too. The fundamental theorem of algebra states that you can always write a polynomial of degree $k$ as the product of $k$ linear factors $\left(x - {x}_{i}\right)$, i.e., every polynomial of degree $k$ has $k$ complex roots.

These roots can be complex but not real, and in that case, you cannot simplify the polynomial in $\setminus m a t h \boldsymbol{R}$.

In your case, the solutions are easy to find, without doing any calculation: you can remember the rule which states that, if you have a quadratic of the form ${x}^{2} - s x + p$, the product of the solutions equals $p$ and the sum of the solutions equals $s$.

So, we have to find two numbers which sum up to $- 6$, and that multiplied one for the other give $- 16$. You can easily convince yourself that these two numbers are $- 2$ and $8$

If $- 2$ and $8$ are roots of our polynomial, for what we said above the following must hold:
x^2−6x−16=(x+2)(x-8)