# How do you write x^2-4x+16 in factored form?

Sep 13, 2015

${x}^{2} - 4 x + 16 = \left(x - 2 - 2 i \sqrt{3}\right) \left(x - 2 + 2 i \sqrt{3}\right)$

#### Explanation:

Let $f \left(x\right) = {x}^{2} - 4 x + 16$. This is of the form $a {x}^{2} + b x + c$, with $a = 1$, $b = - 4$ and $c = 16$.

This has discriminant $\Delta$ given by the formula:

$\Delta = {b}^{2} - 4 a c = {\left(- 4\right)}^{2} - \left(4 \times 1 \times 16\right) = 16 - 64 = - 48$

Since this is negative, $f \left(x\right) = 0$ has no Real roots and no factors with Real coefficients.

The Complex roots are given by the quadratic formula as:

$x = \frac{- b \pm \sqrt{\Delta}}{2 a} = \frac{4 \pm i \sqrt{48}}{2} = 2 \pm 2 i \sqrt{3}$

So $f \left(x\right) = \left(x - 2 - 2 i \sqrt{3}\right) \left(x - 2 + 2 i \sqrt{3}\right)$

Refer to explanation

#### Explanation:

it is ${x}^{2} - 4 x + 16 = 0 \implies {x}^{2} - 4 x + 4 - {\left(\sqrt{12} \cdot i\right)}^{2} = {\left(x - 2\right)}^{2} - {\left(\sqrt{12} \cdot i\right)}^{2} \implies \left(x - 2 + i \cdot 2 \sqrt{3}\right) \cdot \left(x - 2 - i 2 \sqrt{3}\right)$

where ${i}^{2} = - 1$ is a complex number

If you are looking factorization in the domain of reals numbers this cannot be done.