Question

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

Answer 1

`x^2-4x+16 = (x - 2 - 2isqrt(3))(x - 2 + 2isqrt(3))`

Explanation:

Let `f(x) = x^2-4x+16`. This is of the form `ax^2+bx+c`, with `a=1`, `b=-4` and `c = 16`.

This has discriminant `Delta` given by the formula:

`Delta = b^2-4ac = (-4)^2-(4xx1xx16) = 16-64 = -48`

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

The Complex roots are given by the quadratic formula as:

`x = (-b +-sqrt(Delta))/(2a) = (4+-isqrt(48))/2 = 2 +-2isqrt(3)`

So `f(x) = (x - 2 - 2isqrt(3))(x - 2 + 2isqrt(3))`

Answer 2

Refer to explanation

Explanation:

it is `x^2-4x+16=0=>x^2-4x+4-(sqrt12*i)^2=(x-2)^2-(sqrt12*i)^2=>(x-2+i*2sqrt3)*(x-2-i2sqrt3)`

where `i^2=-1` is a complex number

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