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

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Answer 1 · 2015-09-13T14:11:29

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

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 · 2015-09-13T14:15:49

Refer to 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.

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