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

2 Answers
Sep 13, 2015

Answer:

#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:

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.