Question

factor x^2-4x+8

Answer 1

b^2 -4ac = (-4)^2-4(1)(8) =-16 <0
therefore, no solution

Answer 2

there is a solution with complex numbers

Answer 3

then, use (-b+-sqrt(b^2-4ac))/2a
you will get -ve inside the sqrt
sqrt(-1) =i
you can then find the answer

Answer 4

i'm still confused

Answer 5

quadratic equation

Answer 6

(-b+-sqrt(b^2-4ac))/2a
\[=(4\pm \sqrt{16-4(1)(8)}) /2\]
=\[=(4\pm4i)/2\]

Answer 7

x^2-4x=-8
x^2-4x+4=-4
(x-2)^2=-4
\[x-2=\pm \sqrt{-4}\]
\[x-2=\pm 2i\]
\[x=2 \pm 2i\]

Answer 8

The task is to factor x^2-4x+8.

It has been established above that x^2-4x+8 will not factor over the set of Real numbers.

It has been established above that x^2-4x+8 has roots that are not Real.

Depending upon the level of your studies, you may still be expected to factor the expression. Usually, a problem states in the instructions over what set to factor.

It has been established above that the roots are x = 2 + 2i and x = 2 - 2i.

In case you need the factorization over the set of Complex numbers:

x = 2 + 2i and x = 2 - 2i

x - 2 - 2i = 0 and x - 2 + 2i = 0

(x - 2 - 2i) (x - 2 + 2i ) = 0

Factorization is : (x - 2 - 2i) (x - 2 + 2i ) = x^2-4x+8

Answer 9

Good point. Forgot the question wanted the factorization