SOLUTION: I have to factor the polynomial completely (a) over the real numbers, and (b) over the complex numbers. X to the third power + 4x to the second power + 5x I started with x(x sq

Algebra ->  Real-numbers -> SOLUTION: I have to factor the polynomial completely (a) over the real numbers, and (b) over the complex numbers. X to the third power + 4x to the second power + 5x I started with x(x sq      Log On


   

Question 1012085: I have to factor the polynomial completely (a) over the real numbers, and (b) over the complex numbers.
X to the third power + 4x to the second power + 5x
I started with x(x squared +4x+5) i know that x=0 is a real root. But there are no real factors of 5 that add up to 4? So what do i do?
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

x%5E3%2B+4x+%5E2%2B+5x

=x%28x%5E2%2B+4x+%2B+5%29

=x%28x%5E2%2B+4x+%2B+5%29......use quadratic formula to find the roots of x%5E2%2B+4x+%2B+5

x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+

x+=+%28-4+%2B-+sqrt%28+4%5E2-4%2A1%2A5+%29%29%2F%282%2A1%29+

x+=+%28-4+%2B-+sqrt%28+16-20+%29%29%2F2+

x+=+%28-4+%2B-+sqrt%28+-4+%29%29%2F2+

x+=+%28-4+%2B-+2i%29%2F2+

x+=+%28-cross%284%292+%2B-+cross%282%29i%29%2Fcross%282%29+

x+=+%282+%2B-i%29+

roots: x+=+%282+%2Bi%29+ and x+=+%282+-i%29+

use their product rule and you have %28x%2B+%282+%2Bi%29%29 and %28x%2B%282+-i%29%29+

so, to continue with factoring, we have

=x%28x%2B+%282+%2Bi%29%29%28x%2B%282+-i%29%29

=x%28x%2B+2+%2Bi%29%28x%2B2+-i%29