SOLUTION: I don't know how to answer this because as far as I know this is not factorable, but the question seems to say that it is. David’s homework was to factor the polynomial comple

Algebra ->  Complex Numbers Imaginary Numbers Solvers and Lesson -> SOLUTION: I don't know how to answer this because as far as I know this is not factorable, but the question seems to say that it is. David’s homework was to factor the polynomial comple      Log On


   

Question 1163109: I don't know how to answer this because as far as I know this is not factorable, but the question seems to say that it is.
David’s homework was to factor the polynomial completely (a) over the real numbers, (b) over the complex numbers.
x^3+4x^2+5x


He started x(x^2+4x+5) He knows that x=0 is a real root.
But there are no real factors of 5 that add up to 4???? So what does he do?
Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52812) About Me  (Show Source):
You can put this solution on YOUR website!
.

By writing  x^3+4x^2+5x = x(x^2+4x+5),  you just get factoring over real numbers and over integer numbers.

The quadratic polynomial  (x^2+4x+5)  is just NOT factorable further over real numbers and over integer numbers.



Answer by greenestamps(13203) About Me  (Show Source):
You can put this solution on YOUR website!


At the point where you have

x%5E3%2B4x%5E2%2B45+=+x%28x%5E2%2B4x%2B5%29

you are finished with part (a); the remaining quadratic factor does not factor over the real numbers.

However, every quadratic expression factors over the complex numbers; so you can do part (b). Solve the equation

x%5E2%2B4x%2B5+=+0

using the quadratic formula. That gives you the roots of the quadratic equation; therefore it shows you how to factor the quadratic over the complex numbers.

x+=+%28-4+%2B+sqrt%2816-20%29%29%2F2+=+-2%2Bi

and

x+=+%28-4+-+sqrt%2816-20%29%29%2F2+=+-2-i

So

x%5E2%2B4x%2B5+=+%28x-%28-2%2Bi%29%29%28x-%28-2-i%29%29

And so, finally,

x%5E3%2B4x%5E2%2B5x+=+%28x%29%28x-%28-2%2Bi%29%29%28x-%28-2-i%29%29