SOLUTION: How to find the zero's of: P(x) = x^2 - x^2 - 6x and P(x) = x^4 - 3x + 2x Please explain how this is done Thank you

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: How to find the zero's of: P(x) = x^2 - x^2 - 6x and P(x) = x^4 - 3x + 2x Please explain how this is done Thank you      Log On


   

Question 950238: How to find the zero's of:
P(x) = x^2 - x^2 - 6x
and
P(x) = x^4 - 3x + 2x
Please explain how this is done
Thank you
Answer by josgarithmetic(39620) About Me  (Show Source):
You can put this solution on YOUR website!
The two definitions for P(x) are not equal, so what you mean is not clear. The zeros for the two different definitions could also be different.

The first definition is uninvolved and very simple.

The second definition
x%28x%5E3-3%2B2%29
x%28x%5E3-1%29
x%28x-1%29%28x%5E2%2Bx%2B1%29, and you could use polynomial division to uncover this. You'd divide the cubic factor by x-1.

ROOTS: 0, 1, and two complex roots from the quadratic factor.

FINDING THOSE LAST TWO ROOTS:
General Solution of Quadratic Equation
%28-1%2B-+sqrt%281%5E2-4%2A1%2A1%29%29%29%2F2
%28-1%2B-+sqrt%28-3%29%29%2F2
highlight_green%28%28-1%2B-+i%2Asqrt%283%29%29%2F2%29