I need to factor P(X)= 2X^4-6X^3+2X^2?

Question

I HATE factoring...and need the real zeros of this equation. Any help?

Answer 1

Hi,

First step on factoring is always to look for a GCF. In this problem, every term is divisible by 2x^2. Dividing 2x^2 out of each term as the GCF, you are left with 2x^2( x^2 -3x + 1).
We could set the GCF of 2x^2 = 0 and solve to get x = 0. That's your first zero.

Then we need to look at x^2 -3x + 1. This trinomial doesn't factor, so to solve it for zeros, use the quadratic equation,
x = -b +/- sqrt(b/^2 - 4ac)/(2a). Your values are "a" = 1, "b" = -3, and "c" = 1. When you substitute these, your equation becomes x = -(-3) +/- sqrt((-3)^2 - 4*1*1)/(2*1). This simplifies to x = [3 +/- sqrt(5)]/2. The decimal value for these are
x = [3 + sqrt(5)]/2 = 2.618 and x = [3 - sqrt(5)]/2 = .382.

So your 3 zeros are x = 0, 2.618, and .382.


I hope that helps!! :-)

Answer 2

P(X)= 2X^4-6X^3+2X^2

p(x) = 2x^2 * (x^2 - 3x + 1)

p(x) = 0. Hence 2x^2 = 0 (meaning x = 0) or x^2 - 3x + 1 = 0

Now factor the quadratic expression. This doesn't have any integer roots, so use the expression for finding the roots of a quadratic equation. You'll get x = (3 + SQRT(5))/2 and x = (3 - SQRT(5))/2.

hence x = 0, (3 + SQRT(5))/2 and (3 - SQRT(5))/2

Answer 3

P(X) = 2X² (X² - 3X + 1)
Zeros occur when X = 0, X² - 3X + 1 = 0
Consider X² - 3X + 1 = 0
X = [3 ±√(9-4)] / 2
X = [3 ±√(5)] / 2
X = 2.62, 0.38
Zeros occur when X = 0, 0.38, 2.62