How do you factor y= 3x^7 - 9x ^6 + 18x ^5y=3x79x6+18x5?

1 Answer
Dec 16, 2015

y = 3x^7-9x^6+18x^5=3x^5(x^2-3x+6)y=3x79x6+18x5=3x5(x23x+6)

Explanation:

First note that all the terms are divisible by 3x^53x5, so separate that out as a factor:

y = 3x^7-9x^6+18x^5=3x^5(x^2-3x+6)y=3x79x6+18x5=3x5(x23x+6)

The remaining quadratic factor is in the form ax^2+bx+cax2+bx+c with a=1a=1, b=-3b=3 and c=6c=6.

This has discriminant Delta given by the formula:

Delta = b^2-4ac = (-3)^2-(4xx1xx6) = 9-24 = -15

Since this is negative, the quadratic has only Complex zeros and cannot be factored any further with Real coefficients.