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

1 Answer
Dec 16, 2015

Answer:

#y = 3x^7-9x^6+18x^5=3x^5(x^2-3x+6)#

Explanation:

First note that all the terms are divisible by #3x^5#, so separate that out as a factor:

#y = 3x^7-9x^6+18x^5=3x^5(x^2-3x+6)#

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