Question

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

Answer 1

`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.