Question

How do you factor `x^7 - 16x^5 + 5x^4 - 80x^2`?

Answer 1

`x^7-16x^5+5x^4-80x^2`

`= x^2(x+root(3)(5))(x^2-root(3)(5)x+root(3)(25))(x-4)(x+4)`

Explanation:

The sum of cubes identity can be written:

`a^3+b^3=(a+b)(a^2-ab+b^2)`

The difference of squares identity can be written:

`a^2-b^2=(a-b)(a+b)`

Given:

`x^7-16x^5+5x^4-80x^2`

All of the terms are divisible by `x^2`, so that can be separated out as a factor.

Note also that the ratio between the first and second terms is the same as that between the third and fourth terms. So this quadrinomial will factor by grouping:

`x^7-16x^5+5x^4-80x^2`

`= x^2((x^5-16x^3)+(5x^2-80))`

`= x^2(x^3(x^2-16)+5(x^2-16))`

`= x^2(x^3+5)(x^2-16)`

`= x^2(x^3+(root(3)(5))^3)(x^2-4^2)`

`= x^2(x+root(3)(5))(x^2-root(3)(5)x+root(3)(25))(x-4)(x+4)`