Question

Factor `9a^5 -4a^3 -81a^2 +36` completely over the a) intergers c) reals b) rationals d) complex numbers ??

Answer 1

Factor out the first two terms and the next two terms to get an equal factor out.

Explanation:

`9a^5-4a^3-81a^2+36`

Factor the first two an the next two terms out.

`=a^3(9a^2-1)-9(9a^2-4)`

Factor each of the monomials out.

`=(9a^2-1)(a^3-9)`

`=(3a+1)(3a-1)(x-root(3)3)(x^2+root(3)3x+3root(3)3)`

Answer 2

`9a^5 -4a^3 -81a^2 +36`

`a^3(9a^2-4)-9(9a^2-4)`

`(a^3-9)(9a^2-4)`

`a^3-9` cannot be factored over the integers and `9a^2-4` is a difference of squares, so it can be factored over `ZZ`

`(a^3-9)(3a+2)(3a-2)`

`root(3)9` is irrational and so is `root(3)9^2 = 3root(3)3`

Use the difference of cubes to factor.

`(a-root(3)9)(a^2+aroot(3)9+root(3)9^2)(3a+2)(3a-2)`

The other 2 cube roots of `9` are imaginary and can be found by applying the quadratic formula (of complete the square) for

`a^2+aroot(3)9+root(3)9^2 =0`

`a = (-root(3)9 +- sqrt((root(3)9)^2-4(1)(root(3)9^2)))/(2(1))`

`= (-root(3)9 +- sqrt((3root(3)3)-12root(3)3))/2`

`= (-root(3)9 +- sqrt(-9root(3)3))/2`

`= (-root(3)9 +- 3sqrt(root(3)3)i)/2`

`= (-root(3)9 +- 3root(6)3i)/2`

So we can finish factoring over `CC`

`(a-root(3)9)(a-(-root(3)9 +3root(6)3i)/2)(a-(-root(3)9 - 3root(6)3i)/2)(3a+2)(3a-2)`