We are asked to factor and simplify `6x^2(x^2-4)^2-5(x^2-4)^3 ` :
First we factor out the common factor ` (x^2-4)^2 ` to get:
Working within the parentheses, we use the distributive property to get:
Now, x^2-4 is a difference of two squares and will factor further. On the other hand, x^2+20 is prime (irreducible) (at least in the real numbers*). Factoring, we get:
`((x+2)(x-2))^2(x^2+20) ` or
`(x+2)^2(x-2)^2(x^2+20) ` which is the result we seek.
* In the complex numbers, this further factors to: