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:

`(x^2-4)^2(6x^2-5(x^2-4)) `

Working within the parentheses, we use the distributive property to get:

`(x^2-4)^2(6x^2-5x^2+20) `

`(x^2-4)^2(x^2+20) `

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:

`(x+2)^2(x-2)^2(x+2sqrt(5)i)(x-2sqrt(5)i) `

**Further Reading**

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