Question

How do you solve `x^3 + 4x^2 - 20x = 80`?

Answer 1

Subtract `80` from both sides, then factor by grouping to find:

`x = +-2sqrt(5)` or `x = -4`

Explanation:

First subtract `80` from both sides to get:

`x^3+4x^2-20x-80 = 0`

Factor the left hand side by grouping:

`x^3+4x^2-20x-80`

`=(x^3+4x^2)-(20x+80)`

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

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

`=(x-sqrt(20))(x+sqrt(20))(x+4)`

`=(x-sqrt(2^2*5))(x+sqrt(2^2*5))(x+4)`

`=(x-2sqrt(5))(x+2sqrt(5))(x+4)`

So the roots of `x^3+4x^2-20x=80` are `x = +-2sqrt(5)` and `x=-4`