How do you solve x^3 + 4x^2 - 20x = 80x3+4x220x=80?

1 Answer
Nov 28, 2015

Subtract 8080 from both sides, then factor by grouping to find:

x = +-2sqrt(5)x=±25 or x = -4x=4

Explanation:

First subtract 8080 from both sides to get:

x^3+4x^2-20x-80 = 0x3+4x220x80=0

Factor the left hand side by grouping:

x^3+4x^2-20x-80x3+4x220x80

=(x^3+4x^2)-(20x+80)=(x3+4x2)(20x+80)

=x^2(x+4)-20(x+4)=x2(x+4)20(x+4)

=(x^2-20)(x+4)=(x220)(x+4)

=(x-sqrt(20))(x+sqrt(20))(x+4)=(x20)(x+20)(x+4)

=(x-sqrt(2^2*5))(x+sqrt(2^2*5))(x+4)=(x225)(x+225)(x+4)

=(x-2sqrt(5))(x+2sqrt(5))(x+4)=(x25)(x+25)(x+4)

So the roots of x^3+4x^2-20x=80x3+4x220x=80 are x = +-2sqrt(5)x=±25 and x=-4x=4