How do you solve x^3 + 4x^2 - 20x = 80x3+4x2−20x=80?
1 Answer
Nov 28, 2015
Subtract
x = +-2sqrt(5)x=±2√5 orx = -4x=−4
Explanation:
First subtract
x^3+4x^2-20x-80 = 0x3+4x2−20x−80=0
Factor the left hand side by grouping:
x^3+4x^2-20x-80x3+4x2−20x−80
=(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)=(x2−20)(x+4)
=(x-sqrt(20))(x+sqrt(20))(x+4)=(x−√20)(x+√20)(x+4)
=(x-sqrt(2^2*5))(x+sqrt(2^2*5))(x+4)=(x−√22⋅5)(x+√22⋅5)(x+4)
=(x-2sqrt(5))(x+2sqrt(5))(x+4)=(x−2√5)(x+2√5)(x+4)
So the roots of