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

1 Answer
George C.
Nov 28, 2015

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#

Related questions
See all questions in Solving Problems Algebraically and Graphically
Impact of this question
3301 views around the world
You can reuse this answer
Creative Commons License