How do you solve #9x ^ { 4} + 18x ^ { 3} + 9x ^ { 2} = 0#?

1 Answer
Oct 25, 2017

Answer:

#x = -1, -1, 0, 0#

Explanation:

First, begin by factoring as much as possible from each term of the equation on the left hand side:

#9x^4 + 18x^3 + 9x^2 = 0 #

#(9x^2)(x^2 + 2x + 1) = 0#

Now factor the quadratic:

#(9x^2)(x+1)(x+1) = 0#

Since we know the product of the three terms on the left comes out to 0, we know that any one of those terms evaluating to 0 will be a solution of the equation. Thus, we solve each factor separately to determine all possible solutions for #x#:

#9x^2 = 0#

#x^2 = 0#

#x = +- 0 => x = 0 " (Multiplicity 2)" #

#x+1 = 0 => x = -1#

#x + 1 = 0 => x = -1#

We get four roots: #x = -1, -1, 0, 0#.