How do you solve #4x^3 + x^2 - 4x - 1=0# by factoring?

1 Answer
Mr. Hsia
May 21, 2018

#(x - 1)(x+1)(4x+1) = 0#
#x = 1# , #-1# , #-1/4#

Explanation:

#4x^3 + x^2 - 4x - 1 = 0#

Start of Factoring by Grouping:
#(4x^3 + x^2) + (-4x - 1) = 0#
#x^2(4x + 1) - 1(4x + 1) = 0#
#(x^2 - 1)(4x + 1) = 0#

We still have a Difference of Squares #(x^2 - a^2) = (x-a)(x+a)#
#(x - 1)(x+1)(4x+1) = 0#

Set each Factor equal to 0 and Solve for x:
#x - 1 = 0 -> x = 1#
#x + 1 = 0 -> x = -1#
#4x + 1 = 0 -> x = -1/4#

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