How do you find the zeros of #x^3+x^2-4x-4#?

1 Answer
Aug 19, 2016

Answer:

This cubic has zeros: #2, -2, -1#

Explanation:

This cubic factors by grouping then using the difference of squares identity:

#a^2-b^2=(a-b)(a+b)#

with #a=x# and #b=2# as follows:

#x^3+x^2-4x-4#

#=(x^3+x^2)-(4x+4)#

#=x^2(x+1)-4(x+1)#

#=(x^2-4)(x+1)#

#=(x^2-2^2)(x+1)#

#=(x-2)(x+2)(x+1)#

Hence zeros:

#2, -2, -1#