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

1 Answer
Jun 4, 2016

Answer:

Factor by grouping to find zeros:

#x = 2#, #x = -2# and #x = 1#

Explanation:

Notice that the ratio of the first and second terms is the same as that between the third and fourth terms. So this cubic can be factored by grouping:

#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:

#x = 2#, #x = -2# and #x = 1#