How do you find the zeros of x^3-x^2-4x+4x3−x2−4x+4?
1 Answer
Jun 4, 2016
Factor by grouping to find zeros:
x = 2x=2 ,x = -2x=−2 andx = 1x=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+4x3−x2−4x+4
=(x^3-x^2)-(4x-4)=(x3−x2)−(4x−4)
=x^2(x-1)-4(x-1)=x2(x−1)−4(x−1)
=(x^2-4)(x-1)=(x2−4)(x−1)
=(x^2-2^2)(x-1)=(x2−22)(x−1)
=(x-2)(x+2)(x-1)=(x−2)(x+2)(x−1)
Hence zeros:
x = 2x=2 ,x = -2x=−2 andx = 1x=1
