How do you find the zeros of # (-x+1)(x+2)(x-3)^2#?

1 Answer
Mar 26, 2016

Answer:

There are four zero in total: at #x=-2, x=1, x=3 and x=3#

graph{(-x+1)(x+2)(x-3)^2 [-7.116, 10.05, -12.5, 34.3]}

Explanation:

Given: #f(x)=(-x+1)(x+2)(x-3)^2#
Required the zeros?
Solution Strategy: Since #f(x)# is given by a product of three function #h(x), g(x), v(x)# such #h(x)=(-x+1), g(x)=(x+2), v(x)=(x-3)^2#
#f(x)=0# for any #x# that sets either #(h(x) or g(x) or v(x))=0# thus
#h(x) = 0, x=1#
#g(x) = 0, x=-2#
#v(x) = 0, x_(1,2)=3#

There are four zero in total: at #x=-2, x=1, x=3 and x=3#