What are all the possible rational zeros for #f(x)=32x^3-52x^2+17x+3# and how do you find all zeros?
1 Answer
Explanation:
#f(x) = 32x^3-52x^2+17x+3#
By the rational roots theorem, any rational zeros of
That means that the only possible rational zeros are:
#+-1/32, +-1/16, +-3/32, +-1/8, +-3/16, +-1/4, +-3/8, +-1/2, +-3/4, +-1, +-3/2, +-3#
That is rather a lot of possibilities to try, but note that the sum of the coefficients of
#32-52+17+3 = 0#
So
#32x^3-52x^2+17x+3 = (x-1)(32x^2-20x-3)#
We can find zeros of the remaining quadratic factor using an AC method:
Find a pair of factors of
The pair
#32x^2-20x-3#
#=32x^2-24x+4x-3#
#=(32x^2-24x)+(4x-3)#
#=8x(4x-3)+1(4x-3)#
#=(8x+1)(4x-3)#
Hence zeros: