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
Aug 22, 2016

Answer:

#f(x)# has zeros #1#, #-1/8# and #3/4#

Explanation:

#f(x) = 32x^3-52x^2+17x+3#

By the rational roots theorem, any rational zeros of #f(x)# are expressible in the form #p/q# for integers #p, q# with #p# a divisor of the constant term #3# and #q# a divisor of the coefficient #32# of the leading term.

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 #f(x)# is zero. That is:

#32-52+17+3 = 0#

So #f(1) = 0#, #x=1# is a zero and #(x-1)# is a factor:

#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 #AC=32*3 = 96# which differ by #B=20#.

The pair #24, 4# works. Use this pair to split the middle term, then factor by grouping:

#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: #x=-1/8# and #x=3/4#