How do you use the rational roots theorem to find all possible zeros of #f(x) = 2x^3 + 3x^2 – 8x + 3#?

1 Answer
Aug 9, 2016

Answer:

#f(x)# has zeros #1#, #1/2# and #-3#

Explanation:

#f(x) = 2x^3+3x^2-8x+3#

By the rational root 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 #2# of the leading term.

That means that the only possible rational zeros are:

#+-1/2, +-1, +-3/2, +-3#

We find:

#f(1) = 2+3-8+3 = 0#

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

#2x^3+3x^2-8x+3 = (x-1)(2x^2+5x-3)#

Substituting #x=1/2# in the remaining quadratic we find:

#2x^2+5x-3 = 2(1/4)+5(1/2)-3 = 1/2+5/2-3 = 0#

So #x=1/2# is a zero and #(2x-1)# a factor:

#2x^2+5x-3 = (2x-1)(x+3)#

So the final zero is #x=-3#