How do you find the zeros of #f(x) = 2x^3 − 5x^2 − 28x + 15#?

1 Answer
Jul 7, 2016

Answer:

Zeros: #x=1/2#, #x=5#, #x=-3#

Explanation:

#f(x) = 2x^3-5x^2-28x+15#

By the rational root theorem, since #f(x)# has integer coefficients, any rational zeros must be expressible in the form #p/q# for integers #p, q# with #p# a divisor of the constant term #15# and #q# a divisor of the coefficient #2# of the leading term.

That means that the only possible rational zeros of #f(x)# are:

#+-1/2, +-1, +-3/2, +-5/2, +-3, +-5, +-15/2, +-15#

Trying the first one we find:

#f(1/2) = 2/8-5/4-28/2+15 = (1-5-56+60)/4 = 0#

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

#2x^3-5x^2-28x+15#

#=(2x-1)(x^2-2x-15)#

To factor the remaining quadratic, find a pair of factors of #15# which differ by #2#. The pair #5, 3# works, so:

#x^2-2x-15 = (x-5)(x+3)#

So the two remaining zeros are #x=5# and #x=-3#