How do you use the rational root theorem to find the roots of #2x^3 + 7x^2 - 77x - 40#?

1 Answer
May 7, 2016

Answer:

#{(x=-1/2), (x=-8), (x=5) :}#

Explanation:

#f(x) = 2x^3+7x^2-77x-40#

By the rational root theorem, any rational zeros of this polynomial must be expressible in the form #p/q# for integers #p#, #q# with #p# a divisor of the constant term #-40# and #qa# a divisor of the coefficient #2# of the leading term.

That means that the only possible rational zeros are:

#+-1/2#, #+-1#, #+-2#, #+-5/2#, #+-4#, #+-5#, #+-8#, #+-10#, #+-20#, #+-40#

This is rather a lot of possibilities to try, but trying each in turn we soon find:

#f(-1/2) = -1/4+7/4+77/2-40 = (-1+7+154-160)/4 = 0#

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

#2x^3+7x^2-77x-40 = (2x+1)(x^2+3x-40)#

To factor the remaining quadaratic find a pair of factors of #40# that differ by #3#. The pair #8, 5# works.

Hence:

#x^2+3x-40 = (x+8)(x-5)#

So the other two zeros are #x=-8# and #x=5#