What are all the possible rational zeros for #f(x)=2x^4-13x^3-34x^2+65x+120# and how do you find all zeros?

1 Answer
Aug 3, 2017

Answer:

"Possible" rational zeros:

#+-1/2, +-1, +-3/2, +-2, +-5/2, +-3, +-4, +-5, +-6, +-15/2, +-8, +-10, +-12, +-15, +-20, +-24, +-30, +-40, +-60, +-120#

Actual zeros:

#-3/2#, #8#, #+-sqrt(5)#

Explanation:

Given:

#f(x) = 2x^4-13x^3-34x^2+65x+120#

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 #120# 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, +-2, +-5/2, +-3, +-4, +-5, +-6, +-15/2, +-8, +-10, +-12, +-15, +-20, +-24, +-30, +-40, +-60, +-120#

That's quite a few possibilities to try, so let's try a few easier ones to evaluate and deduce where the zeros might be...

#f(-2) = 2(color(blue)(-2))^4-13(color(blue)(-2))^3-34(color(blue)(-2))^2+65(color(blue)(-2))+120#

#color(white)(f(-2)) = 32+102-136-130+120 = -12 < 0#

#f(-1) = 2+13-34-65+120 = 36 > 0#

So #f(x)# has a zero in #(-2, -1)#. Is it the only possible rational one?

#f(-3/2) = 2(color(blue)(-3/2))^4-13(color(blue)(-3/2))^3-34(color(blue)(-3/2))^2+65(color(blue)(-3/2))+120#

#color(white)(f(-3/2)) = 2(81/16)+13(27/8)-34(9/4)-65(3/2)+120#

#color(white)(f(-3/2)) = 1/8(81+351-612-780+960) = 0#

So #x=-3/2# is a zero and #(2x+3)# a factor:

#2x^4-13x^3-34x^2+65x+120 = (2x+3)(x^3-8x^2-5x+40)#

The remaining cubic can be factored by grouping:

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

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

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

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

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

So the other three zeros are #+-sqrt(5)# and #8#