What are all the possible rational zeros for #f(x)=2x^3+x^2+8x-21# and how do you find all zeros?

1 Answer
Sep 9, 2016

Answer:

#f(x)# has rational zero #x=3/2# and other zeros #x = -1+-sqrt(6)i#

Explanation:

#f(x) = 2x^3+x^2+8x-21#

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 #-21# 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, +-7/2, +-7, +-21/2, +-21#

In addition note that the signs of the coefficients are in the pattern #+ + + -#. By Descartes' rule of signs, since there is one change of sign, there is exactly one positive Real zero.

So let's try the positive rational possibilities first:

#f(1/2) = 2(1/8)+(1/4)+8(1/2)-21 = -33/2#

#f(1) = 2+1+8-21 = -10#

#f(3/2) = 2(27/8)+(9/4)+8(3/2)-21 = (27+9+48-84)/4 = 0#

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

#2x^3+x^2+8x-21 = (2x-3)(x^2+2x+7)#

We can factor the remaining quadratic by completing the square and using the difference of squares identity:

#a^2-b^2 = (a-b)(a+b)#

with #a=(x+1)# and #b=sqrt(6)i# as follows:

#x^2+2x+7 = x^2+2x+1+6#

#color(white)(x^2+2x+7) = (x+1)^2-(sqrt(6)i)^2#

#color(white)(x^2+2x+7) = ((x+1)-sqrt(6)i)((x+1)+sqrt(6)i)#

#color(white)(x^2+2x+7) = (x+1-sqrt(6)i)(x+1+sqrt(6)i)#

Hence zeros:

#x = -1+-sqrt(6)i#