What are the possible rational roots of #2x^3+3x^2-8x+3=0# and then determine the rational roots?

1 Answer
Jan 31, 2017

Answer:

The "possible" rational roots are:

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

The actual roots are:

#1, 1/2, -3#

Explanation:

Given:

#2x^3+3x^2-8x+3=0#

By the rational roots theorem, any rational zeros of this cubic 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#

In addition, note that the sum of the coefficients is #0#. That is:

#2+3-8+3 = 0#

Hence #x=1# is a root and #(x-1)# a factor:

#0 = 2x^3+3x^2-8x+3#

#color(white)(0) = (x-1)(2x^2+5x-3)#

To find the zeros of the remaining quadratic we could try each of the other possible rational roots in turn, but I'd rather complete the square:

#0 = 8(2x^2+5x-3)#

#color(white)(0) = 16x^2+40x-24#

#color(white)(0) = (4x)^2+2(4x)(5)+(5)^2-49#

#color(white)(0) = (4x+5)^2-7^2#

#color(white)(0) = ((4x+5)-7)((4x+5)+7)#

#color(white)(0) = (4x-2)(4x+12)#

#color(white)(0) = (2(2x-1))(4(x+3))#

#color(white)(0) = 8(2x-1)(x+3)#

So the remaining two roots are:

#x=1/2" "# and #" "x=-3#