What are all the possible rational zeros for #y=3x^3+10x^2+4x-8# and how do you find all zeros?

1 Answer
Sep 7, 2016

Answer:

Zeros of #y=3x^3+10x^2+4x-8# are #2# and #2/3#.

Explanation:

If we put #x=-2# in #y=3x^3+10x^2+4x-8#, we get

#y=3(-2)^3+10(-2)^2+4(-2)-8#

= #3×(-8)+10×4-8-8#

= #-24+40-8-8=0#

Hence #(x+2)# is a factor of #y=3x^3+10x^2+4x-8# and

#y=3x^3+10x^2+4x-8#

= #3x^3+6x^2+4x^2+8x-4x-8#

= #3x^2(x+2)+4x(x+2)-4(x+2)#

= #(x+2)(3x^2+4x-4)#

= #(x+2)(3x^2+6x-2x-4)#

= #(x+2)(3x(x+2)-2(x+2))#

= #(x+2)(x+2)(3x-2)#

Hence zeros of #y=3x^3+10x^2+4x-8# are #2# and #2/3#.