How do you divide #3x^3-4x^2-17x+6# by 3x-1 and is it a factor of the polynomial?

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Answer 1 · 2016-08-21T16:56:11

#(3x-1)# is a factor of the given poly., and, when the poly. is divided

by #(3x-1)#, the quotient is #(x^2-x-6)#.

Let #p(x)=3x^3-4x^2-17x+6#.

To find whether or not #(3x-1)# is a factor of #p(x)#, we have to

check whether #p(1/3)# is 0 or not.

#p(1/3)=3(1/27)-4(1/9)-17(1/3)+6=1/9-4/9-17/3+6=-1/3-17/3+6=-6+6=0#

Hence, #(3x-1)# is a factor of #p(x)#.

Now, #p(x)=3x^3-4x^2-17x+6#.

#=ul(3x^3-x^2)-ul(3x^2+x)-ul(18x+6)#.

#=x^2(3x-1)-x(3x-1)-6(3x-1)#

#=(3x-1)(x^2-x-6)#

Thus, #(3x-1)# is a factor of #p(x)#, and, when #p(x)# is divided by

#(3x-1)#, the quotient is #(x^2-x-6)#.

Enjoy Maths.!