How do you factor #y= 6x^3+13x-14x+3# ?

1 Answer
Dec 30, 2015

Answer:

Use the rational root theorem to help find the first root and factor, then an AC method to factor the remaining quadratic to find:

#6x^3+13x^2-14x+3 = (3x-1)(2x-1)(x+3)#

Explanation:

I will guess that #13x# should have been #13x^2# in the question.

Let #f(x) = 6x^3+13x^2-14x+3#

By the rational root theorem, any rational zeros of #f(x)# are expressible in lowest terms in the form #p/q#, where #p# and #q# are integers, #p# a divisor of the constant term #3# and #q# a divisor of the coefficient #6# of the leading term.

That means that the only possible rational roots are:

#+-1/6#, #+-1/3#, #+-1/2#, #+-1#, #+-3/2#, #+-3#

Let's try some of these:

#f(1/6) = 6/216+13/36-14/6+3#

#=(1+13-84+108)/36 = 38/36 = 19/18#

#f(-1/6) = (-1+13+84+108)/36 = 204/36 = 17/3#

#f(1/3) = 6/27+13/9-14/3+3#

#=(2+13-42+27)/9 = 0#

So #x=1/3# is a zero and #(3x-1)# is a factor:

#6x^3+13x^2-14x+3 = (3x-1)(2x^2+5x-3)#

Then use an AC method to help factor #2x^2+5x-3#.

Look for a pair of factors of #AC = 2*3 = 6# that differ by #B=5#.

The pair #6, 1# works, hence:

#2x^2+5x-3#

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

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

#= 2x(x+3)-1(x+3)#

#= (2x-1)(x+3)#

Putting it all together:

#6x^3+13x^2-14x+3 = (3x-1)(2x-1)(x+3)#