How do you list all possible roots and find all factors of #6x^3+7x^2-3x-1#?

1 Answer
Jan 17, 2017

Answer:

The "possible" rational zeros are:

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

The factorisation is:

#6x^3+7x^2-3x-1#

#= 3(2x-1)(x+5/6-sqrt(13)/6)(x+5/6+sqrt(13)/6)#

Explanation:

Given:

#f(x) = 6x^3+7x^2-3x-1#

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 #-1# and #q# a divisor of the coefficient #6# of the leading term.

That means that the only possible rational zeros are:

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

We find:

#f(1/2) = 6(1/8)+7(1/4)-3(1/2)-1 = (3+7-6-4)/4 = 0#

So #x=1/2# is a zero and #(2x-1)# a factor:

#6x^3+7x^2-3x-1 = (2x-1)(3x^2+5x+1)#

We can factor the remaining quadratic by completing the square.

To reduce the need to do arithmetic with fractions, I will premultiply by #3*2^2 = 12# and divide by it at the end:

#12(3x^2+5x+1) = 36x^2+60x+12#

#color(white)(12(3x^2+5x+1)) = (6x)^2+2(6x)(5)+25-13#

#color(white)(12(3x^2+5x+1)) = (6x+5)^2-(sqrt(13))^2#

#color(white)(12(3x^2+5x+1)) = ((6x+5)-sqrt(13))((6x+5)+sqrt(13))#

#color(white)(12(3x^2+5x+1)) = (6x+5-sqrt(13))(6x+5+sqrt(13))#

#color(white)(12(3x^2+5x+1)) = 12*3(x+5/6-sqrt(13)/6)(x+5/6+sqrt(13)/6)#

So:

#3x^2+60x+12 = 3(x+5/6-sqrt(13)/6)(x+5/6+sqrt(13)/6)#

Putting it all together:

#6x^3+7x^2-3x-1#

#= 3(2x-1)(x+5/6-sqrt(13)/6)(x+5/6+sqrt(13)/6)#