How do you find the zeroes of # f(x)=6x^4-5x^3-12x^2+5x+6#?

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Answer 1 · 2016-06-24T17:51:00

#6x^4-5x^3-12x^2+5x+6 = (x-1)(x+1)(3x+2)(2x-3)#

#f(x) = 6x^4-5x^3-12x^2+5x+6#

First note that the sum of the coefficients is #0#.

That is:

#6-5-12+5+6=0#

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

#6x^4-5x^3-12x^2+5x+6 = (x-1)(6x^3+x^2-11x-6)#

If you reverse the signs of the terms of odd degree on the remaining cubic factor, then the sum is #0#.

That is:

#-6+1+11-6 = 0#

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

#6x^3+x^2-11x-6=(x+1)(6x^2-5x-6)#

To factor the remaining quadratic expression, use an AC method:

Find a pair of factors of #AC = 6*6 = 36# which differ by #B=5#.

The pair #9, 4# works.

Use this pair to split the middle term and factor by grouping:

#6x^2-5x-6#

#=6x^2-9x+4x-6#

#=(6x^2-9x)+(4x-6)#

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

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

Putting it all together:

#6x^4-5x^3-12x^2+5x+6 = (x-1)(x+1)(3x+2)(2x-3)#