How do you find all the zeros of #f(x)=6x^4-5x^3-12x^2+5x+6# given 1 and -1 as zeros?

1 Answer
Aug 14, 2016

Answer:

The remaining zeros are #-2/3# and #3/2#

Explanation:

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

Since we are told that #1# and #-1# are zeros, there are corresponding factors #(x-1)# and #(x+1)# which we can separate out:

#6x^4-5x^3-12x^2+5x+6#

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

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

To factor the remaining quadratic we can 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)#

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

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

Hence zeros #-2/3# and #3/2#