How do you find all the zeros of #f(x)=12x^4+14x^3-8x^2-14x-4#?

1 Answer
George C.
Mar 5, 2016

Use some observations of sums of coefficients, division and AC Method factoring to find zeros: #1, -1, -2/3, -1/2#

Explanation:

First note that the sum of the coefficients of #f(x)# is zero. That is:

#12+14-8-14-4 = 0#

So #f(1) = 0# and #(x-1)# is a factor.

Also all of the terms are divisible by #2#, so we find:

#12x^4+14x^3-8x^2-14x-4#

#=2(6x^4+7x^3-4x^2-7x-2)#

#=2(x-1)(6x^3+13x^2+9x+2)#

The remaining cubic factor has all positive coefficients, so no zeros for positive values of #x#, but if you reverse the signs of the coefficients of the terms of odd degree the sum is zero. That is:

#-6+13-9+2 = 0#

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

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

The remaining quadratic factor can be factored using an AC method: Find a pair of factors of #AC = 6*2 = 12# whose sum is #B = 7#. The pair #3, 4# works. Then use this pair to split the middle term and factor by grouping:

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

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

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

So the zeros of #f(x)# are #1#, #-1#, #-2/3#, #-1/2#

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