How do you find all the zeros of f(x)=6x^4-5x^3-12x^2+5x+6f(x)=6x4−5x3−12x2+5x+6 given 1 and -1 as zeros?

1 Answer
George C.
Aug 14, 2016

The remaining zeros are -2/3−23 and 3/232

Explanation:

f(x) = 6x^4-5x^3-12x^2+5x+6f(x)=6x4−5x3−12x2+5x+6

Since we are told that 11 and -1−1 are zeros, there are corresponding factors (x-1)(x−1) and (x+1)(x+1) which we can separate out:

6x^4-5x^3-12x^2+5x+66x4−5x3−12x2+5x+6

=(x-1)(6x^3+x^2-11x-6)=(x−1)(6x3+x2−11x−6)

=(x-1)(x+1)(6x^2-5x-6)=(x−1)(x+1)(6x2−5x−6)

To factor the remaining quadratic we can use an AC method:

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

The pair 9, 49,4 works. Use this pair to split the middle term and factor by grouping:

6x^2-5x-66x2−5x−6

=(6x^2-9x)+(4x-6)=(6x2−9x)+(4x−6)

=3x(2x-3)+2(2x-3)=3x(2x−3)+2(2x−3)

=(3x+2)(2x-3)=(3x+2)(2x−3)

Hence zeros -2/3−23 and 3/232

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