How do you find all the zeros of f(x)=12x4+14x3−8x2−14x−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,−23,−12

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:

12x4+14x3−8x2−14x−4

=2(6x4+7x3−4x2−7x−2)

=2(x−1)(6x3+13x2+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)(6x2+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)(6x2+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, −23, −12

Related questions
See all questions in Zeros
Impact of this question
2020 views around the world
You can reuse this answer
Creative Commons License