Question

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

Answer 1

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`