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

Mar 5, 2016

Use some observations of sums of coefficients, division and AC Method factoring to find zeros: $1 , - 1 , - \frac{2}{3} , - \frac{1}{2}$

Explanation:

First note that the sum of the coefficients of $f \left(x\right)$ is zero. That is:

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

So $f \left(1\right) = 0$ and $\left(x - 1\right)$ is a factor.

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

$12 {x}^{4} + 14 {x}^{3} - 8 {x}^{2} - 14 x - 4$

$= 2 \left(6 {x}^{4} + 7 {x}^{3} - 4 {x}^{2} - 7 x - 2\right)$

$= 2 \left(x - 1\right) \left(6 {x}^{3} + 13 {x}^{2} + 9 x + 2\right)$

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 $\left(x + 1\right)$ a factor:

$= 2 \left(x - 1\right) \left(x + 1\right) \left(6 {x}^{2} + 7 x + 2\right)$

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

$= 2 \left(x - 1\right) \left(x + 1\right) \left(6 {x}^{2} + 3 x + 4 x + 2\right)$

$= 2 \left(x - 1\right) \left(x + 1\right) \left(3 x \left(2 x + 1\right) + 2 \left(2 x + 1\right)\right)$

$= 2 \left(x - 1\right) \left(x + 1\right) \left(3 x + 2\right) \left(2 x + 1\right)$

So the zeros of $f \left(x\right)$ are $1$, $- 1$, $- \frac{2}{3}$, $- \frac{1}{2}$