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

Aug 14, 2016

The remaining zeros are $- \frac{2}{3}$ and $\frac{3}{2}$

#### Explanation:

$f \left(x\right) = 6 {x}^{4} - 5 {x}^{3} - 12 {x}^{2} + 5 x + 6$

Since we are told that $1$ and $- 1$ are zeros, there are corresponding factors $\left(x - 1\right)$ and $\left(x + 1\right)$ which we can separate out:

$6 {x}^{4} - 5 {x}^{3} - 12 {x}^{2} + 5 x + 6$

$= \left(x - 1\right) \left(6 {x}^{3} + {x}^{2} - 11 x - 6\right)$

$= \left(x - 1\right) \left(x + 1\right) \left(6 {x}^{2} - 5 x - 6\right)$

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

Find a pair of factors of $A C = 6 \cdot 6 = 36$ which differ by $B = 5$.

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

$6 {x}^{2} - 5 x - 6$

$= \left(6 {x}^{2} - 9 x\right) + \left(4 x - 6\right)$

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

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

Hence zeros $- \frac{2}{3}$ and $\frac{3}{2}$