# How do you find the zeroes of  f(x)=6x^4-5x^3-12x^2+5x+6?

Jun 24, 2016

$6 {x}^{4} - 5 {x}^{3} - 12 {x}^{2} + 5 x + 6 = \left(x - 1\right) \left(x + 1\right) \left(3 x + 2\right) \left(2 x - 3\right)$

#### Explanation:

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

First note that the sum of the coefficients is $0$.

That is:

$6 - 5 - 12 + 5 + 6 = 0$

Hence $x = 1$ is a zero and $\left(x - 1\right)$ a factor:

$6 {x}^{4} - 5 {x}^{3} - 12 {x}^{2} + 5 x + 6 = \left(x - 1\right) \left(6 {x}^{3} + {x}^{2} - 11 x - 6\right)$

If you reverse the signs of the terms of odd degree on the remaining cubic factor, then the sum is $0$.

That is:

$- 6 + 1 + 11 - 6 = 0$

Hence $x = - 1$ is a zero and $\left(x + 1\right)$ a factor:

$6 {x}^{3} + {x}^{2} - 11 x - 6 = \left(x + 1\right) \left(6 {x}^{2} - 5 x - 6\right)$

To factor the remaining quadratic expression, 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:

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

$= 6 {x}^{2} - 9 x + 4 x - 6$

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

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

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

Putting it all together:

$6 {x}^{4} - 5 {x}^{3} - 12 {x}^{2} + 5 x + 6 = \left(x - 1\right) \left(x + 1\right) \left(3 x + 2\right) \left(2 x - 3\right)$