# How do you find the zeros of 8x^4 - 6x^3 - 7x^2 + 6x - 1 = 0?

Feb 18, 2016

Examine the sum of the coefficients and reason a little to find:

$8 {x}^{4} - 6 {x}^{3} - 7 {x}^{2} + 6 x - 1 = \left(x - 1\right) \left(x + 1\right) \left(4 x - 1\right) \left(2 x - 1\right)$

Hence zeros $1$, $- 1$, $\frac{1}{4}$, $\frac{1}{2}$.

#### Explanation:

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

$8 - 6 - 7 + 6 - 1 = 0$

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

$8 {x}^{4} - 6 {x}^{3} - 7 {x}^{2} + 6 x - 1 = \left(x - 1\right) \left(8 {x}^{3} + 2 {x}^{2} - 5 x + 1\right)$

Examining the remaining cubic factor, notice that summing the coefficients with inverted signs on the odd powers also results in $0$. That is:

$- 8 + 2 + 5 + 1 = 0$

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

$8 {x}^{3} + 2 {x}^{2} - 5 x + 1 = \left(x + 1\right) \left(8 {x}^{2} - 6 x + 1\right)$

Finally note that $8 {x}^{2} - 6 x + 1 = \left(4 x - 1\right) \left(2 x - 1\right)$

There are several different ways to find this last factorisation.

Here's a slightly strange one:

The digits $8$, $6$, $1$ reversed form the number:

$168 = {13}^{2} - 1 = 14 \cdot 12$.

Notice that the multiplication of $14 \cdot 12$ involves no carrying of digits. Hence we can reverse all the digits, finding $41 \cdot 21 = 861$ and deduce:

$8 {x}^{2} - 6 x + 1 = \left(4 x - 1\right) \left(2 x - 1\right)$

Hence the two remaining zeros are $x = \frac{1}{4}$ and $x = \frac{1}{2}$