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

##### 1 Answer

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

#8x^4-6x^3-7x^2+6x-1=(x-1)(x+1)(4x-1)(2x-1)#

Hence zeros

#### Explanation:

First note that the sum of the coefficients is

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

Hence

#8x^4-6x^3-7x^2+6x-1=(x-1)(8x^3+2x^2-5x+1)#

Examining the remaining cubic factor, notice that summing the coefficients with inverted signs on the odd powers also results in

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

So

#8x^3+2x^2-5x+1 = (x+1)(8x^2-6x+1)#

Finally note that

There are several different ways to find this last factorisation.

Here's a slightly strange one:

The digits

#168 = 13^2-1 = 14*12# .

Notice that the multiplication of

#8x^2-6x+1 = (4x-1)(2x-1)#

Hence the two remaining zeros are