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

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Answer 1 · 2016-02-18T20:49:23

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 $1$ , $-1$ , $1/4$ , $1/2$ .

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 $(x-1)$ a factor:

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

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

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

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

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

Finally note that $8x^2-6x+1 = (4x-1)(2x-1)$

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:

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

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

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

Hence the two remaining zeros are $x = 1/4$ and $x=1/2$

How do you find the zeros of 8x^4 - 6x^3 - 7x^2 + 6x - 1 = 08x^4 - 6x^3 - 7x^2 + 6x - 1 = 0?