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#