# How do you solve #1/((x - 1)(x - 2)) + 1/((x - 2)(x - 3) )+ 1/((x - 3)(x - 4)) = 1/6#?

##### 1 Answer

#### Explanation:

Given:

#1/((x-1)(x-2))+1/((x-2)(x-3))+1/((x-3)(x-4)) = 1/6#

Multiply through by

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

Multiply out:

#6(x^2-7x+12)+6(x^2-5x+4)+6(x^2-3x+2) = x^4-10x^3+35x^2-50x+24#

#18x^2-90x+108 = x^4-10x^3+35x^2-50x+24#

Subtract the left hand side from the right to get:

#x^4-10x^3+17x^2+40x-84 = 0#

By the rational root theorem, any rational zeros of this polynomial must be expressible in the form

That means that the only possible rational roots are:

#+-1# ,#+-2# ,#+-3# ,#+-4# ,#+-6# ,#+-7# ,#+-12# ,#+-14# ,#+-21# ,#+-28# ,#+-42# ,#+-84#

Substituting

#x^4-10x^3+17x^2+40x-84 = 16-80+68+80-84 = 0#

So

#x^4-10x^3+17x^2+40x-84 = (x-2)(x^3-8x^2+x+42)#

Substituting

#x^3-8x^2+x+42 = -8-32-2+42 = 0#

So

#x^3-8x^2+x+42 = (x+2)(x^2-10x+21)#

To factor and find the zeros of the remaining quadratic, note that

#x^2-10x+21 = (x-3)(x-7)#

So the remaining zeros are

So all the zeros of our quartic polynomial are:

#-2, 2, 3, 7#

Note that the values

So the solutions of the original rational equation are