ANSWER:

x=−2 or x=7

Explanation:

Given:

1(x−1)(x−2)+1(x−2)(x−3)+1(x−3)(x−4)=16

Multiply through by 6(x−1)(x−2)(x−3)(x−4) to get:

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(x2−7x+12)+6(x2−5x+4)+6(x2−3x+2)=x4−10x3+35x2−50x+24

→

18x2−90x+108=x4−10x3+35x2−50x+24

Subtract the left hand side from the right to get:

x4−10x3+17x2+40x−84=0

By the rational root THEOREM, any rational zeros of this POLYNOMIAL must be expressible in the form pq for integers p,q with p a divisor of the constant term −84 and q a divisor of the coefficient 1 of the leading term.

That MEANS that the only possible rational roots are:

±1, ±2, ±3, ±4, ±6, ±7, ±12, ±14, ±21, ±28, ±42, ±84

Substituting x=2 into the quartic we find:

x4−10x3+17x2+40x−84=16−80+68+80−84=0

So x=2 is a zero and (x−2) a factor:

x4−10x3+17x2+40x−84=(x−2)(x3−8x2+x+42)

Substituting −2 into this cubic we find:

x3−8x2+x+42=−8−32−2+42=0

So x=−2 is a zero and (x+2) a factor:

x3−8x2+x+42=(x+2)(x2−10x+21)

To factor and find the zeros of the remaining QUADRATIC, note that 3+7=10 and 3⋅7=21, so:

x2−10x+21=(x−3)(x−7)

So the remaining zeros are x=3 and x=7.

So all the zeros of our quartic polynomial are:

−2,2,3,7

Note that the values 2 and 3 are not solutions of the original rational equation, since they result in zero denominators.

So the solutions of the original rational equation are x=−2 and x=7

hope it helps you