Question 509401
When doing these kinds of problems, always factor everything first! Then see if something cancels out.

6x^2 - 31x + 5 =  (6x+1)(x-5)

(6x-1)(x-5)
-----------
(x-5)

(x-5)s cancel out.

6x-1 is what you are left with. Be sure to state that x cannot equal 5.

For this one you'll have to long divide since the numerator is not factorable.

x-1 |  x^3 + 2x^2 -3x + 2

How many times does x go into x^3?

x^2 <---- Remember

So x^2(x-1) =  x^3 - x^2

x^3 + 2x^2 - 3x + 2  -(x^3 - x^2) =  x^2 -3x + 2

x-1 | 3x^2 -3x + 2

How many times does x go into 3x^2?

3x  <---- Remember

x(x-1) =  3x^2 -3x

3x^2 -3x + 2 - (3x^2 - 3x)

2

How many times does x-1 go into 2?  Doesn't make sense.

So we are left with a remainder of 2 or ... 2/x-1 <-- Remember

Putting it all together we have

{{{expr(x^2 + 3x +  (2/(x-1)))}}}

Be sure to state that x cannot equal 1.