How do you combine #\frac { 3} { x ^ { 2} - 4} - \frac { 2x } { x ^ { 2} - 7x + 10} + \frac { 6} { x ^ { 2} - 3x - 10}#?

Question

617 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-12-10T14:02:52

#(-2x^2+5x-27)/(x^3-5x^2-4x+20)#

First we want to find the least common denominator. We'll need to factor all of the denominators for that:

#x^2-4=(x-2)(x+2)#
#x^2-7x+10=(x-2)(x-5)#
#x^2-3x-10=(x-5)(x+2)#

We can rewrite the original as:

#\frac { 3} { (x+2)(x-2)} - \frac { 2x } {(x-2)(x-5)} + \frac { 6} { (x-5)(x+2)}#

So from the factorizations we can see that the first fraction needs an #(x-5)#, the second needs an #(x+2)#, and the third needs an #(x-2)#:

#\frac { 3(x-5)} { (x+2)(x-2)(x-5)} - \frac { 2x(x+2) } {(x-2)(x-5)(x+2)} + \frac { 6(x-2)} { (x-5)(x+2)(x-2)}#

Now we can clean up the numerator and put it all over one denominator:

#(3x-15-(2x^2+4x)+6x-12)/((x+2)(x-2)(x-5))#

simplifying further gives:

#(-2x^2+5x-27)/((x+2)(x-2)(x-5))#

we can expand the denominator at this point:

#(-2x^2+5x-27)/(x^3-5x^2-4x+20)#