Math final simplifying math final review

Question

simplify

n+3)/(n²-5n+6)  +  (6)/(n²-7n+12)

Answer 1

(n+3)/(n²-5n+6) + 6/(n²-7n +12)

 

First lets see if we can factor the denominators;

n²-5n+6  ... factors of 6 that add to -5 are -3 and -2 so

n²-5n+6 = (n-3)(n-2)

 

n²-7n+12 ... factors of 12 that add to -7 are -4 and -3 so

n²-7n + 12 = (n-3)(n-4)

 

Now we have:

(n+3)/(n-2)(n-3) + 6/(n-3)(n-4)

 

Let's find a common denominator...  this would be (n-2)(n-3)(n-4)

 

((n-4)(n+3))/((n-2)(n-3)(n-4)) + (6(n-2))/((n-2)(n-3)(n-4))

 

= [(n-4)(n+3)) + 6(n-2)]/((n-2)(n-3)(n-4))

 

= (n²-n-12+6n-12)/(n-2)(n-3)(n-4)

 

combining terms in the numerator we have:

 

(n²+5n-24)/((n-2)(n-3)(n-4))

 

Let's now factor the numerator

the factors of -24 that add to 5 are 8 and -3 so

n²+5n-24 = (n+8)(n-3)  so now we have:

 

((n+8)(n-3))/((n-2)(n-3)(n-4))

 

As can be seen numerator and denominator both have the term (n-3) so cancel it out

 

= (n+8)/((n-2)(n-4))