1. ## rational expressions

my question is
2x^2-5x+7 - x^2+3x-8
x^2-2x-15 x^2-2x-15

the common factor is (x-5)(x+3) but do still multiply the numerator with all of the denominator. or do i collect like terms. If i collect like terms i got
3x^4+2x+1. and i cant factor that out

2. Originally Posted by jay123
my question is
2x^2-5x+7 - x^2+3x-8
x^2-2x-15 x^2-2x-15

the common factor is (x-5)(x+3) but do still multiply the numerator with all of the denominator. or do i collect like terms. If i collect like terms i got
3x^4+2x+1. and i cant factor that out
the denominator's are the same, just subtract the numerators and then simplify:

$\displaystyle \frac {2x^2 - 5x + 7}{x^2 - 2x - 15} - \frac {x^2 + 3x - 8}{x^2 - 2x - 15} = \frac {2x^2 - 5x + 7 - x^2 - 3x + 8}{x^2 - 2x - 15}$

now simplify the top, then factorize to see if anything cancels

3. IM HAVING TROUBLE WITH THIS PROMBLEM...

y+2 - y-4
y-5 y+3... i changed the numerator's signs

y-2 + y+4
y-5 y+3 im confused on what i should do when i see that the demonators are different ... do i factor them out or do i mult each one by the num

4. Originally Posted by jay123
IM HAVING TROUBLE WITH THIS PROMBLEM...

y+2 - y-4
y-5 y+3... i changed the numerator's signs

y-2 + y+4
y-5 y+3 im confused on what i should do when i see that the demonators are different ... do i factor them out or do i mult each one by the num
no need to go there just yet. see post #5 here to see a quick method for adding fractions like these.

after reading that, you should understand how i made the first step:

$\displaystyle \frac {y + 2}{y - 5} - \frac {y - 4}{y + 3} = \frac {(y + 2)(y + 3) - (y - 5)(y - 4)}{(y - 5)(y + 3)}$

now, expand the numerator and continue

5. ok i understand that..
i combine like terms and then i get this -4y+26
(y-5)(y+3)
i can factor out -2(2y-12) but i can factor out a 2 now. -2+2(y-6). what do i do with the -2+2 do i multiply it. -4(y-6).

6. Originally Posted by jay123
ok i understand that..
i combine like terms and then i get this -4y+26
(y-5)(y+3)
i can factor out -2(2y-12) but i can factor out a 2 now. -2+2(y-6). what do i do with the -2+2 do i multiply it. -4(y-6).
you made a mistake somewhere. most likely with the signs. your numerator should come out to $\displaystyle 14y - 14$

7. hummm im still not getting the same answer ...i tried it about 6 times.

8. Originally Posted by jay123
hummm im still not getting the same answer ...i tried it about 6 times.
remember the minus sign between the two terms

$\displaystyle \frac {(y + 2)(y + 3) - (y - 5)(y - 4)}{(y - 5)(y + 3)} = \frac { \left( y^2 + 5y + 6 \right) - \left( y^2 -9y + 20 \right)}{(y - 5)(y + 3)}$

$\displaystyle = \frac {y^2 + 5y + 6 - y^2 + 9y - 20}{(y - 5)(y + 3)}$

$\displaystyle = \frac {14y - 14}{(y - 5)(y + 3)}$

$\displaystyle = \frac {14(y - 1)}{(y - 5)(y + 3)}$

9. thanks.

I did this problem but i keep getting it wrong. can you see the problem?
x-3 + x-1
x^2+3x+2 x^2-4

x-3 + x-1 = (x-3)(x-2) + (x-1)(x+1) = x^2-5x+x^2-1
(x+1)(x+2) (x-2)(x+2) (x+1)(x+2)(x-2) (x+1)(x+2)(x-2)

= 2x^2-5x-7

10. Originally Posted by jay123
thanks.

I did this problem but i keep getting it wrong. can you see the problem?
x-3 + x-1
x^2+3x+2 x^2-4

x-3 + x-1 = (x-3)(x-2) + (x-1)(x+1) = x^2-5x+x^2-1
(x+1)(x+2) (x-2)(x+2) (x+1)(x+2)(x-2) (x+1)(x+2)(x-2)...
there should be a +6 in the numerator of the last fraction that isn't there

and how did you get your last line to be like that?

11. isnt difference of squares x^2-4 (x-2)(x+2)

x^2+3x2 + x^2-4 = isnt this diff. of squares..
(x+2)(x+1)

12. Originally Posted by jay123
isnt difference of squares x^2-4 (x-2)(x+2)

x^2+3x2 + x^2-4 = isnt this diff. of squares..
(x+2)(x+1)
i don't understand what you're saying. please be more clear.

13. x^2-4. this is difference of squares right? (x-2)(x+2)= x^2-4

14. Originally Posted by jay123
x^2-4. this is difference of squares right? (x-2)(x+2)= x^2-4
yes, but that was not the issue here. that's not where you went wrong