SOLUTION: How would I solve for 3x^2+10x-8/3x^2+x-2? I got 4, is that correct.

Question 946600: How would I solve for 3x^2+10x-8/3x^2+x-2? I got 4, is that correct.
Found 2 solutions by richard1234, MathLover1:
Answer by richard1234(7193) (Show Source): You can put this solution on YOUR website!
You cannot "solve" 3x^2+10x-8/3x^2+x-2. You can find other things, e.g. asymptotes, or domain (set of all possible x-values for which the expression is defined), but the word "solve" in this case makes no sense.

Post your question exactly as written, with all information.
Answer by MathLover1(20849) (Show Source): You can put this solution on YOUR website!
....factor completely; write in numerator as , and in denominator as

...group

...factor out

since denominator cannot be equal to zero, we will have if or
since we can cancel ) like

then cannot be equal to

since numerator is and it will be zero if , so
solution is:

RELATED QUESTIONS
please help me solve. 1) (2x-3)(x-4) for an answer I got 2x^2-11x+12=0 is this... (answered by chiefman)
I got 2 for this problem. Is it correct, if not can someone please assist.... (answered by nerdybill)
Find X. 3X^2+10X+8/-X-2=-2 I already factorized and got: (X-2)(3X-4)/-X-2=-2 (answered by josgarithmetic)
Solve algebraically for the variable: x + 2 = 4 (3x -2) I got this: x + 2 = 12x... (answered by Alwayscheerful)
Can you please tell me if i got these correct? Im doing a practice test to study for my... (answered by Edwin McCravy,solver91311)
-(3*7 < = -4 power 2 -8(x-2)+7x < = -(12*3) _______... (answered by jim_thompson5910)
2/3x+8>12 What i did was: 2/3x+8-8>12-8 = 2/3x>4 2/3-2/3x > 4/1*3/2 = x>6 this is... (answered by actuary)
A rectangular picture has a length of 2x+2 and a width of x-4 inches. You decide to frame (answered by ankor@dixie-net.com)