Posted by Ashley on .
X^24x+3/x^2+3x18X621/x^2+10x+24
can someone work this problem for me. I have tried it does seem to work right.

algebra 
MathMate,
The problem itself is not right. It is hard to interpret, much less to attempt if you don't indicate what you are supposed to do with it (simplify, order in decreasing powers, ...). The part indicated in bold is ambiguous. Is X the same as x?
X^24x+3/x^2+3x18X621/x^2+10x+24
If X62 is meant to be x^2, and X is the same as x, then the expression can be simplified to:
x² 4x +3/x^2 +3x 18 x² 1/x^2 +10x +24
Order in decreasing powers of x, and group like terms (terms with the same variable raised to the same power):
x² x² 4x +3x +10x 18 +24 +3/x^2 1/x^2
which when like terms are added (or subtracted) gives:
9x +6 +2/x^2 
algebra 
Ashley,
x^2 minus 4x+3 over x^2 olus 3x18 x^2minus 1 over x^2 plus 10x plus 24 =

algebra : Parentheses()!! 
MathMate,
Wouldn't it be simpler if parentheses are used?
(x^24x+3)/(x^2+3x18)  (x^21)/(x^2+10x +24) = ?
This problem can be solved the same way one would solve
32/2413/15
=23/(3*8)  13/(3*5)
=(23*513*8)/(3*5*8)
=11/120
Now here's the parallel, assuming you have learned to factorize quadratic expressions.
First factorize the numerator and denominator of each term:
(x^24x+3)/(x^2+3x18)  (x^21)/(x^2+10x +24)
=(x3)(x1)/((x3)(x+6))  (x+1)(x1)/((x+6)(x+4))
Notice that there is a common factor of (x+6) in the denominator, so the common denominator is
(x+6)(x3)(x+4)
and the numerator would then be
(x3)(x1)(x+4)(x+1)(x1)(x3)
Again, since each term of the numerator has the common factor of (x3)(x1), we can group the terms to simplify the numerator:
(x3)(x1)[(x+4)(x+1)]
=3(x3)(x1)
Now we write out the whole expression:
3(x3)(x1)/((x+6)(x3)(x+4))
and cancelling out the common factor (x3), we get
3(x1)/((x+6)(x+4))
Either leave that as is, or you can multiply out to give
(3x3)/(x²+10x+24)