The problems is:
2x/ x^2+9x+20 times(*) x^2-4x-32/3x^2
SO I MULTIPLIED THE NUMERATORS 2x(x^2-4x-32) AND GOT 2x^3-8x^2-64x, THEN I TOOK OUT A 2x AND GOT 2x(x^2-4x-32)
THEN I FACTORED & GOT 2(x-8)(x+4)
Then I multiplied the denominators and got 3x^4+27x^3+60x^2 THEN I TOOK OUT A 3x^2 AND GOT 3x^2(x^2+9x+20) THEN FACTORED & GOT 3x(x+5)(x+4)
Then the i canceled out the (x+4) {both} AND ENDED WITH
2(X-8) OVER 3X(X+5)
DID I GET IT RIGHT?
-THANK YOU ALL WHO HELPED ME WITH MY OTHER PROBLEMS... I LOVE YOU ALL
2007-02-03 16:37:36 · 6 answers · asked by Becky 1 in Science & Mathematics ➔ Mathematics
You did everything right.
2007-02-03 16:49:14 · answer #1 · answered by bruinfan 7 · 0⤊ 0⤋
you got the correct answer
don't multiply anything until you factor. then you can use cancellation.
2x/(x+5)(x+4) * (X-8)(X+4)/3X^2
then the x+4 just cancels and the x^2 becomes x
By multplying you are creating all of these high ordered terms
Factor everything first. You will be amazed at how often most of the stuff just falls out.
2007-02-04 00:44:50 · answer #2 · answered by trichbopper 4 · 0⤊ 0⤋
You must have got it right! I got the same answer and I am a 15-year-old with an IQ of 131! Right on!
2007-02-04 00:47:02 · answer #3 · answered by lambdamuomega1 2 · 0⤊ 0⤋
Your work is fine. I looked at it again and saw that you already canceled out the x. Yes, that's right.
2007-02-04 00:43:54 · answer #4 · answered by Chris S 5 · 0⤊ 0⤋
good job, and that 15 year old probably took a online tickle test :D
2007-02-04 01:37:47 · answer #5 · answered by austinblnd 4 · 0⤊ 0⤋
Yep, thats the right answer :)
2007-02-04 00:43:24 · answer #6 · answered by sarajschryver 2 · 0⤊ 0⤋