9x^2
------
9x^2-3x
I don't get this next one either. . .
6t^2+2t
----------
3t^2-5t-2
2007-01-31 09:13:30 · 3 answers · asked by snape_fan_2005 2 in Science & Mathematics ➔ Mathematics
You are trying to reduce terms, like the fraction 8/12 - both the numerator and denominator have a common factor, 4, so this can be reduced to 2/3, so that the numerator and denominator are relatively prime.
Ok, so look at the second one:
(6t^2 +2t) / (3t^2 -5t - 2) Lets factor each:
2t(3t + 1) / (3t + 1)(t-2)
Both have a common factor, 3t +1, so:
2t/(t-2) is an equivalent expression.
2007-01-31 09:24:14 · answer #1 · answered by John T 6 · 0⤊ 0⤋
Factor the numerator and denominator when possible and then cancel. You must cancel only when it's multiplication/division and not with addition. What I mean by this is that the 9x^2's do NOT cancel in the first problem. Factoring out the denominator, you get 3x(3x-1). So the 3x in the denominator cancels with one of the three's and one of the x's in the numerator, so you have left:
3x/(3x-1).
2007-01-31 09:18:03 · answer #2 · answered by Professor Maddie 4 · 0⤊ 0⤋
9x^2=3x*3x
-------
9x^2-3x=3x(3x-1)
3x*3x
---------
3x(3x-1)
cancel out 3x from top and bottom
3x
---
3x-1
2007-01-31 09:31:26 · answer #3 · answered by Bill F 6 · 0⤊ 0⤋