Multiply the following fractions.
2x^2 - x - 3 / 3x^2 + 7x + 4
x
3x^2 - 11x – 20 / 4x^2 - 9
2007-03-08 11:50:00 · 5 answers · asked by Mr.Archie G 2 in Science & Mathematics ➔ Mathematics
first simplify each part (make the quadratic into a two part multiplying bracket)
(2x-3)(x+1) / (3x+4)(x-5)
x
(3x+4)(x+1) / (2x-3)(2x+3)
now you can simplify the (2x-3)'s, (x+1)'s and the (3x+4)'s, so you're left with just
(x-5)/(2x+3)
2007-03-08 11:52:47 · answer #1 · answered by Zuri 3 · 0⤊ 0⤋
Ok, I'll do it, step by step :
2x^2 - x - 3 = (2x -3)(x+1)
3x^2 + 7x + 4 = (3x + 4)(x+1)
Numerator : (2x -3)(x+1)*(3x + 4)(x+1)
3x^2 -11x - 20 = (3x +4)(x-5)
4x^2 - 9 = (2x+3)(2x-3)
Denominator : (3x +4)(x-5)*(2x+3)(2x-3)
Numerator / Denominator = (x+1)^2 / (x-5)(2x+3)
Hope that helped you
2007-03-08 19:57:52 · answer #2 · answered by anakin_louix 6 · 0⤊ 0⤋
(2x-3)(x+1) / (3x+4)(x+1)
(3x+4)(x-5) / (2x-3)(2x+3)
cancel out the 2 3x+4 and 2x-3 and x+1
Answer(x-5)/(2x+3)
2007-03-08 19:59:46 · answer #3 · answered by Dave aka Spider Monkey 7 · 0⤊ 0⤋
Factor the polynomials in each fraction. (I hope you know how to do that.) Simplify each fraction, then cross-simplify on the diagonals. Then multiply straight across, and simplify if you haven't already. Nothing your algebra teacher shouldn't have taught you how to do already, assuming you have an algebra teacher. (Incidentally, we just learned this in my algebra class. I'm pretty good at it.)
2007-03-08 20:04:18 · answer #4 · answered by dark_load1 2 · 0⤊ 0⤋
I don't like math or scince but use the calcalator. I can't though my mom n dad
2007-03-08 19:55:32 · answer #5 · answered by Miss Simple 1 · 0⤊ 0⤋