ok so ive been doing this problem forever, can i get some help, i need to SIMPLIFY
x^2-2x-35/2x^3-3x^2 X 4x^3-9x/7x-49
2007-08-21 11:03:08 · 4 answers · asked by johnalds_1990 2 in Science & Mathematics ➔ Mathematics
(x - 7)(x + 5) / x²(2x - 3) X [x (4x² - 9) / 7(x - 7)]
(x + 5)(x)(2x - 3) (2x + 3) / [x²(2x - 3)(7) (x - 7)]
[(x + 5)(2x + 3) ] / [ 7x (x - 7) ]
2007-08-21 11:32:17 · answer #1 · answered by Como 7 · 2⤊ 0⤋
[(x^2-2x-35) / (2x^3-3x^2)] * [(4x^3-9x) / (7x-49)]
Factor:
= [(x-7)(x+5) / (x^2)(2x - 3)] * [(x)(4x^2 - 9) / (7(x-7))]
= [(x-7)(x+5) / (x^2)(2x - 3)] * [(x)(2x-3)(2x+3) / (7(x-7))]
Multiply numerators and denominators and divide:
= [(x-7)(x+5)(x)(2x-3)(2x+3)] / [(x^2)(2x-3)(7)(x-7)]
Cancel out terms that are common in the numerator and denominator:
= [(x+5)(2x+3)] / [(x)(7)]
= (x+5)(2x+3) / 7x
2007-08-21 18:09:15 · answer #2 · answered by whitesox09 7 · 0⤊ 0⤋
The trick to these is to factor, and then cancel out like terms.
(x-7)(x+5)/x^2(2x-3) * (2x-3)(2x+3)/7(x-7)
(x+5)/(x^2) * x(2x+3)/7(x-7)
(x+5)(2x+3)/7x
2007-08-21 18:06:05 · answer #3 · answered by de4th 4 · 0⤊ 1⤋
I've hated math ever since I scored abnormal low on that part of the SAT as a kid. At least my verbal was above average.
2007-08-21 18:07:10 · answer #4 · answered by Anonymous · 0⤊ 1⤋