4x^2 - 23x + 15/ 8x - 6 divided by x^2 - 25/6x + 30
2006-10-23 16:59:56 · 5 answers · asked by Tiffany B 1 in Science & Mathematics ➔ Mathematics
[ (4x^2 - 23x + 15) / (8x-6) ] / [ (x^2 - 25) / (6x + 30) ]
= [(4x^2 - 23x + 15)*(6x + 30)] / [(8x - 6)(x^2 - 25)]
= [6(4x - 3)(x - 5)(x + 5)] / [2(4x - 3)(x + 5)(x - 5)]
= 3
2006-10-23 17:06:22 · answer #1 · answered by James L 5 · 0⤊ 0⤋
((4x^2 - 23x + 15)/(8x - 6)) / ((x^2 - 25)/(6x + 30))
((4x^2 - 23x + 15)/(8x - 6)) * ((6x + 30)/(x^2 - 25))
((4x^2 - 23x + 15)(6x + 30))/((x^2 - 25)(8x - 6))
(6(x - 5)(4x - 3)(x + 5))/(2(x^2 - 25)(4x - 3))
(6(x^2 - 25)(4x - 3))/(2(x^2 - 25)(4x - 3))
ANS : 3
2006-10-23 17:09:12 · answer #2 · answered by Sherman81 6 · 0⤊ 0⤋
well after long devision you get a 2 quationt and a remainder like
(ax + b)/(x^2 - 25/6 x + 30) ok. then just break the denominater in to its partial fractions and simplify it
2006-10-23 17:13:22 · answer #3 · answered by Raven 2 · 0⤊ 0⤋
4x^2-23x+15
=4x^2-20x-3x+15
=4x(x-5)-3(x-5)
=(x-5)(4x-3)
8x-6=2(4x-3)
x^2-25=(x+5)(x-5)
6x+30
=6(x+5)
now substituting
[(x-5)(4x-3)/2(4x-3)]
/[(x+5)(x-5)/6(x+5)]
[(x-5)/2][6/(x-5)]
=3 (ans.)
2006-10-23 17:11:04 · answer #4 · answered by raj 7 · 0⤊ 0⤋
First multiply out the denominator (2b + a million)^3 = (2b + a million)(2b + a million)(2b + a million) = (4b^2 + 4b + a million)(2b + a million) you'll locate that the first time period of it is an same by technique of using truth the numerator, so that they cancel; [4b^2 + 4b + a million] / [(2b + a million)^3] = [4b^2 + 4b + a million] / [(4b^2 + 4b + a million)(2b + a million)] = a million / [2b + a million] desire that permits :)
2016-12-05 04:08:37 · answer #5 · answered by Anonymous · 0⤊ 0⤋