Can anyone factor y=15x^3+5x^2-38x-4?
Anyone? Without use cubic equation...
2006-12-11 12:52:14 · 3 answers · asked by mandeep s 1 in Science & Mathematics ➔ Mathematics
Hm....well the way to go about this would be to find a factor by trial and error...this appears not be be a whole number solution
through trial and error i have found the first factor to be approx 1.4886999
Now divide (x-1.4886999) into 15x^3+5x^2-38x-4 you will get a quadratic, solve this using the quadratic formula and u wil have your 3 factors
2006-12-11 13:07:24 · answer #1 · answered by Anonymous · 0⤊ 0⤋
factoring a third power is quite hard. here I would reccomend synthetic division to lose the x^3 as the last number is only 4, there are only 6 whole number possiblities. positive 1,2 and 4 and negative 1,2 and 4
2006-12-11 20:56:17 · answer #2 · answered by spiffo 3 · 0⤊ 0⤋
Synthetic division...or even better, just use a graphing calculator.
2006-12-11 21:10:51 · answer #3 · answered by bibimbapbambina 3 · 0⤊ 0⤋