2006-12-11 16:37:37 · 8 answers · asked by k* * 1 in Education & Reference ➔ Homework Help
15x^5-12x^4+27x^3-3x^2
3x^2 * (5x^3-4x^2+8x-1)
What exactly do you need to do? This simplifies it a bit...
2006-12-11 16:41:26 · answer #1 · answered by jbchild788 2 · 0⤊ 0⤋
By factoring the equation, we have:
15x^5 -12x^4 + 27x^3 - 3x^2
= 3x^2 ( 5x^3 - 4x^2 + 9x - 1)
2006-12-11 17:28:20 · answer #2 · answered by chocolate_gurl 2 · 0⤊ 0⤋
You can factor it to 3x²(5x^3 - 4x^2 + 9x - 1), so you know 2 of the solutions are 0. The factor in parentheses is cubic with a positive leading coefficient, so you know there's at least 1 real solution, and if there are complex solutions, they occur in conjugate pairs. The only possible rational solutions are 1, -1, 1/5, and -1/5, and synthetic division quickly shows you those don't work. So you're left with using your graphing calculator, unless you class is into Cardano's methods.
2006-12-11 17:08:21 · answer #3 · answered by Philo 7 · 0⤊ 0⤋
it will be an equation and not in the form of a single term. just put the number of the power to which u raise x to in front of it . just write the equation as it is and just put the power on top of x by removing the^ sign. that will be the answer
2006-12-11 16:42:28 · answer #4 · answered by minks m 1 · 0⤊ 0⤋
You should do your own howework, Im not sure about the answer
Justin
2006-12-11 20:24:54 · answer #5 · answered by Anonymous · 0⤊ 0⤋
with a financial calculator and your algebra book
2006-12-11 16:38:54 · answer #6 · answered by jit bag 4 · 0⤊ 0⤋
you solve it by studying real hard
2006-12-11 16:39:57 · answer #7 · answered by Anonymous · 0⤊ 0⤋
factor it.
2006-12-11 16:39:23 · answer #8 · answered by devildawg200218 2 · 0⤊ 0⤋