Question

Find a polynomial equation with integer coefficients that has the given numbers as solutions and the indicated degree. -3, 6, 5; degree 3

A) X^3+8x^2+3x+90=0
B) x^3 -8x^2 -3x +90 =0
C) x^3 -8x^2 -3x - 90 =0
D) x^3 - 8x^2 +3x - 90=0
E) x^3 +8x^2 -3x+90=0

Answer 1

well, it looks like theve given you all the number values to play with, so its just a matter of determing signs

Answer 2

the simplest way to do this would be to use synthetic division to see what zeros out; or simply apply the the values to see what fits

Answer 3

descartes sign rules says that all +s means no postive roots exists

all negative signs means the no negative roots exist, since we have a collection of pos and neg roots we can cancel out those with all like signs

Answer 4

one negative root would imply we have one sign change in the negative version

Answer 5

so what's the answer...because I've never been taught long division.

Answer 6

i think we can narrow it down to b or e

Answer 7

of course you could just rebuild the polynomial from the roots given to get an exacting result

Answer 8

how do I do that??

Answer 9

you go in the opposite direction of finding roots. The idea is, if you can go in one direction, you should be able to turn around and go back again

Answer 10

you should do this one for me...I'm sorry, Im not getting this

Answer 11

umm, nah. im not really up for that option.

Answer 12

if your not getting it, then you should take the time to review your material on this subject to get better acquianted with it

Answer 13

the simplest polynomial with roots of -3, 6, 5 with a degree of 3 is
f(x)=(x+3)(x-6)(x-5)
if you expand this, you should get what you're looking for

Answer 14

B) \[x^3-8 x^2-3 x+90=(x-6) (x-5) (x+3) \]