Question

How do I find the possible zeros for the polynomial h(x)=2x^3+5x^2-31x-15?

Answer 1

was there a specific method you are supposed to use?

Answer 2

I don't think so, this is the actual question

Answer 3

we're assuming you've covered the "rational root test" already ?

Answer 4

yes sir!

Answer 5

You need to apply the Rational Root Theorem.

Take the constant term in the polynomial (without the sign) and find all possible factors.
Take the coefficient of the leading term and list all possible factors.

Then the possible zeros (or roots) of the polynomial are: plus/minus constant term factors / leading coefficient factors.

Answer 6

there are no zeros for this function as far as i can find.

Answer 7

then just factor the leading term coefficient and the constant, the possible answers will be their fraction, the constant factor's as numerators, and the leading coefficient as denominator

Answer 8

You pick the term without x, in your case -15, and the highest grade cofficent, in that case, 2.
then you make a list of every divisor of each
15: 1, 3, 5, 15
2: 1, 2

Finally you divide EACH element of the first list for each element in the second list, and place behind every result a +/- sign.
in this case: \[\pm {1\over 1}; \pm {1 \over 2}; \pm {3 \over 1}; \pm {3 \over 2}; \pm {5 \over1}; \pm {5 \over 2}; \pm {15 \over 1} ; \pm {15 \over 2}\]
Then try to divide the polinomy by ( x - every item on the list) to lower the grade.

Answer 9

\(\bf h(x)=2x^3+5x^2-31x-15\\ \quad \\
15\implies 3\times 5\times 1\qquad \qquad 2\implies 2\times 1\)

and the possible answers, is as @Paounn just described, those fractions

Answer 10

So the possible zeros would be 3/2 or -3/2?

Answer 11

Wait, the possible zeros would be all of those fractions, right?

Answer 12

yeap

Answer 13

I'm sorry, I was thinking of the bounds of the zeros!

Answer 14

Each one is a candidate to be a zero. I can tell for sure it has at least ONE real zero.

Answer 15

Thank you so much guys for explaining this! :)

Answer 16

So just to clarify, the possible zeros are all of the fractions that Paounn listed?

Answer 17

Yes. Every one of them. 8 of them positive and 8 of them negative. Total 16 fractions.

Answer 18

Thank you! :)

Answer 19

You could also have irrational zeros (zeroes) which, of course, aren't in that list, but there are no "quick" way to find them

Answer 20

yw. Give the medal to Paounn as he/she solved the whole thing for you.

Answer 21

I see, but I wouldn't need the irrational ones since the problem did not specifically ask for that, correct? @Paounn & of course, Paounn definitely deserves a medal! I wish I could give more than one :p @ranga

Answer 22

Irrational ones are quite a pain to get. As a rule of thumb, you'd start praying at least one zero is in that list, so you lower the grade to 2, which has an immediate formula to give you roots.

Answer 23

oh i see, thank you for all of your help!!! :)