Question

Possible zeros identified?


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Answer 1

use +-p/q, where p is the last number and q is the leading coefficient

Answer 2

ok

Answer 3

@genius12 I'm not sure if that is right.. what do you think?

Answer 4

also I think you use the leading coefficient..

Answer 5

\[\pm1, \pm \frac{ 1 }{ 3 }, \pm \frac{ 7 }{ 3 }, \pm7, \pm3\]

Answer 6

I think that's it..

Answer 7

Think about what happens when you multiply polynomials together. The lowest order term is always formed by the products of the coefficients of the lowest order terms in the products, and the highest order term is formed by the products of the coefficients of the highest order terms in the products. However, because you can't tell if 4 is produced by 2*2 or -2 * -2, you've got to throw in the +/- business. Similarly, you can't tell if 12 came about by 2*6 or 3*4 or 1*12, etc. so you have to try all the factors. It'd be too easy otherwise :-)

Answer 8

If you have a choice, I think dividing out the biggest possible roots first (to narrow down the remaining factor options) may be better...

Answer 9

i got A but i am not sure

Answer 10

No, A is incorrect. You might have a pair of negative roots. For example,

\[(x^2-4x+4) = (x-2)(x-2)\]

Answer 11

is there a way to make sure your answer is correct?

Answer 12

Practice :-)

Answer 13

Frankly, this is sort of an artificial question. Really, you just want to make sure you've tried all of the possible roots, and probably don't care if you waste a little time trying a potential root or two that actually isn't possible.

Answer 14

Better to try a few more than necessary than to miss a few, in my opinion.

Answer 15

D

Answer 16

Looking at B vs. C: C has a bunch of roots that are clearly not possible: +/- 2 and +/- 8. So C isn't the right answer. A can be ruled out because we do have some negative coefficients in the polynomial, and A doesn't have any. Remember that a polynomial with a given set of roots r1, r2, r3, etc. can be written (x-r1)(x-r2)(x-r3) etc. and if we have a negative coefficient on an odd-order term there must have been some negative roots to generate it (see Descartes' Rule of Signs). That leaves B&D as contenders. Because the leading coefficient of the polynomial is not 1, it has to be D, because it has the only fractions.

Answer 17

Very informative

Answer 18

do you always give extremely informative answers?

Answer 19

are you a teacher?

Answer 20

No, but I like helping people understand :-)

Answer 21

Excellent Thank you

Answer 22

Hang on, there's a bit more to come :-)

Answer 23

I have 3 more questions that i need help with, you think you can help me here or shall i open a new question?

Answer 24

I do not want to waste too much of your time

Answer 25

open a new question, I'll finish up here and maybe someone else will answer some of the others while I'm doing this.

Answer 26

Finish what?

Answer 27

the "there's more to come" part :-)

Answer 28

Perfect :)

Answer 29

One of the things that you have to do when using this theorem is evaluate the polynomial a bunch of times. The straightforward way is not the efficient way, however!

If you have a polynomial P(x) = ax^3+bx^2+cx+d and you want to evaluate it at x = k, here's a better way to write it:

P(x) = ((ax + b) x + c) x + d

Now we substitute in x= k and we do the following:
\[P(k) = ((ak + b)k + c)k + d = (ak^2 + bk + c)k + d = ak^3 + bk^2 + ck + d\]

We get the same result, but with much less work! Let's compare:

P(k) the hard way:
\(a * k * k * k + b * k*k + c * k + d\) is 6 multiplications and 3 additions

P(k) the easy way:
\(((a*k + b) *k + c)* k + d\) is 3 multiplications and 3 additions

We also have smaller numbers to handle as we go along, especially if there are alternating signs. We never have to raise x to large powers.

There's one thing to be careful about when doing this, however: if you have some missing terms, you still have to incorporate them, except the coefficient would be 0.

For example, \[P(x) = 2x^3 + 4x + 5 = 2x^3 + 0x^2 + 4x + 5\]would be
\[P(x) = ((2x + 0)x + 4)x + 5\]

As the polynomials get longer, and the values of x get larger, this approach is a bigger win...

Answer 30

i will save this

Answer 31

It's called Horner's method, named after William George Horner, a British mathematician who lived in the late 18th and early 19th century.