snow? Options

[Testm0nkey]

snow here in south overton

[James]

No snow on northwest loop.

[schwab]

not near university...but then again you posted a long ass time ago

[woody]

shitty

[Jwilly]

the snow will end once i use head and shoulders!!!

[Billy]

You'd solve that one by factoring by grouping.  Notice that the first two have a common factor of y^2 and the latter two have a common factor of -1, so you can factor it as (y^2-1)(y-6) but then you have difference of two squares, so you have (y+1)(y-1)(y-6)

Alternatively, you could use Descartes' Rule of Signs along with the Rational Zero Theorem.

Lastly, you could try using a numerical method by hand, but that would just be a pain and might as well use a calculator.

Thank you james =)

And can you do one more just so I can fully understand what you're saying:



(Just replace the dashes with a power. I'm sure you remember diffeq)

[kcroxyoursox]

its snowing on campus right now
you cant really see it because its pretty light
but maybe if it keeps up long enough...

[schwab]

its snowing on campus right now
you cant really see it because its pretty light
but maybe if it keeps up long enough...

yeah when i was walking home i noticed that

[James]

Thank you james =)

And can you do one more just so I can fully understand what you're saying:



(Just replace the dashes with a power. I'm sure you remember diffeq)

Alright. This one will be a bit longer since it involves a lot more work than factoring by grouping.

Consider the polynomial function: f(x)=x^3 - 7y^2 + 7y + 15

First, apply Descartes' Rule of Signs to determine the number of possible positive and negative zeros. Descartes' Rule of Signs states that the number of sign changes in f(x) indicates the maximum number of positive sign changes, including decrements of 2n (i.e. if there are 5 sign changes, then there are 5, 3, or 1 positive roots. if there there are 4 sign changes, then there are 4, 2, or 0 positive roots. Furthermore, it states that the number of sign changes in f(-x) indicates the maximum number of negative sign changes, including decrements of 2n.

The number of sign changes in f(x) is 2 (there's one from x^3 to -7y^2 and then one more from -7y^2 to 7y), so there are 2 or 0 positive roots.

f(-x)=-x^3 - 7y^2 - 7y + 15

The number of sign changes in f(-x) is 1 (there's one from -7y to 15), so there is exactly 1 negative root (This is NICE!)

Now apply the Rational Zero Theorem. It states that every polynomial, written generally as f(x)=a_n x^n + a_n-1 x^n-1 + ... + a_1 x^1 + a_0 has potential rational zeros at the ratio of the factors of a_0 over a_n. For instance, if a_0 had factors of 1, 2, -1, and -2 and a_n had factors of 1, 3, -1, and -3, then all possible rational zeros would be: 1/1, 1/3, 1/-1, 1/-3, 2/1, 2/3, 2/-1, 2/-3.

In our problem, a_0 is 15, so its factors are 1, 3, 5, 15, -1, -3, -5, and -15. Additionally, a_n is 1, so its factors are 1 and -1

Thus, all possible rational zeros are the following: 1, 3, 5, 15, -1, -3, -5, -15

Now, recall from earlier that we know we have exactly one negative root. If we're lucky enough for this to be a doctored problem, then that root will be rational, so let's try the negative numbers first. You should be able to tell by just looking at it that -15 and -5 will cause it to blow up too much, so don't bother trying those first. Let's try f(-1). If you get f(-1)=0, then you found a root. f(-1)=-1 - 7 - 7 + 15 = 0. Therefore, -1 is a root. That means x+1 is a factor of our polynomial.

Now, you can do one of two things. You can do polynomial division, which always works in reducing your polynomial by one degree, or you can try your other possible rational zeros. I prefer polynomial division as it's not that difficult.

Evaluate (x^3 - 7x^2 + 7x + 15) / (x+1) and you'll get (x^2 - 8x + 15) which you know factors to (x-5)(x-3).

Therefore, the factors of this polynomial are (x+1)(x-5)(x-3).

[schwab]

what the fuck deos this have to do with snow

[James]

Billy asked.

[Billy]

Wow James. Thanks a bunch.

[James]

No problem.