Factoring problem that has me stumped!


by crookesm
Tags: factoring, stumped
crookesm
crookesm is offline
#1
May26-04, 03:25 AM
P: 5
Given:

[tex] s(t) = 1/2t^3-5t^2+3t+6 [/tex]

I'm trying to find all values of t where s(t) = -30

My first thought is to solve for 0 hence:

[tex] 1/2t^3-5t^2+3t+36=0 [/tex]

I know the answers are t=4 and t=8.196 but I can't get to it...I'm assuming I need to factor this down but I'm can't see it. Any help/hints would be most appreciated as I've been banging my head against a brick wall for some time now...
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arildno
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#2
May26-04, 03:29 AM
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You know that 4 is a root of the polynomial in your last equation.
Use polynomial division to compete the factoring.
chris_tams
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#3
May26-04, 02:55 PM
P: 18
As above, but divide (x-4) into 1/2t^3-5t^2+3t+36

Then when you get the second polynomial use the quadratic solution to get t=8.196

crookesm
crookesm is offline
#4
May28-04, 05:28 AM
P: 5

Factoring problem that has me stumped!


Thanks for the hints - reading up on polynomial division (which I wasn't familiar with) I have found that the factors are:

[tex] (t-4)(1/2t^2-3t-9) [/tex]

However, (and I realise I'm getting slightly off topic here) how would I even arrive at (t-4) being one of the original factors. Using GCF I can easily see that:

[tex] (t)(1/2t^2-5t+3)+36 = 0 [/tex]

But the jump to t-4 has got me stumped!

Any pointers to threads/websites on factoring polynomials/finding roots of polynomials would be most appreciated.
arildno
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#5
May28-04, 08:19 AM
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I believe Fermat developed various techniques in order to make good, reasoned guesses for the roots of polynomials.

However, when meeting a polynomial in a textbook that you don't immediately recognize the roots of, remember that the constant term is the product of the roots.

Therefore, one way of arriving at 4 as a root is to examine the integer factors of 36, and see which of these (if any!) might be a root.
jatin9_99
jatin9_99 is offline
#6
May30-04, 11:22 AM
P: 15
putting the equation equal to -30 u can make a cubic equation which will be then easy to solve for
mccp
mccp is offline
#7
May30-04, 02:58 PM
P: 1
Use descarte's rule of signs to find the nature of the roots (positive, negative, or complex).

After that take X minus all factors (positive and negative) of the coefficient of the highest degree of your variable over the factors of your constant.

3X^3 + 5X^2 - 2X + 8 =0

so your possible (real) factors are going to be (3/8), (3/4), (3/2), 3, (1/8), (1/4), (1/2), 1 and their inverses.

Well, now you're left with 14 possible roots, you can a) try them all algorithmically, or b) apply some heuristics. Graph the polynomial (preferably on a calculator) and look at which of the roots seem to be true (where the funtion intersects the X-Axis)--After that, try each of your roots until you find the root you're looking for.

Divide the equation by your new-found root and find the other two from your quadratic.

Hope I helped.


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