do you know much about polynomial functions? pls help...
By lonely_f16
@lonely_f16 (2146)
Philippines
September 4, 2007 9:42am CST
I really suck in this subject...sorry..here are the questions and thanks to those who find time...
1.) When the polynomial P(x) is divided by x-5, the remainder is 2. What is P(5) ?
2.) If P(-1) =3, what is the remainder when P(x) is divided by x + 1?
3.) If P(2) = 0, what is one factor of P(x)?
4.) If x + 2 is a factor of P(x), what is the remainder when P(x) is divided by x + 2?
5.) How do you find the other factors of P(x) if the depressed polynomial is a quadratic?
thanks really...God Bless!
1 person likes this
4 responses
@stvasile (7306)
• Romania
4 Sep 07
Most of your questions are linked to a relation called "Bezout's Theorem" if I remember correctly. It says that if the number a is a root for the polynom (meaning that P(a)=0) that polynom can be divided by x-a (the remainder being zero).
So, to answer your questions the best I can:
1) P(5)= 2
2) The remainder is 3
3) One factor is x-2
4) The remainder is zero
and 5) I don't know... sorry!
@stvasile (7306)
• Romania
4 Sep 07
I am not very familiar with math terms in English so I don't really know what you mean at point 5.
If I understand correctly, you're asking how to find the second solution if you know a solution of a second degree equation.
If so, you need Viete's equations for a second degree equation. This equations, generalized by Girard say that, for a second-degree polynomial,
P(X)= a*X^2 + b*X + c, if you have two solutions of this polynomial, r1 and r2,
r1 + r2 = -(b/a)
and r1*r2 = c/a.
So, if you know a solution and the equation you can easily find out the second solution.
If you want more on these formulas, you can go to
http://mathworld.wolfram.com/VietasFormulas.html
You will find what I just told you, written in a different way, as the relations (6), (7) and (8) on that site.
I hope this is what you needed.
1 person likes this
@lonely_f16 (2146)
• Philippines
4 Sep 07
thank you very much! though the number 5 is not yet answered
1 person likes this
@lonely_f16 (2146)
• Philippines
5 Sep 07
thank you for sharing your views and answering them..thanks really...!!! :)
1 person likes this
@nonew3 (1941)
• United States
4 Sep 07
Hi! May I first ask this question: What math class are you in? I took college level precalculus 1 (college algebra), but it has been a few years.
This looks like it must either college algebra or calculus. If it is college algebra, I could look that up in my precalculus textbook that I still have from when I took college algebra. If it's calculus, I won't be able to help you because I haven't even taken trigonometry yet.
@nonew3 (1941)
• United States
5 Sep 07
Eeeks...I just had a look. It is WAY beyond my math skills level!
I am really glad I asked what math course this involves before trying to tackle this one.
In college precalculus 1 (college algebra), we barely touched the surface of polynomial functions, and I do mean barely.
My curiosity is piqued, but maybe I'll figure this out in a future college calculus course or beyond.
You are a math genius compared to little ol' newbie me. LOL!
Good luck with it.
1 person likes this
@lonely_f16 (2146)
• Philippines
5 Sep 07
no i'm not a math genius lolz...i don't even know how to answer this...hehehe
@peaceful (3294)
• United States
4 Sep 07
I haven't used polynomials in several thousand years, but there are people who use the stuff every single day! :)
Takes a trip over to Hot Math:
http://hotmath.com/
and see if they can help... good luck! :)
@pawan_tiwari (8)
• India
7 Sep 07
remainder theorem states that when a polynomial in x{p(x)} is divided by a linear equation in x {x-a}then the remainder is p(a).
and when p(a)=0, x-a is a factor of p(x).
apply these concepts to get the correct answers
1) p(5)=2
2) remainder is 3
3) one factor is x-2
4) remainder is 0
you can use the synthetic division method to get the remainder or apply the remainder theorem. see if p(b)=0 if yes then x-b is a factor
1 person likes this
@lonely_f16 (2146)
• Philippines
7 Sep 07
thank you for replying..you're a great help too...thanks...
