Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

**Randy123****Guest**

Suppose f is a polynomial such that f(0) = 47, f(1) = 32, f(2) = -13, and f(3)=16. What is the sum of the coefficients of f?

**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,167

hi Randy123

A polynomial has the form:

You only have 4 constraints so you'd have to assume that a5, a6, .... are all zero.

You can then form 4 equations like this:

I got that one by putting x=2

You'll get three more by putting x=0, x=1 and x=3.

Then you need to use simultaneous equation methods to solve for all the 'a's.

If you can get those equations, but cannot solve them, then post again with your 4 equations.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Offline

**thickhead****Member**- Registered: 2016-04-16
- Posts: 1,086

The sum of coefficients , obviously is

**{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha{Gods rejoice at those places where ladies are respected.}**

Offline

**thickhead****Member**- Registered: 2016-04-16
- Posts: 1,086

The problem is just a hoax, simply to test presence of mind.

*Last edited by thickhead (2017-06-15 21:19:11)*

**{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha{Gods rejoice at those places where ladies are respected.}**

Offline

**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,167

Hi thickness,

Oh yes! Silly me. . Obviously, this reveals I have no presence of mind. In fact, maybe no mind at all. And it doesn't even have to be cubic

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Offline

**thickhead****Member**- Registered: 2016-04-16
- Posts: 1,086

I am sorry, bob bundy. I did not mean that.You have had so many wonderful solutions.But when we are preoccupied with some problem, we can't differentiate between tea and coffee. I only hoped you could put x=1 instead of x-2 in your post.

**{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha{Gods rejoice at those places where ladies are respected.}**

Offline

**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,167

It's ok. I wasn't offended. Just cross with myself for missing that. There's a phrase for it here in the UK: can't see the wood for the trees.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Offline

**mathicINDIA****Member**- Registered: 2017-06-03
- Posts: 7

If 1 & -1 are the zeroes of the polynomial p(x)=Lx^4+Mx^3+Nx^2+Rx+P=0, prove that L+M+P=M+R=0. [JEE mains 2014, AIEEE 2006, IIT 1998]

Offline

**thickhead****Member**- Registered: 2016-04-16
- Posts: 1,086

mathicINDIA wrote:

If 1 & -1 are the zeroes of the polynomial p(x)=Lx^4+Mx^3+Nx^2+Rx+P=0, prove that L+M+P=M+R=0. [JEE mains 2014, AIEEE 2006, IIT 1998]

I think it should be

If 1 & -1 are the zeroes of the polynomial

(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha

{Gods rejoice at those places where ladies are respected.}

Offline

Pages: **1**