What is the Remainder When a Polynomial is Divided by a Product of Linear Terms?

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Homework Help Overview

The discussion revolves around a polynomial \( p(x) \) that leaves specific remainders when divided by linear factors \( (x-a) \), \( (x-b) \), and \( (x-c) \). Participants explore the implications of these conditions on the form of the remainder when \( p(x) \) is divided by the product of these linear terms.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the application of the remainder theorem and the implications of the polynomial's degree on the remainder's form. There is an exploration of the assumption that \( p(a) = a \), \( p(b) = b \), and \( p(c) = c \) leads to a quadratic remainder, which raises questions about the number of roots a quadratic can have.

Discussion Status

The discussion is ongoing, with participants questioning the assumptions made regarding the polynomial's divisibility and the nature of the remainder. Some suggest reconsidering the implications of the polynomial's degree and the specific values of \( a \), \( b \), and \( c \) as roots.

Contextual Notes

There is a noted confusion regarding the conditions under which \( p(x) \) is divisible by the linear factors, as well as the maximum degree of the remainder polynomial in relation to the cubic divisor.

sbhit2001
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Homework Statement


If a , b, c are distinct and p(x) is a polynomial in x which leaves remainders a,b,c on division by (x-a),(x-b),(x-c) respectively. Then the remainder on division of p(x) by(x-a)(x-b)(x-c) is

Homework Equations


As it is given that p(x) gives remainder a when divided by (x-a), so p(a) should be equal to a by remainder theorem.Similarl p(b) = b and p(c)=c.

The Attempt at a Solution


As (x-a)(x-b)(x-c) is a cubic polynomial, remainder can be max quadratic so I assume it to be px^2 + qx + r.Again by remainder theorem we will get 3 equations for p,q,r by using a,b,c. As we see that p(a) = a, p(b)=b,p(c)=c ; Then we can say that a,b,c will be roots of px^2 +x(q-1) + r. But a quadratic polynomial can have max 2 roots. Can u please tell me what did I do wrong here?
 
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Then we can say that a,b,c will be roots of px^2 +x(q-1) + r.
How did you come to that conclusion?
 
mfb said:
How did you come to that conclusion?
Because (x-a)(x-b)(x-c) will be 0 at the values ie a,b,c so remainder at each of these values will be p(a),p(b),p(c) ie a,b,c. So, eg if we take a we get pa^2 + qa + r = a , ie pa^2 +a(q-1) + r = 0 . In all a,b,c we get same expression. So I think that means a , b ,c will be roots of px^2 + x(q-1) + r.
 
sbhit2001 said:

Homework Statement


If a , b, c are distinct and p(x) is a polynomial in x which leaves remainders a,b,c on division by (x-a),(x-b),(x-c) respectively. Then the remainder on division of p(x) by(x-a)(x-b)(x-c) is
The remainder is what?
sbhit2001 said:

Homework Equations


As it is given that p(x) is divisible by (x-a)
This is NOT given. It does NOT say that p(x) is divisible by x - a, or x - b, or x - c.
sbhit2001 said:
, so p(a) should be equal to a by remainder theorem.Similarl p(b) = b and p(c)=c.

The Attempt at a Solution


As (x-a)(x-b)(x-c) is a cubic polynomial,

remainder can be max quadratic so I assume it to be px^2 + qx + r.Again by remainder theorem we will get 3 equations for p,q,r by using a,b,c. As we see that p(a) = a, p(b)=b,p(c)=c ; Then we can say that a,b,c will be roots of px^2 +x(q-1) + r. But a quadratic polynomial can have max 2 roots. Can u please tell me what did I do wrong here?
 
Id like you to consider this eg: let p(x)=x and when you divide this by x-a,x-b,x-c you get a,b,c as remainders and when you divide this by their combined product you get x ie p(x) itself as the remainder.So I think your remainder will be p(x) in your question ie the degree of p(x) is < 3.
 

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