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

AI Thread Summary
The discussion centers on finding the remainder when a polynomial p(x) is divided by the product of linear terms (x-a)(x-b)(x-c), given that p(x) leaves remainders a, b, and c when divided by (x-a), (x-b), and (x-c) respectively. It is established that p(a) = a, p(b) = b, and p(c) = c, leading to the assumption that the remainder can be expressed as a quadratic polynomial. However, a quadratic can only have a maximum of two roots, which raises questions about the validity of this assumption. The conclusion drawn is that the degree of p(x) must be less than three, suggesting that the remainder could simply be p(x) itself. The key takeaway is that the remainder when dividing by the cubic polynomial is determined by the behavior of p(x) at the specified values.
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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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