Polynomial. Again?

Algebra Level 3

Suppose a polynomial $p(x)$ such that it has a the following properties:
- Leaves a remainder $a$ on division by $(x-a)$ ,
- Leaves a remainder $b$ on division by $(x-b)$ ,
- Leaves a remainder $c$ on division by $(x-c)$ .

Find the remainder when $p(x)$ is divided by $(x-a)(x-b)(x-c)$ .

x+a+b

x+b

x+a

x

x+c

x+a+b+c

x+b+c

x+a+c

2 solutions

Steven Yuan
Jun 10, 2015

From the Remainder Theorem, we have $p(a) = a, p(b) = b, p(c) = c$ . Define $q(x) = p(x) - x.$ Then, $q(a) = q(b) = q(c) = 0$ i.e. $a, b, c$ are roots of $q(x)$ . Thus, $q(x)$ can be factored as $(x - a)(x - b)(x - c)r(x)$ , for some polynomial $r(x)$ . Substituting back, we have $p(x) - x = (x - a)(x - b)(x - c)r(x)$ , or $p(x) = (x - a)(x - b)(x - c)r(x) + x.$ This, when divided by $(x - a)(x - b)(x - c)$ , has remainder $\boxed{x}$ .

Moderator note:

Nice standard approach. Can you generalize this? That is to say:

Prove that for positive integers $k=1,2,3,\ldots,n$ , suppose a polynomial $p(x)$ has the properties that it leaves a remainder $a_k$ when divided by $(x-a_k)$ , then $\displaystyle \prod_{k=1}^n (x-a_k)$ divides $(p(x) - x )$ .

Thanks For the solution! $+0!$ from me :D

Mehul Arora - 6 years ago

You're welcome. $\ddot \smile$

Steven Yuan - 6 years ago

Challenge Master:

I'll assume that the $a_k$ are real numbers, not functions. From the Remainder Theorem, we have that $p(a_k) = a_k.$ Let $q(x) = p(x) - x.$ Then, since $q(a_k) = p(a_k) - a_k = 0,$ some of the roots of $q(x)$ are the $a_k.$ Thus, $q(x) = r(x)\prod_{k =1}^{n} (x - a_k),$ for some polynomial $r(x).$ Replacing the left hand side, we get $p(x) - x = r(x)\prod_{k =1}^{n} (x - a_k).$ Therefore, $\prod_{k =1}^{n} (x - a_k)$ divides $p(x) - x$ evenly.

Steven Yuan - 6 years ago

Andy B
Jun 17, 2015

One can pretty quickly realize that such a polynomial p(x) can equal x. The remainder of x/(x-a)(x-b)(x-c) is clearly x, and that's your answer.

Polynomial. Again?

2 solutions

Moderator note:

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