1

Proof.
$p(x) = (x-a)q(x) + r(x)$
Rearranging, $r(x) = p(x) - (x-a)q(x)$
Plug $x=a$, $r(a) = p(a) - (a-a)q(x)$
$r(a) = p(a)$

To me this seems like it only proves that $r(a) = p(a)$ for the specific instance where $x=a$. The textbook that I am using also states that this is true. It states that a proof will have to include the fact that $r(a)$ is a constant as $(x-a)$ is a first degree polynomial, and therefore, the remainder will always equal to $p(a)$. This makes sense to me, however, it still seems like it only applies for the specific instance where $x=a$, and does not prove the theorem.

Any explanation will be much appreciated. For reference I was using an A-level textbook.

QuantumSuperfield
  • 2,661
useruser334455
  • 13
  • $r(a)$ and $p(a)$ are constants, they are not dependent on $x$. – lulu Jan 19 '24 at 14:09
  • 1
    @lulu I don't see how that addresses OP's question. I actually agree with OP here. All the proof in the question does is show that $r(a)=p(a),$ which is kind of obvious, and doesn't prove that $p(x)=(x-a)q(x) + p(a).$ For example, the "proof" in the question does not explain why we cannot have $p(x) = x^2 = 2(x-a) +r(x)\ $(so here $q(x)=2)).$ The conclusion of the proof in the question is that $p(a)=r(a),$ which is true, but we still don't know what $r(x)$ is, sothe remainder theorem has not been proved... – Adam Rubinson Jan 19 '24 at 14:59
  • 1
    If however, we assume from the outset that $r(x)$ is a constant, then the remainder theorem has been proved. Either OP has not put the entire proof from the book in the body of the question, or the book's proof lacks details. – Adam Rubinson Jan 19 '24 at 15:00
  • Also, it doesn't surprise me this comes from an A-Level textbook. They always use hand-wavy proofs and lack details. – Adam Rubinson Jan 19 '24 at 15:09
  • @AdamRubinson I took the question as asking just about the proof that $r(a)=p(a)$, the argument appears to assume the remainder theorem at the start...it certainly isn't a proof of that result. Of course I might have misunderstood...the OP should clarify. – lulu Jan 19 '24 at 15:11
  • Well based on the title and OP's question, I think he was expecting a proof of the remainder theorem. – Adam Rubinson Jan 19 '24 at 15:17
  • Hi @lulu, sorry if the question was unclear, but basically the author was using r(a)=p(a) as a proof of the theorem, their exact line was, "Since the divisor (x − a) is a 1st-degree polynomial, r(x) must be a 0th-degree polynomial, i.e. a constant. We showed above that the remainder is sometimes (i.e. at least once) equal to p(a). So, being a constant, it is always equal to p(a).". This did not make sense to me, hence my question. – useruser334455 Jan 19 '24 at 15:21
  • Are you looking for a proof that $r(x)$ is a constant or $0$th-degree polynomial, which is outside the quoted proof? – peterwhy Jan 19 '24 at 15:23
  • As stated, there is no reason for $r(x)$ to be constant. You could have anything at all for $q(x)$, and then define $r(x)$ as $p(x)-(x-a)q(x)$. But you will always have $p(a)=r(a)$ – lulu Jan 19 '24 at 15:58

1 Answers1

0

Let's start with the theorem and its proof.

The remainder theorem states, if you divide a polynomial $P(x)$ by $x−a$, the remainder would be P(a).

And here's the proof.

Step 1. And polynomial P(x) can be written as

$$P(x) = (x-a) \cdot q(x) + r$$

Here $r$ is a constant. Because the remainder will be of a degree smaller than the degree of the divisor. And the divisor, $x - a$, is of degree 1.

Step 2. Substituting $a$ in place of $x$, we get

$$\begin{align*} P(a) &= (a - a) \cdot q(x) + r \\[0.3cm] &= 0 \cdot q(x) + r \\[0.3cm] &= r \end{align*}$$

Now, it's important to understand that just because we used $x = a$ in this step, doesn't mean the value of r depends on $a$. Remember $r$ is a constant. Our substitution is just a trick to get rid of the other term by making it $0$.

Step 2 is simply where we do the calculation. The value of $r$ is fixed in step 1, where we used $x - a$ as the divisor. And that's something in the very statement of the theorem.

So again, as long as the divisor is $x - a$, the remainder would be $P(a)$.

Haris
  • 3,409