Several difficult problems on polynomial remainder/factor theorems

[stfz]

Homework Statement

I am currently working through a chapter on Polynomial Remainder and Factor Theorems in my book, Singapore College Math, Syllabus C.

There were a few problems which I got stuck on:

25) The positive or zero integer $r$ is the remainder when the positive integer, $n$ is divided by the positive integer $p$. Show that $p$ is an exact divisor of $(n^p-n) - (r^p-r)$. Factorize $(r^p-r)$ in each of the special cases, $p = 3, 5 , 7,$ and deduce that in these cases $p$ is an exact divisor of $n^p-n$.

19) Show that $a+b$ is a factor of $abc-(a+b+c)(bc+ca+ab)$ and factorize this expression completely.

Example) Prove that $(a+b)$ is a factor of $a^2(b+c)+b^2(c+a)+c^2(a+b)+2abc$
and write down the other two factors.

Proof
Let $f(x) = x^2(b+c) + b^2(c+x)+c^2(a+b)+2xbc$. When f(x) is divided by x+b, the remainder is: ... by remainder theorem... 0.
So it's a factor. Now the part I have a problem with:
-----------------------------------------------------
since $(a+b)$ is a factor of $f(x)$
∴ $f(x) = (a+b)Q(x)$, Q(x) = quotient.
Substituting $x = a$, we have
$a^2(b+c) + b^2(c+a) + c^2(a+b) + 2abc = (a+b)Q(a)$

The other factors are $b+c$ and $c+a$
-----------------------------------------------------

How do they know that (b+c) and (c+a) are factors?
I can see them in the equation, but how does it work?

Homework Equations

Factor/remainder theorems and long polynomial division (that's what the chapter was on)

The Attempt at a Solution

25) - This one I have no idea on how to tackle. $(n^p-n)/p)$ is obvious, but that I think is not part of the solution.

19) - The second one - I don't know how to do this either :(

Example) I understand everything here to the last step - getting the other factors. It didn't show how it was done. It just gave the conclusion. Were the factors found by long division, or were they taken out of the equation because of some rule?

 

[maajdl]

25)

n-r is divisible by p, by assumption.
Factor the polynomial and jump to the conclusion.

 

[Curious3141]

Another Singaporean, I see.

Homework Statement

I am currently working through a chapter on Polynomial Remainder and Factor Theorems in my book, Singapore College Math, Syllabus C.

There were a few problems which I got stuck on:

25) The positive or zero integer $r$ is the remainder when the positive integer, $n$ is divided by the positive integer $p$. Show that $p$ is an exact divisor of $(n^p-n) - (r^p-r)$. Factorize $(r^p-r)$ in each of the special cases, $p = 3, 5 , 7,$ and deduce that in these cases $p$ is an exact divisor of $n^p-n$.

$n = kp + r$ where k is some integer, and $0 \leq r \leq (p-1)$.

Now evaluate $n^p - n = (kp + r)^p - r$ using Binomial Theorem. What do you notice about the binomial coefficients? Which of them are multiples of $p$? Rearrange to get the required result.

For the next part, do the factorisation. Note the possible values that $r$ can take in each case of $p$. Make an argument in each of those cases that show that exactly one of the factors is a multiple of $p$. This way you can prove that $r^p - r$ is a multiple of $p$ in each of these cases, and you can go back to the originally proven result to immediately deduce that $n^p-n$ is a multiple of $p$ in these cases.

19) Show that $a+b$ is a factor of $abc-(a+b+c)(bc+ca+ab)$ and factorize this expression completely.

Use the same method used in the example. Start by replacing $a$ with $x$ then put $x = -b$ to establish that $(x+b)$ is a factor of the resulting polynomial. Finally, substitute $x = a$ back into the polynomial to deduce that $a+b$ is a factor of the original expression.

To save yourself a lot of work, you can appeal to the cyclic symmetry of the original expression. Note that switching from $a \to b$, $b \to c$ and $c \to a$ will leave the expression unchanged. Which means automatically that if $(a+b)$ is a factor, then so are $(b+c)$ and $(c+a)$. Just do the cyclic switching to see why this is so.

However, if you can't immediately have this insight you can plod along by doing these steps:

Now, repeat by replacing $b$ with $y$ and then putting $y = -c$. What can you deduce from this?

Finally, do it one last time by replacing $c$ with $z$ then putting $z = -a$. Make the deduction.

You've found three linear factors. Those are the only three factors dependent on one or more of the variables that can be present because of the order (3) of the original expression. There's a minus sign in the expression, which means that there has to be a minus sign appended to the factors, so put a minus sign before the result you got (the minus sign can't go anywhere else without disrupting the cyclic symmetry). Now you can expand out your result to show that it's equivalent to the original expression if you like, but it's not strictly necessary for the proof (plus it can be tedious).

Example) Prove that $(a+b)$ is a factor of $a^2(b+c)+b^2(c+a)+c^2(a+b)+2abc$
and write down the other two factors.

Proof
Let $f(x) = x^2(b+c) + b^2(c+x)+c^2(a+b)+2xbc$. When f(x) is divided by x+b, the remainder is: ... by remainder theorem... 0.
So it's a factor. Now the part I have a problem with:
-----------------------------------------------------
since $(a+b)$ is a factor of $f(x)$
∴ $f(x) = (a+b)Q(x)$, Q(x) = quotient.
Substituting $x = a$, we have
$a^2(b+c) + b^2(c+a) + c^2(a+b) + 2abc = (a+b)Q(a)$

The other factors are $b+c$ and $c+a$
-----------------------------------------------------

How do they know that (b+c) and (c+a) are factors?
I can see them in the equation, but how does it work?

Same principles I've already explained (observing the cyclic symmetry is probably what they did, too).

 

[Curious3141]

Now evaluate $n^p - n = (kp + r)^p - r$ using Binomial Theorem.

This line should read: "Now evaluate $n^p - n = (kp + r)^p - (kp + r)$ using Binomial Theorem. Just reading through my posts and noticed this error, and it was too late to edit.