...and Higher

Division of polynomials

A polynomial of degree can be divided by another polynomial of degree iff . What we get is

or

where and are known as the quotient and remainder, respectively and

Example: Here's an interesting application of the division algorithm. Let's prove the following:

A positive integer is divisible by 11 if and only if the sum of the odd-position digits minus the sum of the even-position digits is a multiple of 11.

By way of an example, this statement is saying that you can easily determine that the number 845,221,729 is divisible by 11 because

8 - 4 + 5 - 2 + 2 - 1 + 7 - 2 + 9 = 22 which is divisible by 11.

Here's a proof using the division algorithm.

Consider any positive integer

where each

is a digit from 0 to 9. Now define

We wish to show that is divisible by 11 if and only if is divisible by 11.

By the division algorithm above we can write

so

Now plug in

Now plug in

From this equation we can see that is divisible by 11 if and only if is divisible by 11.

So, for what it's worth, this theorem tells us that the number 438,241,518 must be divisible by 11 because

4 + 8 + 4 + 5 + 8 - (3 + 2 + 1 + 1) = 29 - 7 = 22

and 22 is divisible by 11.

At this point you should be able to do the...........Skill Set, Higher Order Polynomials, No.1 attached below.

Gold Question: (P. Zeitz) Find the remainder when is divided by .

Purple Question: Given that

is an integer, for how many values of will

also be an integer.

Orange Question: Prove that an integer is divisible by 9 iff the sum of its digits is divisible by 9.

Brown Question: The five-digit number 9T67U, where T and U are digits, is divisible by by 36. Determine all possible values for T and U.

Roots and factors

Red Question: (P. Zeitz) Find a polynomial with integer coefficients that has a root of . Hint: Try thinking what factors the polynomial could have in addition to the obvious ones.

Blue Question: (P. Zeitz) Factor into the product of two smaller-degree polynomials whose forms you can guess.

Yellow Question: (P. Zeitz) Solve by first factoring it into the product of smaller polynomials whose forms you can guess.

STEP 2000, Math III, #6: Given that

express

, , and in terms of , , and .

Show also that

is a root of the cubic equation

is a root in the case , , .

Hence or otherwise express

STEP 2000, Math I, #2: Prove that if is a factor of the polynomial , then . Prove a corresponding

result if

is a factor of .

Given that the polynomial

has a factor of the form , find .

, and for it to be possible to write the quartic expression

in the form

, for some choice of values of , , and .

Show further that this condition holds if and only if it is possible to write the quartic expression in the

form

, for some choice of values of , and .

Find the roots of the quartic equation .

STEP 2008, Math I, #5: the polynomial is given by

are fixed real numbers and . Let be the greatest value of for . Then Chebyshev's

theorem states that

.

(i) Prove Chebyshev's theorem in the case

and verify that Chebyshev's theorem holds in the following cases:

(ii) Use Chebyshev's theorem to show that the curve has at least one turning point in

the interval

.

Symmetric Polynomials

Symmetric polynomials are ones in which the coefficients read the same left to right and right to left. Like this one:

The coefficients here are 1 , -3 , 2 , -3 , 2 , 1.

We can take advantage of this symmetry and use a clever substitution which reduces the degree of the equation and hopefully allows us to solve it.

First, divide everything by .

Now regroup the terms

(*)

Now we employ the substitution

By squaring this equation we find that

So plugging this into (*) we get

so

or

Now use these values of

to find .

or

or

or

Here is a more complicated example.

Solve .

Divide everything by

.

Regroup.

Now letting

and therefore

we get

The solutions of this are

so

or

This is about to get very messy, so we can try to cut down on the confusion by assigning

and

Then we have

or

Simplifying these we get

or

so that

or (**)

Now just looking at

and we see that

so

and

so that

I can now put these into (**).

You can try the above trick on symmetric polynomials of even degree and the technique always reduces the polynomial to one of

lesser degree. But what about polynomials of odd degree? Like this one...

What do we divide by? If we divide by then we get

which doesn't allow us to regroup into useful terms.

If we divide the original equation by we get

We are able to regroup into symmetric terms but it isn't clear what substitution would be useful. The substitution

doesn't help because

The technique doesn't seem to work.

However, we can do something. In order to see what this is we need to make two observations.

The coefficients of come in the form

So they come in pairs. Furthermore, for each pair, one is the coefficient of an even power and the other is a coefficient of an

odd power. This leads to the fact that

must be a root of the equation because

is a root of the equation, we can divide out by the factor using Synthetic Division to get

The quotient is symmetric too!

So now we can use our previous technique on that remaining quotient and thereby solve the equation.

Should this always be the case that the quotient will itself be symmetric?

A general proof is messy but we can easily see in the case of a fifth degree polynomial that this will happen.

Because, if our original polynomial is

,

you can verify using Synthetic Division that the quotient will be