Lesson Divisibility of polynomial f(x) by a binomial (x-a) and the Remainder theorem

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Divisibility of a polynomial f(x) by a binomial (x-a) and the Remainder theorem

In this lesson you will learn a theorem that helps you easily solve the problems that are difficult to solve otherwise.
The theorem is called the remainder theorem. It is presented below in three statements. The quantity

here is an arbitrary real number.

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|   Theorem   (the remainder theorem)
|
|   1. The remainder of division the polynomial    by the binomial    is equal to the value    of the polynomial.
|
|   2. The binomial    divides the polynomial    if and only if the value of    is the root of the polynomial , i.e. .
|
|   3. The binomial    factors the polynomial    if and only if the value of    is the root of the polynomial , i.e. .
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Proof of the Statement 1 (The remainder of division the polynomial f%28x%29

by the binomial x-a

is equal to the value f%28a%29

of the polynomial)

If f%29x%29

is the polynomial of the degree

then the formula of division the polynomial f%28x%29

by the binomial x-a

has the form

f%28x%29

. (1)

Here g%28x%29

is the polynomial of degree n-1

and

is the remainder which is the polynomial of degree 0, i.e. the constant.

Substitute x=a

to both sides of this equation. You get

f%28a%29

, i.e.

, or

.

This is exactly what the Statement 1 says.

Proof of the Statement 2 (The binomial x-a

divides the polynomial f%28x%29

if and only if the value of

is the root of the polynomial f%28x%29

, i.e.

)

a) If the binomial x-a

divides the polynomial f%29%28x%29

then

for some polynomial g%28x%29

. Substitute x=a

to both sides of this equation. You get

f%28a%29

which means that x=a

is the root of the polynomial f%29%28x%29

.
So, we proved that if the binomial x-a

divides the polynomial f%28x%29

then the value of

is the root of the polynomial f%28x%29

.

b) Conversely, let us assume that the value of

is the root of the polynomial f%28x%29

, i.e.

.
As before, the general formula of division the polynomial f%28x%29

by the binomial x-a

has the form

f%28x%29

. (2)

Here g%28x%29

is the polynomial of degree n-1

and

is the remainder which is the real number (same as the formula (1)).
Since f%28a%29=0

, we have

after substituting x=a

to both sides of the equation (2). Then

f%28x%29

which means divisibility the polynomial f%28x%29

by the linear binomial x-a

.
Thus the Statement 2 is proved.

Proof of the Statement 3 (The binomial x-a

factors the polynomial f%28x%29

if and only if the value of

is the root of the polynomial f%28x%29

, i.e.

)

Statement 3 is an equivalent re-statement (re-formulation) of the Statement 2.
Since the Statement 2 is proved (see above), the Statement 3 is proved too.

Example 1

and

are the roots of the quadratic polynomials f%28x%29+=+ax%5E2%2Bbx%2Bc

, then the polynomial is divided by x-x%5B1%5D

and

and is equal to f%28x%29

.

Solution
This is the direct consequence of the remainder theorem.

Example 2

Using the remainder theorem prove that the binomial x%5En-a%5En

is divisible by the binomial x-a

for any integer index

.

Solution
Note that this fact was proved in the lesson Factoring the binomials x%5En-a%5En

, which contains also examples and the detailed discussion.
Here we only check the divisibility using the remainder theorem.

Indeed, let us calculate the value of the binomial f%28x%29

. It is equal to

f%28x%29

.

So, in accordance with the remainder theorem, the binomial x%5En-a%5En

is divisible by the binomial x-a

Example 3

Using the remainder theorem prove that the binomial x%5En%2Ba%5En

is divisible by the binomial x%2Ba

for any odd integer index

.

Solution
Note that this fact was proved in the lesson Factoring the binomials x%5En%2Ba%5En

for odd degrees, which contains also examples and the detailed discussion.
Here we only check the divisibility using the remainder theorem.

Indeed, let us calculate the value of the binomial f%28x%29

. It is equal to

f%28x%29

for the odd integer index

.
So, in accordance with the remainder theorem, the binomial x%5En%2Ba%5En

is divisible by the binomial x%2Ba

for any odd integer index

Example 4

Using the remainder theorem prove that the binomial x%5En%2Ba%5En

is not divisible by the binomial x%2Ba

for any even integer index

unless the value of

is not equal to

.

Solution
Note that this fact was proved in the lesson Factoring the binomials x%5En%2Ba%5En

for odd degrees, which contains also examples and the detailed discussion.
Here we only check the fact of non-divisibility using the remainder theorem.

Indeed, let us calculate the value of the binomial f%28x%29

. It is equal to

f%28x%29

for the even integer index

. It is not equal to

unless the value of

is not equal to

.
Therefore, in accordance with the remainder theorem, the binomial x%5En%2Ba%5En

is not divisible by the binomial x%2Ba

for any even integer index

unless the value of

is not equal to

Example 5

Find the remainder of division the polynomial f%28x%29+=+%28x%2B1%29%5E99%2B1

by the binomial x-1

.

Solution
In accordance with the remainder theorem, this remainder is equal to f%281%29

, i.e.

, or

.

Answer The remainder of division the polynomial %28x%2B1%29%5E99%2B1

by the binomial x-1

is equal to 2%5E99%2B1

Example 6

Find the remainder of division the polynomial f%28x%29+=+x%5E100-x%5E99%2Bx%5E98-x%5E97-+ellipsis+-x%2B1

by the binomial x-1

.

Solution
In accordance with the remainder theorem, this remainder is equal to f%281%29

, i.e.

1%5E100-1%5E99%2B1%5E98-1%5E97-+ellipsis+-1%2B1

, or

.

Answer The remainder of division the polynomial f%28x%29+=+x%5E100-x%5E99%2Bx%5E98-x%5E97-+ellipsis+-x%2B1

by the binomial x-1

is equal to

Example 7

Prove that the polynomial f%28x%29+=+%28x%2B1%29%5E100+-+%28x-3%29%5E100

is divisible by the binomial x-1

.

Solution
Let us calculate the value of the polynomial %28x%2B1%29%5E100+-+%28x-3%29%5E100

at the value of x=1

.

It is equal to %281%2B1%29%5E100+-+%281-3%29%5E100

= 0.

Hence, the polynomial f%28x%29+=+%28x%2B1%29%5E100+-+%28x-3%29%5E100

is divisible by the binomial x-1

, in accordance with the remainder theorem.

OK, now you know a lot about divisibility of a polynomial f%28x%29

by a binomial x-a

.
What about divisibility by a binomial x%2Ba

?

Note that x%2Ba

.
Therefore, divisibility of a polynomial f%28x%29

by a binomial x%2Ba

is the same as divisibility of a polynomial f%28x%29

by a binomial x-%28-a%29

.
Moreother, the entire remainder theorem can be re-stated for the binomial x%2Ba

as follows:

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|   Theorem   (the remainder theorem for the binomial )
|
|   1. The remainder of division the polynomial    by the binomial    is equal to the value    of the polynomial.
|
|   2. The binomial    divides the polynomial    if and only if the value of    is the root of the polynomial , i.e. .
|
|   3. The binomial    factors the polynomial    if and only if the value of    is the root of the polynomial , i.e. .
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This Theorem does not require the special proof because everything was just proved for the binomial x-a

.

The Example 3 and Example 4 above illustrate this Theorem.

My other lessons on the Remainder theorem in this site are

    - Typical problems on the Remainder thoerem
    - Advanced problems on the Remainder theorem
    - Finding unknown coefficients of a polynomial having given info about its polynomial divisors
    - Finding unknown coefficients of a polynomial based on some given info about its roots
    - Nice_Olympiad_level_problems_on_divisibility_of_polynomials
    - OVERVIEW of lessons on Divisibility of polynomial f(x) by binomial (x-a) and the Remainder theorem

Use this file/link ALGEBRA-I - YOUR ONLINE TEXTBOOK to navigate over all topics and lessons of the online textbook ALGEBRA-I.

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