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# Remainder Theorem – Methods & Examples

A polynomial is an algebraic expression with one or more terms in which an addition or a subtraction sign separates a constant and a variable.

The **general form of a polynomial** is ax^{n} + bx^{n-1 }+ cx^{n-2 }+ …. + kx + l, where each variable has a constant accompanying it as its coefficient. The different types of polynomials include; binomials, trinomials, and quadrinomial.

**Examples of polynomials are**; 3x + 1, x^{2} + 5xy – ax – 2ay, 6x^{2} + 3x + 2x + 1 etc.

The procedure of dividing a polynomial by another polynomial can be lengthy and cumbersome. For instance, the polynomial long division method and synthetic division involve several steps in which one can easily make a mistake and thus ending up getting a wrong answer.

*Let’s briefly take a look at an example of the polynomial long division method and synthetic division.*

- Divide 10x⁴ + 17x³ – 62x² + 30x – 3 by (2x² + 7x – 1) using the polynomial long division method;

__Solution__

- Divide 2x
^{3 }+ 5x^{2 }+ 9 by x + 3 using synthetic method.

__Solution__

Reverse the sign of constant in the divisor x + 3 from 3 to -3 and bring it down.

_____________________

x _{+ }3 | 2x^{3} + 5x^{2} + 0x + 9

-3| 2 5 0 9

Bring down the coefficient of the first term in dividend. This will be our first quotient.

-3 | 2 5 0 9

________________________

2

Multiply -3 by 2 and add 5 to the product to get -1. Bring -1 down;

-3 | 2 5 0 9

-6

________________________

2 -1

Multiply -3 by -1 and add 0 to the result to get 3. Bring 3 down.

-3 | 2 5 0 9

-6 3

________________________

2 -1 3

Multiply -3 by 3 and add -9 to the result to get 0.

-3 | 2 5 0 9

-6 3 -9

________________________

2 -1 3 0

Therefore, (2x^{3 }+ 5x^{2 }+ 9) ÷ (x + 3) = 2x^{2}– x + 3

To avoid all these difficulties when dividing polynomials by either using the long division or synthetic division method, the Remainder Theorem is applied.

**The remainder theorem is useful because it helps us find the remainder without the actual polynomials division.**

Consider, for example, a number 20 is divided by 5; 20 ÷ 5 = 4. In this case, there is no remainder or the remainder is zero, 2o is the dividend when 5 and4 are the divisor and quotient, respectively. This can be expressed as:

Dividend = (Divisor × Quotient) + Remainder

i.e.20 = (5 x 4) + 0

Consider another case where a polynomial x^{2} + x – 1 is divided by x + 1 to get 4x-3 as the quotient and 2 as the remainder. This can also be expressed as:

4x^{2} + x – 1= (x + 1) * (4x-3) + 2

## What is the Remainder Theorem?

Given two polynomials p(x) and g(x), where p(x) > g(x) in terms of degree and g(x) ≠0, if p(x) is divided by g(x) to get q(x) as quotient and r(x) as remainder, then we can represent this statement as:

Dividend = (Divisor × Quotient) + Remainder

p(x) = g(x) * q(x) + r(x)

p(x) = (x – a) * q(x) + r(x),

But if r(x) = r

p(x) = (x – a) * q(x) + r

Then;

p(a) = (a – a) * q(a) + r

p(a) = (0) *q(a) + r

p(a) = r

According to the **Remainder Theorem**, when a polynomial, f (x), is divided by a linear polynomial, x – a the remainder of the division process is equivalent to f (a).

## How to use the Remainder Theorem?

Let’s see a few examples below to learn how to use the Remainder Theorem.

* *

*Example 1*

Find the remainder when the polynomial x^{3 }– 2x^{2 }+ x+1 is divided by x – 1.

__Solution__

p(x) = x^{3 }– 2x^{2 }+ x + 1

Equate the divisor to 0 to get;

x – 1 = 0

x = 1

Substitute the value of x into the polynomial.

⟹ p (1) = (1)^{3} – 2(1)^{2} + 1 + 1

= 2

Therefore, the remainder is 2.

* *

*Example 2*

What is the remainder when 2x^{2 }− 5x −1 is divided by x – 3

__Solution__

Given the divisor = x-3

∴ x – 3 =0

x = 3

Substitute the value of x in the dividend.

⟹ 2(3)^{2 }− 5(3) −1

= 2 x 9 − 5 x 3 − 1

= 18 – 15 − 1

= 2

*Example 3*

Find the remainder when 2x^{2 }− 5x − 1 is divided by x – 5.

__Solution__

x – 5 = 0

∴ x = 5

Substitute the value x = 5 in the dividend.

⟹ 2(5)^{2 }− 5(5) − 1 = 2 x 25 – 5 x 5 − 1

= 50 – 25 −1

= 24

* *

*Example 4*

What is a remainder when (x^{3} – ax^{2} + 6x – a) is divided by (x – a)?

__Solution__

Given the dividend; p(x) = x^{3} – ax^{2} + 6x – a

Divisor = x – a

∴ x – a = a

x = a

Substitute x = a in the dividend

⟹ p(a) = (a)^{3} – a(a)^{2} + 6a – a

= a^{3} – a^{3} + 6a – a

= 5a

* *

*Example 5*

What is the remainder of (x^{4} + x^{3} – 2x^{2} + x + 1) ÷ (x – 1).

__Solution__

Given the dividend = p(x) = x^{4} + x^{3} – 2x^{2} + x + 1

Divisor = x – 1

∴ x – 1 = 0

x = 1.

Now substitute x = 1 into the dividend.

⟹ p (1) = (1)^{4} + (1)^{3} – 2(1)^{2} + 1 + 1 = 1 + 1 – 2 + 1 + 1 = 2.

Hence, 2 is the remainder.

* *

*Example 6*

Find the remainder of (3x^{2} – 7x + 11)/ (x – 2).

__Solution__

Given the dividend = p(x) = 3x^{2} – 7x + 11;

Divisor = x – 2

∴x – 2 =0

x = 2

Substitute x = 2 in the dividend

p(x) = 3(2)^{2 }– 7(2) + 11

= 12 – 14 + 11

= 9

* *

*Example 7*

Find out whether 3x^{3} + 7x is a multiple of 7 + 3x

__Solution__

Take p(x) = 3x^{3} + 7x as the dividend and 7 + 3x as the divisor.

Now apply the Remainder Theorem;

⟹ 7 + 3x = 0

x = -7/3

Substitute x = -7/3 in the dividend.

⟹ p(x) = 3x^{3} + 7x = 3(-7/3)^{3} + 7(-7/3)

⟹-3(343/27) – 49/3

⟹ -(345 – 147)/9

= -490/9

Since the remainder – 490/9 ≠ 0, therefore 3x^{3} + 7x is NOT a multiple of 7 + 3x

* *

*Example 8*

Use the Remainder theorem to check if 2x + 1 is a factor of 4x^{3} + 4x^{2} – x – 1

__Solution__

Let the dividend be 4x^{3} + 4x^{2} – x – 1 and the divisor be 2x + 1.

Now, apply the Theorem;

⟹ 2x + 1 = 0

∴ x = -1/2

Substitute x = -1/2 in the dividend.

= 4x^{3} + 4x^{2} – x – 1 ⟹ 4( -1/2)^{3} + 4(-1/20^{2} – (-1/2) – 1

= -1/2 + 1 + ½ – 1

= 0

Since, the remainder=0, then 2x + 1 is a factor of 4x^{3} + 4x^{2} – x – 1