**Division of a polynomial** by another polynomial is one of the important concept in Polynomial expressions. In this article explained about basic phenomena of diving polynomial algorithm in step by step process

## Algebra division| Dividing Polynomials Long Division

Contents

Before going to algebra divisions observe the normal* numerical division algorithm*

When we divide 137 by 5 we get the quotient 27 and remainder 2

We can write 137 as

137 = (5 x 27) + 2 (Note : Here remainder 2 and it is less than divisor 5)

**i.e Dividend = Divisor x Quotient + Remainder**

The same division algorithm of number is also applicable for **division algorithm of polynomials.**

i.e When a polynomial divided by another polynomial

Dividend = Divisor x Quotient + Remainder, when remainder is zero or polynomial of degree less than that of divisor

A polynomial f(x) is divided by another polynomial g(x) we get quotient q(x) and remainder p(x) such that

f(x) = g(x) . q(x) + p(x)

Where p(x) = 0 or degree of p(x) < degree of g(x)

### Polynomial long division examples with solution

#### Dividing polynomials by monomials

Take one example

**Example -1 :** Divide the polynomial 2x^{4} +3x^{2} +x by x

Here

= 2x^{3} + 3x +1

So we write the polynomial 2x^{4} +3x^{2} +x as product of x and 2x^{3} + 3x +1

2x^{4} +3x^{2} +x = (2x^{3} + 3x +1) x

It means x & 2x^{3} + 3x +1 are factors of 2x^{4} +3x^{2} +x

**Example – 2** : Divide the polynomial 3x^{3} + 9x^{2} + 5 by 3x

Here variable “x” in denominator so it is not a polynomial.

So polynomial of 3x^{3} + 9x^{2} + 5 can be written as

3x^{3} + 9x^{2} + 5 = 3x (x^{2} + 3x) +5

Here we can say (x^{2} + 3x) is the quotient, 3x is the divisor and 5 is the remainder. And also we can say that the reminder is not zero, 3x is not a factor of 3x^{3} + 9x^{2} + 5

#### Dividing Polynomials by Binomials

**Example – 3:** Divide the polynomial 4x^{2}-3x +x^{3} +10 by x+4 and verify the remainder with zero of the divisor

**Solution:** Say f(x) = 4x^{2}-3x +x^{3} +10 and g(x)= x+4

For this division, we have using the steps as follows

**Step: 1 –** Arrange the terms of the dividend f(x) and the divisor g(x) in descending order of their degree

f(x)= x^{3} +4x^{2}-3x +10 and g(x) = x+4

**Step:2 –** We get first term of quotient by dividing the first term of dividend (i.e x^{3} ) with first term of divisor (i.e x)

So

**Step:3** – Now the first term of quotient x^{2} multiplying the divisor g(x) and that term subtract from dividend

x^{2} (x +4) = x^{3} +4x^{2}

**Step : 4 – ** Now repeat the step 2 again

i.e To get second term of quotient by dividing the first term of get remainder in previous step (i.e 3x ) with first term of divisor (i.e x)

So

**Step:5** – Then second term of quotient 3 multiplying the divisor g(x) and that term subtract from remainder got in step 3

3(x +4) = 3x + 12^{
}

Here the quotient is (x^{2} – 3) and remainder is 22

Here we can write this division process as

**Dividend = Divisor x Quotient + Remainder**

f(x) = (x^{2} – 3) g(x) + 22

x^{3} +4x^{2}-3x +10 = (x^{2} – 3) ( x+4) + 22

By using the **remainder theorem** we obtaining the remainder without performing the above process

*Note : Remainder theorem applicable only for divisor have linear polynomial with one variable*

Consider zero of the polynomial of g(x) is – 4

g(x) = x +4

g(-4) = -4 +4 = 0

Now find the value of f(x) at x= -4

f(4) = 4 (-4)^{2 }– 3(-4) + (-4)^{3} + 10 = 22

#### Dividing Polynomials by Trinomials

**Example -4: **Divide the polynomial 5x^{5 }+ x^{2} + 3x^{3} – 4x^{4}– 10 by 2 + x^{2 }– x

**Solution: ** Say f(x) = 5x^{5 }+ x^{2} + 3x^{3} – 4x^{4}– 10 and g(x)= 2 + x^{2 }– x

Now divide f(x) by g(x) by the following steps

**Step: 1** – Arrange the terms of the dividend f(x) and the divisor g(x) in descending order of their degree

f(x)= 5x^{5 }– 4x^{4 }+ 3x^{3} + x^{2} – 10 and g(x) = x^{2 }– x + 2

**Step:2** – We get first term of quotient by dividing the first term of dividend (i.e 5x^{5} ) with first term of divisor (i.e x^{2})

So

**Step : 3** – Now the first term of quotient 5x^{3} multiplying the divisor g(x) and that term subtract from dividend

5x^{3} (x^{2 }– x + 2) = 5x^{3} – 5x^{4 }+ 10x^{3}

**Step : 4** – Now repeat the step 2 again

i.e To get second term of quotient by dividing the first term of get remainder in previous step (i.e x^{4} ) with first term of divisor (i.e x^{2})

So

**Step:5 –** Then second term of quotient x^{2} multiplying the divisor g(x) and that term subtract from remainder got in step 3

x^{2} (x^{2 }– x + 2) = x^{4 }– x^{3} + 2x^{2}^{
}

**Step : 6** – Now repeat the step 4 again

i.e To get third term of quotient by dividing the first term of get remainder in previous step (i.e -6x^{3} ) with first term of divisor (i.e x^{2})

So

**Step:7 –** Then third term of quotient -6x multiplying the divisor g(x) and that term subtract from remainder got in step 5

-6x (x^{2 }– x + 2) = -6x^{3 } + 6x^{2} – 12x^{
}

**Step : 8 – ** Now repeat the step 6 again

i.e To get forth term of quotient by dividing the first term of get remainder in previous step (i.e -7x^{2} ) with first term of divisor (i.e x^{2})

So

**Step:9** – Then forth term of quotient -7 multiplying the divisor g(x) and that term subtract from remainder got in step 7

-7 (x^{2 }– x + 2) = -7x^{2 } + 7x – 14^{
}

Here the division process is completed because degree of (5x + 4) = 1 which is less than the degree of the divisor (x^{2 }– x + 2)

So the quotient is (5x^{3} + x^{3} -6x – 7) and remainder is 5x + 4

Here we can write this division process as

Dividend = Divisor x Quotient + Remainder

f(x) = (5x^{3} + x^{3} -6x – 7) g(x) + ( 5x + 4)

(5x^{5 }+ x^{2} + 3x^{3} – 4x^{4}– 10) = (5x^{3} + x^{3} -6x – 7) (x^{2 }– x + 2) + ( 5x + 4)

#### Dividing Polynomials by Quadratics

**Example:** Divide the polynomial 1 + x + 5x^{5 }-4x^{4} + 3x^{3} – 2x^{2} by 2 + x + x^{2 }+ x^{3}

Solution: Say f(x) = 1 + x + 5x^{5 }-4x^{4} + 3x^{3} – 2x^{2} and g(x)= 2 + x + x^{2 }+ x^{3}

Now divide f(x) by g(x) by the following steps

**Step: 1** – Arrange the terms of the dividend f(x) and the divisor g(x) in descending order of their degree

f(x)= 5x^{5 }-4x^{4} + 3x^{3} – 2x^{2} + x +1 and g(x) = x^{3 }+ x^{2 }+ x + 2

**Step:2** – We get first term of quotient by dividing the first term of dividend (i.e 5x^{5} ) with first term of divisor (i.e x^{3})

So

**Step:3** – Now the first term of quotient 5x^{2} multiplying the divisor g(x) and that term subtract from dividend

5x^{2} (x^{3 }+ x^{2 }+ x + 2) = 5x^{5} + 5x^{4 }+ 5x^{3 }+ 10x^{2}

**Step : 4 –** Now repeat the step 2 again

i.e To get second term of quotient by dividing the first term of get remainder in previous step (i.e -9x^{4} ) with first term of divisor (i.e x^{3})

So

**Step:5** – Then second term of quotient -9x multiplying the divisor g(x) and that term subtract from remainder got in step 3

-9x (x^{3 }+ x^{2 }+ x + 2) = -9x^{4 }– 9x^{3} – 9x^{2 }– 18x

**Step : 6** – Now repeat the step 4 again

i.e To get third term of quotient by dividing the first term of get remainder in previous step (i.e 7x^{3} ) with first term of divisor (i.e x^{3})

So

**Step:7** – Then third term of quotient 7 multiplying the divisor g(x) and that term subtract from remainder got in step 5

7 (x^{3 }+ x^{2 }+ x + 2) = 7x^{3 }+ 7 x^{2 }+ 7x + 14

Here the division process is completed because degree of (- 10x^{2} + 12x -13 ) = 2 which is less than the degree of the divisor (x^{3 }+ x^{2 }+ x + 2)

So the quotient is (5x^{2} -9x + 7) and remainder is -10x^{2} + 12x -13

Here we can write this division process as

Dividend = Divisor x Quotient + Remainder

f(x) = (5x^{2} -9x + 7) g(x) + ( -10x^{2} + 12x -13 )

(5x^{5 }-4x^{4} + 3x^{3} – 2x^{2} + x +1) = (5x^{2} -9x + 7) (x^{3 }+ x^{2 }+ x + 2) + ( -10x^{2} + 12x -13 )

I Hope you liked this article about** division of polynomials examples with solutions**. Give feed back, comments and please don’t forget to share it.

**Related Topics**

Polynomial Basic Concepts | Types of polynomials | Algebraic Expressions

Remainder and Factor Theorem Proof | Remainder and Factor Theorem Test

Remainder Theorem Tough Questions for Competitive Exams | Aptitude Questions