WBBSE Solutions For Class 8 Maths Algebra Chapter 4 Division Of Polynomials

WBBSE Solutions For Class 8 Maths Algebra Chapter 4 Division Of Polynomials

Division of Polynomials Introduction

In class 7 you have studied the method of division of a polynomial by a single term divisor. In this chapter, we shall extend our idea on this topic. Here we shall discuss the method of division of a polynomial by a divisor having more than one term.

Rules for division

In order to divide one polynomial by another, first of all, we arrange both dividend and divisor in the ascending or descending power of the alphabetic symbol. Here we should remember only the index law, x^m ÷ xⁿ = x^m-n

At first, the first term of the dividend is divided by the first term of the divisor. This is taken as the first term of the quotient. All the terms of the divisor are now multiplied by this term and each term of the product is placed under the corresponding term of the dividend. Subtracting this product from the dividend we get the second step of the dividend. Now dividing the first term of this new dividend by the first term of the divisor, we get the second term of the quotient. By this term, we again multiply the whole divisor and we subtract this product from the second step of the dividend. Repeating this process as far as possible we get the final quotient.

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Algebra Chapter 4 Division Of Polynomials Some examples of division

Example 1

Divide 6x⁴ – 11 x³ – 2X² + 4x + 1 by 2x² -3x -1.

Solution : 

Given:-

6x⁴ – 11 x³ – 2X² + 4x + 1 And 2x² -3x -1.

The required quotient = 3x² – x – 1

Example 2

Divide – 6x⁵ + 7x⁴ – 4x³ – 2x² + 9x + 2 by – 2x² + x + 2.

Solution:

Given:-

– 6x⁵ + 7x⁴ – 4x³ – 2x² + 9x + 2 And – 2x² + x + 2.

Example 3

Divide x⁸ + x⁴y⁴ + y⁸ by x⁴ + x²y² + y⁴.

Solution :

Given:-

x⁸ + x⁴y⁴ + y⁸ And x⁴ + x²y² + y⁴.

Example 4

Divide x³ + y³ + z³ – 3xyz by x + y + z.

Solution :

Given:-

x³ + y³ + z³ – 3xyz And x + y + z.

Example 5 

Divide a⁴+ a2b2 + b⁴ by a² -ab + b²

Solution:

Given:-

a⁴+ a2b2 + b⁴ And a² -ab + b²

Example 6

Divide a⁴x⁴+ 4 by a²x² – 2ax + 2.

Solution:

Given

a⁴x⁴+ 4 And a²x² – 2ax + 2.

Example 7

Divide a³ + 8a² + 11a -6 by a² + 2a -1.

Solution:

Given

a³ + 8a² + 11a -6 And a² + 2a -1.

 

Example 8

Divide x⁶+ x⁵+ x by x² + x +1.

Solution:

Given

x⁶+ x⁵+ x And x² + x +1.

Inexact division

In arithmetic when we divide a number by another number there may or may not be any remainder. For example when we divide 40 by 5 then there is no remainder i.e., 40 is exactly divisible by 5. But when we divide 38 by 5 then the quotient is 7 and the remainder is 3. Likewise, in algebra, when we divide one expression by another there may or may not be any remainder. Thus, when we divide an algebraic expression by another and there is a remainder then such division is called an inexact division. If q is the quotient and r is the remainder when b is divided by a,

then the complete quotient = q + r/a.

WBBSE Board Class 8 Math Solution

Example 1

Divide 2x⁶– 3x⁶+ 7x³ -16x +15 by x⁴– 2x² +4.

Solution:

Given:-

2x⁶– 3x⁶+ 7x³ -16x +15 And x⁴– 2x² +4.

Example 2

Divide 12a⁴+ 5a³ -33a² -3a + 16 by 4a² – a – 5.

Solution:

Given:

12a⁴+ 5a³ -33a² -3a + 16 And 4a² – a – 5.

Division by the method of ‘detached coefficients

There is another method of division of two polynomials. In this method, we only write successive coefficients of the dividend and divisor and then divide as before. After obtaining the quotient and remainder in terms of coefficients the alphabetic symbol is Written with successive power. If any term containing a particular power is absent either in the dividend or in the divisor, then the corresponding coefficient is taken as zero.

Example 1

Divide 6x⁴ – 7x³ + 10x² + 8x – 5 by 3x² + x – 1.

Solution :

Given 6x⁴ – 7x³ + 10x² + 8x – 5 And 3x² + x – 1.

Let us first detach the coefficients of the dividend and divisor.

Obviously, the highest power of x in the quotient is x². Hence, the required quotient is 2x² – 3x + 5.

6x⁴ – 7x³ + 10x² + 8x – 5 / 3x² + x – 1 = 2x² – 3x + 5.

Example 2

Divide 12a⁴ + 5a³ – 33a² – 3a + 16 by 4a² – a – 5.

Solution:

Given:

12a⁴ + 5a³ – 33a² – 3a + 16 And 4a² – a – 5.

Let us first detach the coefficients of the dividend and divisor.

Quotient = 3a² + 2a – 4, Remainder = 3a – 4.