WBBSE Solutions For Class 8 Maths Algebra Chapter 3 Multiplication Of Polynomials

WBBSE Solutions For Class 8 Maths Algebra Chapter 3 Multiplication Of Polynomials

Multiplication of Polynomials Introduction

You know that when two or more expressions are related to each other by the sign of addition or subtraction then a polynomial is formed and each expression of that polynomial is called a term. In this chapter, our aim is to multiply two polynomials each with more than two terms. So, in fact, this chapter is nothing but an extension of the same topic which we studied in class VII.

Rules for multiplication

Here we shall discuss the rules for multiplication in brief. The following four rules are common in any type of algebraic multiplication.

1.Commutative law : a x b = b x a.

2.Associative law : a x (b x c) = (a x b) x c.

3. Distributive law : 

1. a x (b + c) = a x b + a x c.

2. (a + b) x c = a x c + b x c.

4. Index law :

1. x^m x xⁿ = x^m+n

2. (X^m)ⁿ = x^mn

3. xº = 1.

When we multiply two polynomials we generally arrange the terms of the multiplicand and those of the multiplier in the ascending or descending powers of a variable (which is expressed by a letter of the English alphabet.)

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Algebra Chapter 3 Multiplication Of Polynomials Some examples of multiplication

Example 1

Multiply : a² – ab + b² by b²+ ab + a².

Solution:

Given:-

a² – ab + b² And b²+ ab + a².

When arranged in the descending powers of a multiplicand

= a² – ab + b² and multiplier = a² + ab + b².

Now, a² – ab + b²

The required product = a⁴+a²b²+b⁴

a² – ab + b² X b²+ ab + a² = a⁴+a²b²+b⁴

Example 2

Multiply ax + by – cz by ax – by + cz.

Solution :

Given:-

ax + by – cz And ax – by + cz.

The required product = a²x² – b²y² – c²z² + 2bcyz

ax + by – cz X ax – by + cz. = a²x² – b²y² – c²z² + 2bcyz

Example 3

Multiply : 4a – 2a² + 1 + 3a³ by 2 – 2a² + a.

Solution:

Given

4a – 2a² + 1 + 3a³ And 2 – 2a² + a.

Arranging in the descending powers of a,

we get,

multiplicand = 3a³ – 2a² + 4a + 1,

multiplier = – 2a² + a + 2.

Now, 3a³ – 2a² + 4a + 1

Example 4

Multiply: p² + g² – pq + p + q + 1 by p + q -1.

Solution :

Given

p² + q² – pq + p + q + 1 And p + q -1.

The required product = p³ + q³ -1 + 3pq

p² + g² – pq + p + q + 1 And p + q -1.= p³ + q³ -1 + 3pq

Example 5

Multiply x² + y² + z² – xy – yz – zx by  +y+ z.

Solution :

Given

x² + y² + z² – xy – yz – zx And  +y+ z.

The required product = x³ +y³ +z³ – 3xyz

x² + y² + z² – xy – yz – zx by  +y+ z. = x³ +y³ +z³ – 3xyz

Example 6

Multiply : x² + 4x + 8 by x² – 4x + 8.

Solution:

Given:-

x² + 4x + 8 And x² – 4x + 8.

The required product = x⁴ + 64

x² + 4x + 8 X x² – 4x + 8. = x⁴ + 64

Example 7

Multiply: 2x² – 3x – 1 by 3x² – x – 1.

Solution :

Given

2x² – 3x – 1 And 3x² – x – 1.

Example 8

Multiply: \(x^3-3 x^2 y^{-\frac{1}{3}}+3 x y^{\frac{2}{3}}-y^{-1} \text { by } x-y^{-\frac{1}{3}}\)

Solution:

Given

\(x^3-3 x^2 y^{-\frac{1}{3}}+3 x y^{\frac{2}{3}}-y^{-1} \text { And } x-y^{-\frac{1}{3}}\)

The required product = \(x^4-4 x^3 y^{-\frac{1}{3}}+6 x^2 y^{-\frac{2}{3}}-4 x y^{-1}+y^{-\frac{4}{3}}\)

Example 9

Multiply : a²x² – 2ax + 2 by a²x² + 2ax + 2.

Solution:

Given

a²x² – 2ax + 2 by a²x² + 2ax + 2.

The required product = a⁴x⁴ + 4

a²x² – 2ax + 2 X a²x² + 2ax + 2. = a⁴x⁴ + 4

Example 10

Multiply : 3x³ – 2x² + 2 by 2x² – x + 1.

Solution:

Given

3x³ – 2x² + 2 And 2x² – x + 1.

The required product = 6x⁵– 7x⁴ + 5x³ + 2x² – 2x + 2

3x³ – 2x² + 2 X 2x² – x + 1 = 6x⁵– 7x⁴ + 5x³ + 2x² – 2x + 2

Multiplication of more than two polynomials

In order to multiply more than two polynomials first of all we multiply any two polynomials by the method already discussed. After that, we take this product as the multiplicand and the third polynomial as the multiplier. We then find the new product. In the next stage, we multiply this product by the fourth polynomial (if any). Proceeding in this way we complete the process of multiplication.

Example 1

Find the continued product of : (a² + ab + b²), (a – b), and (a³ + b³).

Solution :

Given

(a² + ab + b²), (a – b), And (a³ + b³).

Example 2

Multiply : 2x³ – 4X² – 5 by 3X² + 4x – 2.

Solution:

Given:-

2x³ – 4X² – 5 And 3X² + 4x – 2.

Let us first detach the coefficients of the multiplicand and multiplier.

Since the term containing x is absent in the multiplicand we take the co-efficient of x as 0.