Multiplication

Algebraic Expression of Class 7

One number is said to be multiplied by another when we do to the former what is done to unity to obtain the latter.

4 = 1 + 1 + 1 + 1 or 4x = x + x + x + x

The number multiplied is called the multiplicand (28xy) and the number by which it is multiplied is called the multiplier; (4) the result is called the product (112xy)

e.g.: 28xy × 4 = 112xy

The law of signs

(+ a) × (+ b) = + ab                    (– a) × (– b) = + ab

(+ a) × (– b) = – ab                    (– a) × (+ b) = – ab

Laws of multiplication

• a × b = b × a
• (a × b) × c = a × (b × c)
• a × 1 = a
• a (b + c) = ab + ac
• (a × a × ……… m times) × (a × a………. n times) = am × an = am + n

Product of monomials = Product of their numericals coefficients × product of their literal coefficients

e.g.: 2x3y2 × 4x5y6 = 2 × 4 × x3 + 5 y2 + 6 = 8x8y8

Multplication of a polynomial by a monomial

We multiply each term of the polynomial by the monomial

Multiplication of two binomials (horizontal method)

Suppose (a + b), (c + d) are two binomials

By using distributive law of multiplication over addition twice we may find their product as shown below (a + b) × ( c + d) = a × (c + d) + b × (c + d) = ac + ad + bc + bd

This is known as horizontal method

Multiply each term of the first binomial with each term of the second and add the like terms in the product.

Multiplication of binomials using column method:

We can write two binomials one below the other and multiply. Multiply each term of first binomial by each term of the second and then add like terms

Multiplication of a binomial and a trinomial (a + b) (x + y + z)

Multiply each term of the binomial with every term of the trinomial and add all the products so obtained.

Basic Concept: When an expression ‘f’ is divided by the expression ‘g’, the quotient is defined to be that when multiplied by ‘g’ produces ‘f’. This operation of division is denoted by fg, or f/g. The statement may be also be expressed verbally as follows,

quotient × divisor = dividend

Questions:

question 1.Multiply: (3x – 2) (2×2+ 3x – 5)

question 2.(a × b) × c = ……………

question 3.Find the value of(6m – 4n) (4m – 6n) at m = 1, n = -1

question 4.If x + y = 12 and xy = 14, find the value of x2 + y2.

question 5.Prove that : 2a2 + 2b2 + 2c2 – 2ab – 2bc – 2ca = (a – b)2 + (b – c)2 + (c – a)2

Solution:1.6×3+ 5×2– 21x + 10 2. a × (b × c)

3.100 4.116

Solution:2.Multiply: (3x – 2) (2×2+ 3x – 5)

3x (2×2 + 3x – 5) – 2 (2×2 + 3x – 5) = 6×3 + 9×2 – 15x – 4×2 – 6x + 10

= 6×3 + 5×2 – 21x + 10

Solution:3.Find the value of(6m – 4n) (4m – 6n) at m = 1, n = -1

We have,

(6m – 4n) (4m – 6n)

= 6m (4m – 6n) – 4n (4m – 6n)

= 6m 4m – 6m 6n – 4n 4m + 4n 6n

= 24m2 – 36mn – 16mn + 6n2

= 24m2 – 52mn + 24n2

When m = 1, n = -1, we get

(6m – 4n) (4m – 6n)

= 24m2 – 52mn + 24n2

= 24 (1)2 – 52 1 (-1) + 24 (-1)2 = 24 + 52 + 24 = 100

Solution:4.If x + y = 12 and xy = 14, find the value of x2 + y2.

We have,

(x + y)2 = x2 + y2 + 2xy

Putting the values of x + y and xy, we obtain

122 = x2 + y2 + 2 14

144 = x2 + y2 + 28

144 – 28 = x2 + y2

x2 + y2 = 116

Solution:5.Prove that :

2a2 + 2b2 + 2c2 – 2ab – 2bc – 2ca = (a – b)2 + (b – c)2 + (c – a)2

We have,

LHS= 2a2 + 2b2 + 2c2 – 2ab – 2bc – 2ca

= (a2 – 2ab + b2) + (b2 – 2bc + c2) + (c2 – 2ca + a2)

[Re-arranging the terms]

= (a – b)2 + (b – c)2 + (c – a)2

= R.H.S.

Hence, 2a2 + 2b2 + 2c2 – 2ab – 2bc – 2ca = (a – b)2 + (b – c)2 + (c – a)2