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) = a^{m} × a^{n} = a^{m + n}
Product of monomials = Product of their numericals coefficients × product of their literal coefficients
e.g.: 2x^{3}y^{2} × 4x^{5}y^{6} = 2 × 4 × x^{3} ^{+ 5} y^{2} ^{+ 6} = 8x^{8}y^{8}
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 x^{2} + y^{2}.
question 5.Prove that : 2a^{2} + 2b^{2} + 2c^{2} – 2ab – 2bc – 2ca = (a – b)^{2} + (b – c)^{2} + (c – a)^{2}
Answers:
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
= 24m^{2} – 36mn – 16mn + 6n^{2}
= 24m^{2} – 52mn + 24n^{2}
When m = 1, n = -1, we get
(6m – 4n) (4m – 6n)
= 24m^{2} – 52mn + 24n^{2}
= 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 x^{2} + y^{2}.
We have,
(x + y)^{2} = x^{2} + y^{2} + 2xy
Putting the values of x + y and xy, we obtain
12^{2} = x^{2} + y^{2} + 2 14
144 = x^{2} + y^{2} + 28
144 – 28 = x^{2} + y^{2}
x^{2} + y^{2} = 116
Solution:5.Prove that :
2a^{2} + 2b^{2} + 2c^{2} – 2ab – 2bc – 2ca = (a – b)^{2} + (b – c)^{2} + (c – a)^{2}
We have,
LHS= 2a^{2} + 2b^{2} + 2c^{2} – 2ab – 2bc – 2ca
= (a^{2} – 2ab + b^{2}) + (b^{2} – 2bc + c^{2}) + (c^{2} – 2ca + a^{2})
[Re-arranging the terms]
= (a – b)^{2} + (b – c)^{2} + (c – a)^{2}
= R.H.S.
Hence, 2a^{2} + 2b^{2} + 2c^{2} – 2ab – 2bc – 2ca = (a – b)^{2} + (b – c)^{2} + (c – a)^{2}