Page 142 - ICSE Math 7

Multiplication of monomials

The product of two or more monomials is a monomial whose numerical coefficient is equal to the
product of the numerical coefficients of the given monomials, and literal coefficient is the product of
the literal coefficients of the given monomials.
\ Product of monomials = Product of the numerical coefficients × Product of the literal coefficients
Example:

(a) 5x × 4y = (5 × 4) × (x × y) = 20xy
3 2
2
2 2
2 2
(b) –2a × 5a b = (–2 × 5) × (a × a b ) = (–10)(a 1+2 × b ) = –10a b
Multiplication of a polynomial and a monomial
Multiply each term of the polynomial by the given monomial and then add the products to get the
answer. Exampl:

2
2
(a) (6ab) × (3y – 2xz + z ) = (6ab × 3y) + {6ab × (–2xz)} + (6ab × z )
2
= 18aby + (–12abxz) + 6abz = 18aby – 12abxz + 6abz 2
2
2
2
2
(b) (2x + y – 5xy) × (–3xy) = {2x × (–3xy)} + {y × (–3xy)} + {(–5xy) × (–3xy)}
2 2
3
3
= –6x y – 3xy + 15x y
Multiplication of two polynomials
Multiply each term of one polynomial by each term of the other polynomial and then add the products
to get the answer. Alternatively, we can multiply polynomials by column method. For this, write one
polynomial in the first row and the other polynomial in the second row. Multiply each term of the
polynomial in the first row by each term of the polynomial in the second row. To get the result, add
the products obtained by writing the like terms in the same column. Example:
2
2
2
3
2
2
2
(a) (x + y ) × (x + y) = x (x + y) + y (x + y) = x + x y + xy + y 3
(b) (x + y) × (x – y + 3) = x(x – y + 3) + y(x – y + 3)
2
2
2
2
= x – xy + 3x + xy – y + 3y = x – y + 3x + 3y
Example 7: Find the following products.
2
3
2 2
(a) (3xy – 7x y + 5x ) × (–3xy ) (b) (4a – 5b + 3c) × (8abc)
2 2
2
3
Solution: (a) (3xy – 7x y + 5x ) × (–3xy )
2 2
3
2
2
2
= {3xy × (–3xy )} + {(–7x y ) × (–3xy )} + {5x × (–3xy )}
2 3
3 4
4 2
= –9x y + 21x y – 15x y
(b) (4a – 5b + 3c) × (8abc) = (4a × 8abc) + {(–5b) × 8abc} + (3c × 8abc)
2
2
= 32a bc – 40ab c + 24abc 2
2
2
Example 8: Evaluate (2 – x + 3x ) × (4x – 6 + 2x ).
2
2
Solution: (2 – x + 3x ) × (4x – 6 + 2x )
2
2
2
2
= {2 × (4x – 6 + 2x )} + {(–x) × (4x – 6 + 2x )} + {3x × (4x – 6 + 2x )}
2
4
2
2
3
3
= (8x – 12 + 4x ) + (–4x + 6x – 2x ) + (12x – 18x + 6x )
2
2
2
3
3
= 8x – 12 + 4x – 4x + 6x – 2x + 12x – 18x + 6x 4
2
3
4
= 6x + 10x – 18x + 14x – 12
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