Page 99 - Start Up Mathematics_8 (Non CCE)

3. Multiply the following and verify the result for x = 2, y = 1 and z = 3.

3 2 2 2 2 2
(a) y(10y x – 100yx ) (b) 6yz ¥ (x y + 2yz) (c) 6xy ¥ (3z – 4yz)
2
2
3
2
2
2
2
(d) 9x (2x – 5x ) (e) 2.5x y(1.5z + 0.5y )
4. Add the following products:
ˆ
2
2
2
(a) (3x y) ¥ (2x + 7y), 7x y ¥ (5y – 3x) (b) Ê Á Ë 3 xy (2x – 3yz), 5 yz x( - 2yz)
˜
¯
2
2
Ê 3 ˆ - 1 Ê 3 ˆ 6 4 2 4 3 4
(c) 2x Á Ë 4 x - 10y , 2 y Á 2 x + 8 (d) 5x (x – 3x ), 8x (3x – 4x )
˜
˜
¯
Ë
¯
5. Subtract:
2
2
2
3
2
(a) 2a b(ab + b ) from (3b) ¥ (ab – a b) (b) 5xy(x + y – 5) from x(6x – 7y + 5) + xy(x + y)
2
2
(c) (6x ) ¥ (3x + 3) from (2x) ¥ (4x – 1) (d) 6m(m – 13) + 5m(3 – m) from 9m(m – 12)
Multiplication of a binomial by a binomial
Multiplication of a binomial by another binomial involves the use of distributive property of multiplication of
literals over their addition. Thus, if (x + y) and (z + r) are two binomials, then
(x + y) ¥ (z + r) = x ¥ (z + r) + y ¥ (z + r)
= (x ¥ z + x ¥ r) + (y ¥ z + y ¥ r) = xz + xr + y + yr
Multiplication of two binomials can also be done by column method, as explained in the examples.
Ê 3 2 ˆ Ê 2 2 ˆ
2
Example 19: Multiply: (a) (2.5l – 0.5m) and (2.5l + 0.5m) (b) Á Ë 4 a + 3b ˜ ¯ and 4 a - 3 b ˜ (NCERT)
2
Á
Ë
¯
Solution: (a) (2.5l – 0.5m) ¥ (2.5l + 0.5m)
= (2.5l) ¥ (2.5l + 0.5m) – (0.5m) ¥ (2.5l + 0.5m)
= (2.5l ¥ 2.5l) + (2.5l ¥ 0.5m) – {(0.5m ¥ 2.5l) + (0.5m ¥ 0.5m)}
2
2
= 6.25l + 1.25lm – (1.25lm + 0.25m )
2
= 6.25l + (1.25lm – 1.25lm) – 0.25m 2
2
2
2
= 6.25l + 0 – 0.25m = 6.25l – 0.25m 2
Ê 3 2 ˆ Ê 2 2 ˆ Ê 3 2 ˆ Ê 8 2 ˆ
(b) Á 4 a + 3b ˜ ¯ ¥ 4 a - 3 b ˜ ¯ = Á 4 a + 3b ˜ ¯ ¥ Á Ë 4a - 3 b ˜ ˜
2
2
2
2
Á
¯
Ë
Ë
Ë
Ê 3 2 ˆ Ê 8 2 ˆ Ê 8 2 ˆ
2
2
2
= Á 4 a ˜ ¯ ¥ Á Ë 4a - 3 b ˜ + ( 3b ) ¥ Á 4a - 3 b ˜ ¯
¯
Ë
Ë
Ê 3 2 ˆ Ê 3 8 2 ˆ Ê 8 2 ˆ
2
2
2
2
2
-
= Á 4 a ¥ 4a ˜ Á 4 a ¥ 3 b ˜ ¯ + ( 3b ¥ 4a ) - Á Ë 3b ¥ 3 b ˜
¯
Ë
¯ Ë
2 2
2 2
= 3a 2 + 2 – 2a b + 12a b – 8b 2 + 2
2 2
2 2
4
= 3a + (–2a b + 12a b ) – 8b 4
2 2
4
= 3a + 10a b – 8b 4
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