Page 101 - Start Up Mathematics_8 (Non CCE)

2
2
(b) (p + 2p – 3) ¥ (q – pq – 4)
2
2
2
2
= p (q – pq – 4) + 2p(q – pq – 4) – 3(q – pq – 4)
2
2
2
2
2
= (p ¥ q ) – (p ¥ pq) – (p ¥ 4) + (2p ¥ q ) – (2p ¥ pq) – (2p ¥ 4)
2
– {(3 ¥ q ) – (3 ¥ pq) – (3 ¥ 4)}
2
2 2
2
2
2
= p q – (p ¥ p)q – 4p + 2pq – 2 ¥ (p ¥ p)q – (2 ¥ 4)p – {3q – 3pq – 12}
2
2
2 2
3
2
2
= p q – p q – 4p + 2pq – 2p q – 8p – 3q + 3pq + 12
This product can also be obtained by using the column method.
EXERCISE 5.5
1. Multiply the following binomials:
2
2
2
2
2
2
2
2
4
4
(a) (2a – 5b ) and (a + 3b ) (b) (x – y ) and (x + y ) (c) (a – b ) and (a – b)
2
2
2 2
2
2
2
2
2
(d) (3x + 2xy ) and (8x – 4y ) (e) (2 + 6yz) and (x y + 2yz) (f) (x – y ) and (3x + 5y )
ˆ
2
2
(g) (9a b – 6ab) and (3ab + 2a b) (h) Ê Á Ë 4 xy - 3yz and Ê Á Ë 2 xy - xz ˆ ˜ ¯
2
2
˜
¯
9
9
2. Using the column method evaluate the following products:
3
2
3
2
2
2
2
2
(a) (2.2x – 3y ) ¥ (1.6x + 0.4y ) (b) (2pq – 3p + 5) ¥ (p – q ) (c) (2x – 3x + 4) ¥ (x – x + 5x)
3. Simplify the following:
2
(a) 9a + 7a(3a – 2b) + 5ab (b) (2x + y)(4x + 3y) – 6x(x + y)
2
3
2
2
2
2
2
(c) p q(p – q ) + pq (4pq – 2p ) – p q(1 – 2q) (d) (2l + 3)(l – 5l) – (l – 2)(3l + 4l)
2
2
2
2
2
(e) (a + 2)(a – 2a + 4) – (a – 2)(a + 2a + 4) (f) (x – 2x + 3)(3x + 5x – 7) – (3x – 2x + 5)(2x + 5)
4. Find the following products and also find its value for p = 1, q = –1, r = 2.
2
2
2 2
2
2
2
(a) (p – 8p + 4) ¥ (5p + 2) (b) 1 (4q + 5r ) ¥ (4q – 5r ) (c) 12p q ¥ (3p – 2r) ¥ (2q – p)
2
5. From the product of (2l + 3m) and (3l + 4m) subtract the product of (7l + 3m) and (l + 2m).
3
2
6
2
4
2
3
6. Add the product of 9x (x – 2), 5x (2x + 3) and (x – 3x )(4x – 3x ).
Algebraic Identities
An algebraic identity is a statement of equality between two algebraic expressions that is satisfied for all the
values of the variables.
2
For example, (x – 3)(x + 2) ∫ x – x – 6 is satisfied for all values of x. The sign ‘∫’ is used to distinguish an
identity from an equation.
Standard Identities
2
2
I. (p + q) = p + 2pq + q 2
In other words,
2
2
(Sum of the two terms) = (First term) + 2 ¥ (First term) ¥ (Second term) + (Second term) 2
2
Proof: (p + q) = (p + q)(p + q)
= p(p + q) + q(p + q) (Distributive property of multiplication over addition)
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