Expansion formulae
Class-8-Mathematics-Chapter-5-Maharashtra Board
Notes
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Topics to be learn :
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Recall :
(i) (a + b)2 = a2 + 2ab + b2,
Examples :
(a) (x + 2y)2 = x2 + 4xy + 4y2
(b) (101)2 = (100 + 1)2
= (100)2 + 2 x 100 x 1 + 12
= 10000 + 200 +1 = 10201
(ii) (a - b)2 = a2 - 2ab + b2,
Examples :
(a) (2x - 5y)2 = 4x2 - 10xy + 25y2
(b) (98)2 = (100 - 2)2
= (100)2 - 2 x 100 x 2 + (-2)2
= 10000 - 400 + 4 = 9604
(iii) (a + b) (a - b) = a2 - b2
Example :
(5m + 3n)(5m - 3n) = (5m)2 – (3n)2 = 25m2 – 9n2
Activity : Expand (x + a)(x + b) using formulae for areas of a square and a rectangle.

(x + a)(x + b) = x2 +ax + bx + ab = x2 + (a +b)x + ab
Expansion of (x + a) (x +b) :
(x + a) and (x + b) are binomials with one term x in common, a and b are unequal terms.
Let us multiply them.
(x + a) (x + b) = x(x + b) + a(x + b)
= x2 + bx + ax + ab
= x2 + (b + a)x + ab
i.e. x2 + (a + b)x + ab.
Examples :
(1) (x + 2)(x + 3) = x2 + (2 + 3)x + (2 × 3) = x2 + 5x + 6
(2) (y + 4)(y - 3) = y2 + (4 - 3)y + (4) × (-3) = y2 + y - 12
(3) (x - 3)(x - 7) = x2 + (-3 - 7)x + (-3)(-7) = x2 - 10x + 21
Expansion of (a +b)3 :
(a + b)3 = (a + b)(a + b)(a + b)
= (a + b)(a + b)2
= (a + b)(a2 + 2ab + b2)
= a(a2 + 2ah + b2) + b(a2 + 2ab + b2)
= a3 + 2a2b + ab2 + a2b + 2ab2 + b3
= a3 + 3a2b + 3ab2 + b3.
(a + b) 3 = a3 + 3a2b + 3ab2 + b3
Remember :
- The first and the fourth terms in the expansion are the cubes of the first and the second terms.
- The second term is three times the product of the first term squared and the second term.
- The third term is three times the product of the first term and the square of the second term.
Examples :
(1) (x + 3)3
We know that (a + b)3 = a3 + 3a2b + 3ab2 + b3
In the given example, a = x and b = 3
∴ (x + 3)3 = (x) 3 + 3 × x2 × 3 + 3 × x × (3)2 + (3)3 = x3 + 9x2 + 27x +27
(2) (3x + 4y)3 = (3x) 3 + 3(3x)2(4y) + 3(3x)(4y)2 + (4y) 3
= 27x3 + 3 x 9 x2 x 4y + 3 × 3x × 16y2 + 64y3
= 27x3+ 108x2y + 144xy2 + 64y3
Expansion of (a - b)3 :
∴ (a - b)3 = (a - b) (a - b) (a - b) = (a - b)(a - b)2
= (a - b)(a2 - 2ab + b2)
= a(a2 -2ab + b2) - b(a2 - 2ab + b2)
= a3 - 2a2b + ab2 - a2b + 2ab2 - b3
= a3 - 3a2b + 3ab2 - b3
∴ (a - b) 3 = a3 - 3a2b + 3ab2 - b3
In words :
(first term - second term)3 =(first term)3 -3 x (first term)2 x (second term) +3 x (first term) x (second term)2 - (second term)3.
Examples :
(1) Expand (x - 2)3
(a - b) 3 = a3 - 3a2b + 3ab2 - b3 Here taking , a = x and b = 2,
(x - 2)3 = (x) 3 - 3 × x2 × 2 + 3 × x × (2)2 - (2) 3
= x3 - 6x2 + 12x - 8
(2) Find cube of 99 using the expansion formula.
(99)3 = (100 - 1)3 = (100)3 - 3 × (100)2 × 1 + 3 × 100 × (1)2 - 13
= 1000000 - 30000 + 300 - 1 = 9,70,299
Expansion of (a + b + c)2 :
(a + b + c)2 = (a + b + c) × (a + b + c)
= a (a + b + c) + b (a + b + c) + c (a + b + c)
= a2 + ab + ac + ab + b2 + bc + ac + bc + c2
= a2 + b2 + c2 + 2ab + 2bc + 2ac
∴ (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ac
Examples :
(1) Expand: (p + q + 3)2
= p2 + q2 + (3) 2 + 2 × p × q + 2 × q × 3 + 2 × p × 3
= p2 + q2 + 9 + 2pq + 6q + 6p = p2 + q2 + 2pq + 6q + 6p + 9
(2) Fill in the boxes with appropriate terms in the steps of expansion.
(2p + 3m + 4n)2
= (2p)2 + (3m)2 + (4n)2 + 2 × 2p × 3m + 2 × 3m × 4n + 2 × 2p × 4n
= 4p2 + 9m2 + 16n2 + 12pm + 24mn + 16pn
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