Factorisation of Algebraic Expressions
Class-8-Mathematics-Chapter-6-Maharashtra Board
Notes
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Topics to be learn :
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Recall :
Factorising a binomial :
ax + ay = a(x + y)
a2 - b2 = (a + b)(a – b)
Example,
(1) 4xy + 8xy2 = 4xy(1 + 2y)
(2) p2 - 9q2 = (p)2 - (3q)2 = (p + 3q)(p - 3q)
Factors of a Quadratic Trinomial :
The general form of a quadratic trinomial is ax2 + bx + c.
(x + a)(x + b) = x2 + (a + b)x + ab
(1) To find the factors of x2 + 12x + 20, compare it with x2 + (a + b)x + ab,
We get, a + b = 12 and ab = 2 x 10 = 20.
10 + 2 = 12 and 10 x 2 = 20.
∴ x2 + 12x + 20 = x2 + (10 + 2)x + 10 × 2
= x2 + 10x + 2x + 10 × 2
= x(x + 10) + 2(x + 10)
= (x + 10)(x + 2)
Examples :
(1) Factorise : 2x2 - 9x + 9.
Solution:
First we find the product of the coefficient of the square term and the constant term. Here the product is 2 × 9 = 18.
Now, find factors of 18 whose sum is -9, that is equal to the coefficient of the middle term.
18 = (-6) × (-3) ; (-6) + (-3) = -9
Write the term -9x as -6x - 3x
2x2 - 9x + 9
= 2x2 - 6x - 3x + 9
= 2x (x - 3) - 3(x - 3)
= (x - 3)(2x - 3)
(2) Find the factors of 2y2 - 4y - 30.
Solution :
2y2 - 4y - 30
= 2(y2 - 2y - 15) ........ taking out the common factor 2
= 2(y2 - 5y + 3y - 15) ........ [-15 = -5 x 3; (-5) + (3) = -2]
= 2 [y(y - 5) + 3(y - 5)]
= 2(y - 5)(y + 3)
Factors of a3 + b3 :
We know that, (a + b)3 = a3 + 3a2b + 3ab2 + b3, which we can write as
(a + b)3 = a3 + b3 + 3ab(a + b)
Now, a3 + b3 + 3ab(a + b) = (a + b) 3 .......... interchanging the sides.
∴ a3 + b3 = (a + b) 3 - 3ab(a + b)
= [(a + b)(a + b)2] - 3ab(a + b)
= (a + b)[(a + b)2 - 3ab] = (a + b) (a2 + 2ab + b2 - 3ab)
= (a + b) (a2 - ab + b2)
∴ a3 + b3 = (a + b) (a2 - ab + b2)
Examples :
(1) x3 + 27y3
= x3 + (3y) 3
= (x + 3y) [x2 - x(3y) + (3y)2]
= (x + 3y) [x2 - 3xy +9y2]
(2) 8p3 + 125q3
= (2p) 3 + (5q) 3
= (2p + 5q) [(2p)2 - 2p × 5q + (5q)2]
= (2p + 5q) (4p2 - 10pq + 25q2)
Factors of a3 - b3 :
(a - b)3 = a3 - 3a2b + 3ab2 - b3 = a3 - b3 -3ab(a - b)
Now, a3 - b3 - 3ab (a - b) = (a - b) 3
∴ a3 - b3 = (a - b)3 + 3ab (a - b)
= [(a - b)(a - b)2 + 3ab (a - b)]
= (a - b) [(a - b)2 + 3ab]
= (a - b) (a2 - 2ab + b2 + 3ab)
= (a - b) (a2 + ab + b2)
∴ a3 - b3 = (a - b)(a2 + ab + b2)
Examples :
(1) x3 - 8y3
= x3 - (2y)3
∴ x3 - 8y3 = x3 - (2y)3
= (x - 2y) (x2 + 2xy + 4y2)
(2) Simplify : (2x + 3y)3 - (2x - 3y)3
Solution :
Using the formula a3 - b3 = (a - b) (a2 + ab + b2)
∴ (2x + 3y)3 - (2x - 3y)3
= [(2x + 3y) - (2x - 3y)] [(2x + 3y)2 + (2x + 3y) (2x - 3y) + (2x - 3y)2]
= [2x + 3y - 2x + 3y] [4x2 + 12xy + 9y2 + 4x2 - 9y2 + 4x2 - 12xy + 9y2]
= 6y(12x2 + 9y2) = 72x2y + 54y3
Rational algebraic expressions :
If A and B are two algebraic expressions, B ≠ 0, then \(\frac{A}{B}\) is called a rational algebraic expression.
To simplify a rational algebraic expression, follow the operations as performed on rational numbers.
Examples :
(1) Simplify : \(\frac{a^2+5a+6}{a^2-a-12}\)×\(\frac{a-4}{a^2-4}\)
Solution :
\(\frac{a^2+5a+6}{a^2-a-12}\)×\(\frac{a-4}{a^2-4}\)
= \(\frac{(a+3)(a+2)}{(a-4)(a+3)}\)×\(\frac{a-4}{a^2-4}\)
= \(\frac{1}{a-2}\)
(2) Simplify : \(\frac{a^2-9b^2}{a^3-27b^3}\)
Solution :
\(\frac{a^2-9b^2}{a^3-27b^3}\)
= \(\frac{(a+3b)(a-3b)}{(a-3b)(a^2+3ab+9b^2)}\)
= \(\frac{a+3b}{a^2+3ab+9b^2}\)
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