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Expansions Summary
(a + b) 2 = a 2 + 2ab + b 2 (a + b) 2 = a 2 + 2ab + b 2 (a - b) 2 = a 2 - 2ab + b 2 In the above two formulae the middle term = (a + b) (a - b) = a 2 - b 2 (x + a) (x + b) = x 2 + (a + b)x + ab. (a + b + c) 2 = a 2 + b 2 + c 2 + 2ab + 2bc + 2ac (a + b) 3 =..
(a + b) 2 = a 2 + 2ab + b 2 (a + b) 2 = a 2 + 2ab + b 2 (a - b) 2 = a 2 - 2ab + b 2 In the above two formulae the middle term = (a + b) (a - b) = a 2 - b 2 (x + a) (x + b) = x 2 + (a + b)x + ab. (a + b + c) 2 = a 2 + b 2 + c 2 + 2ab + 2bc + 2ac (a + b) 3 =..Factorization
Writing a polynomial as the product of two or more polynomials is called factorisation. If A = B x C, B and C are called factors of A. Most of the polynomials can be factorised by grouping the terms suitably and taking out the common factors. Identities studied in the previous chapter also help i..
Trinomials
Expressions of the form ax 2 + bx + c are called trinomial..
Factorization
If a polynomial can be written as the product of two or more expressions, then each expression is called the factor of the given polynomial.If a polynomial can be written as the product of two or more expressions, then each expression is called the factor of the given polynomi..
Type (ii) By expressing the polynomial as the difference of two squares
121x 2 - 25y 2 = (11x) 2 - (5y) 2 = (11x + 5y) (11x - 5y) [Using the identity a 2 -b 2 =(a-b)(a+b)] Factorise: (5a + 6b) 2 - 49b 2 Let x = 5a + 6b Then the given expression = (x) 2 - (7b) 2 = (x + 7b) (x - 7b) Re-substituting the value of x, we get = [(5a + 6b + 7b)] [(5a + 6b) - 7b]..
121x 2 - 25y 2 = (11x) 2 - (5y) 2 = (11x + 5y) (11x - 5y) [Using the identity a 2 -b 2 =(a-b)(a+b)] Factorise: (5a + 6b) 2 - 49b 2 Let x = 5a + 6b Then the given expression = (x) 2 - (7b) 2 = (x + 7b) (x - 7b) Re-substituting the value of x, we get = [(5a + 6b + 7b)] [(5a + 6b) - 7b]..Factorising Trinomials
When the coefficient of the highest power is 1. i.e., ax 2 bx c, when a = 1 and b and c are integers. When two binomials are multiplied the product is a trinomial. Thus (x + 4) (x + 5) = x 2 + 9x + 20 (1) (x - 4) (x - 5) = x 2 - 9x + 20 (2) In this chapter we try to express a trinomia..
When the coefficient of the highest power is 1. i.e., ax 2 bx c, when a = 1 and b and c are integers. When two binomials are multiplied the product is a trinomial. Thus (x + 4) (x + 5) = x 2 + 9x + 20 (1) (x - 4) (x - 5) = x 2 - 9x + 20 (2) In this chapter we try to express a trinomia..Second Method:
x 2 - 7x - 8x + 56 Write -15x as -7x and -8x = x (x - 7) - 8 (x - 7) = (x - 7) (x - 8) Now, we consider a case where the third term of the trinomial is negative. Resolve into factors: x 2 + 3x - 28 Since the third term of the trinomial is -28, find two factors of 28 which differ by 3. The gr..
x 2 - 7x - 8x + 56 Write -15x as -7x and -8x = x (x - 7) - 8 (x - 7) = (x - 7) (x - 8) Now, we consider a case where the third term of the trinomial is negative. Resolve into factors: x 2 + 3x - 28 Since the third term of the trinomial is -28, find two factors of 28 which differ by 3. The gr..Factorising a3
b3
The product of a + b and a 2 - ab + b 2 is a 3 + b 3 . Hence when a 3 + b 3 is factorised, we get: a 3 + b 3 = (a + b) (a 2 - ab + b 2 ) Similarly, a 3 - b 3 = (a - b) (a 2 + ab + b 2 ) Factorise: x 3 + 8 x 3 + 8 = (x) 3 + (2) 3 = (x + 2) (x 2 - 2x + 4) Factorise 64x 3 - 125. ..
b3
The product of a + b and a 2 - ab + b 2 is a 3 + b 3 . Hence when a 3 + b 3 is factorised, we get: a 3 + b 3 = (a + b) (a 2 - ab + b 2 ) Similarly, a 3 - b 3 = (a - b) (a 2 + ab + b 2 ) Factorise: x 3 + 8 x 3 + 8 = (x) 3 + (2) 3 = (x + 2) (x 2 - 2x + 4) Factorise 64x 3 - 125. ..Summary
In a given sentence, the variables are replaced by letters and the different operations are replaced by their mathematical symbols. We now arrive at a formula. In a given sentence, the variables are replaced by letters and the different operations are replaced by their mathematical symbols. We n..
Result
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