Summary
If all the terms of the polynomial have a common factor, we take out the common factor and factorise . If all the terms of the polynomial have a common factor, we take out the common factor and factorise . If the polynomial can be expressed as the difference of two squares, we use a 2 - b 2..
If all the terms of the polynomial have a common factor, we take out the common factor and factorise . If all the terms of the polynomial have a common factor, we take out the common factor and factorise . If the polynomial can be expressed as the difference of two squares, we use a 2 - b 2..Methods of Factorisation
(i) Common factors (ii) By expressing as difference of squares (iii) By grouping (iv) Trinomials (v) Sum or difference of cub..
Second Method:
x 2 + 8x + 15 = x 2 + 5x + 3x + 15 (after noticing that 5 + 3 = 8 and 5 3 = 15) = x(x +5) + 3(x + 5) = (x + 5) (x + 3) Resolve into factors: x 2 - 15x + 56 x 2 - 15x + 56 Take factors of 56 having their sum = -15 They are -8, -7. \ x 2 - 15x + 56 = x 2 - 7x - 8x + 56 \ x 2 - 15x +..
x 2 + 8x + 15 = x 2 + 5x + 3x + 15 (after noticing that 5 + 3 = 8 and 5 3 = 15) = x(x +5) + 3(x + 5) = (x + 5) (x + 3) Resolve into factors: x 2 - 15x + 56 x 2 - 15x + 56 Take factors of 56 having their sum = -15 They are -8, -7. \ x 2 - 15x + 56 = x 2 - 7x - 8x + 56 \ x 2 - 15x +..Framing of Formulae
We use alphabets like x, y, z etc., to denote variables. For example the length of a rectangle is denoted by 'l'. It takes different values in different rectangles. A formula is a relation between different variables formed using mathematical symbols. Any given condition can be translated into a ..
(i) An arrow diagram
R = {(1, 1), (2, 3), (4, 6)}..
R = {(1, 1), (2, 3), (4, 6)}..Domain
The set of all the first elements of the ordered pairs of a function is called the domai..
Types of Functions (Mapping)
The following are the Types of Functions or Mapping: (1) One-one function (2) Many-one function (3) Onto function (4) Into functio..
1. One-one function
There is one-one correspondence between the elements of the set A and the set B..
There is one-one correspondence between the elements of the set A and the set B..Suggested answer:
Right hand limit at x = 1 = h + (-h+1) = 1 Left hand limit at x = 1 =1 f(1) = |1 - 1| + |1 - 2| = 1 The function is continuous at x =..
Right hand limit at x = 1 = h + (-h+1) = 1 Left hand limit at x = 1 =1 f(1) = |1 - 1| + |1 - 2| = 1 The function is continuous at x =.. Result
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