(2x + 3y)(3x – 4y) 4(x - 2)(x + 3) (2y - 1)(2y + 1) (3 + 5x)(4 – x) Two Special Cases Perfect Square: Difference of two squares: (x + a)2 = (x + a)(x + a) = x2 + 2ax + a2 (x - a)(x + a) = x2 – a2 (2x - 3)2 = (2x – 3)(2x – 3) = 4x2 – 12x + 9 (x - 3)(x + 3) = x2 – 32 = x2 – 9 EXERCISE B Multiply out 1. (x - 1)2 2. (3x + 5)2 3. (7x - 2)2 4. (x + 2)(x - 2) 5. (3x + 1)(3x - 1)
Factorise each of the following 1) 3x + xy 2) 4x2 – 2xy 3) pq2 – p2q 4) 3pq - 9q2 5) 2x3 – 6x2 6) 8a5b2 – 12a3b4 7) 5y(y – 1) + 3(y – 1) Factorising quadratics Simple quadratics: Factorising quadratics of the form The method is: Step 1: Form two brackets (x … )(x … ) Step 2: Find two numbers that multiply to give c and add to make b. These two numbers get written at the other end of the brackets. Example 1: Factorise x2 – 9x – 10. Solution: We need to find two numbers that multiply to make -10 and add to make -9. These numbers are -10 and 1. Therefore x2 – 9x – 10 = (x – 10)(x + 1). General quadratics: Factorising quadratics of the form The method is: Step 1: Find two numbers that multiply together to make ac and add to make b. Step 2: up the bx term using the numbers found in step 1. Step 3: Factorise the front and back pair of expressions as fully as possible. Step 4: There should be a common bracket. Take this out as a common factor. Example 2: Factorise 6x2 + x – 12. Solution: We need to find two numbers that multiply to make 6 × -12 = -72 and add to make 1. These two numbers are -8 and 9. Therefore, 6x2 + x – 12 = 6x2 - 8x + 9x – 12 = 2x(3x – 4) + 3(3x – 4) (the two brackets must be identical) = (3x – 4)(2x + 3) Difference of two squares: Factorising quadratics of the form Remember that = (x + a)(x – a). Therefore:
8) 9) 10) 11) 12) 13) 14) Chapter 5: CHANGING THE SUBJECT OF A FORMULA We can use algebra to change the subject of a formula. Rearranging a formula is similar to solving an equation – we must do the same to both sides in order to keep the equation balanced. Example 1: Make x the subject of the formula y = 4x + 3. Solution: y = 4x + 3 Subtract 3 from both sides: y – 3 = 4x Divide both sides by 4; So is the same equation but with x the subject. Example 2: Make x the subject of y = 2 – 5x Solution: Notice that in this formula the x term is negative.
Exercise C Make x the subject of these formulae: 1) 2) 3) 4) Chapter 6: SOLVING QUADRATIC EQUATIONS A quadratic equation has the form . There are two methods that are commonly used for solving quadratic equations: * factorising * the quadratic formula Note that not all quadratic equations can be solved by factorising. The quadratic formula can always be used however. Method 1: Factorising Make sure that the equation is rearranged so that the right hand side is 0. It usually makes it easier if the coefficient of x2 is positive. Example 1 : Solve x2 –3x + 2 = 0 Factorise (x –1)(x – 2) = 0 Either (x – 1) = 0 or (x – 2) = 0 So the solutions are x = 1 or x = 2 Note: The individual values x = 1 and x = 2 are called the roots of the equation. Example 2: Solve x2 – 2x = 0 Factorise: x(x – 2) = 0 Either x = 0 or (x – 2) = 0
So x = 0 or x = 2 Method 2: Using the formula Recall that the roots of the quadratic equation are given by the formula: Example 3: Solve the equation Solution: First we rearrange so that the right hand side is 0. We get We can then tell that a = 2, b = 3 and c = -12. Substituting these into the quadratic formula gives: (this is the surd form for the solutions) If we have a calculator, we can evaluate these roots to get: x = 1.81 or x = -3.31 If you need more help with the work in this chapter, there is an information booklet downloadable from this web site: EXERCISE 1) Use factorisation to solve the following equations: a) x2 + 3x + 2 = 0 b) x2 – 3x – 4 = 0 c) x2 = 15 – 2x 2) Find the roots of the following equations: a) x2 + 3x = 0 b) x2 – 4x = 0
c) 4 – x2 = 0 3) Solve the following equations either by factorising or by using the formula: a) 6x2 - 5x – 4 = 0 b) 8x2 – 24x + 10 = 0 4) Use the formula to solve the following equations to 3 significant figures. Some of the equations can’t be solved. a) x2 +7x +9 = 0 b) 6 + 3x = 8x2 c) 4x2 – x – 7 = 0 d) x2 – 3x + 18 = 0 e) 3x2 + 4x + 4 = 0 f) 3x2 = 13x – 16 Chapter 7: INDICES Basic rules of indices . 4 is called the index (plural: indices), power or exponent of y.