Further factorising, simplifying, completing the square and algebraic proof

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Further factorising, simplifying, completing the square and algebraic proof 8 CHAPTER 8. Further factorising Quadratic epressions of the form b c were factorised in Section 8.

1 Further factorising, simplifying, completing the square and algebraic proof 8 CHAPTER 8. Further factorising Quadratic epressions of the form b c were factorised in Section 8. by finding two numbers whose product is c and whose sum is b. For eample factorising 7 gives ( 4)( 3) since the two numbers which have a product of and a sum of 7 are 4 and 3 This method is now etended to factorise more general quadratic epressions of the form a b c where a, b and c are numbers and a a is the coefficient of, b is the coefficient of and c is the constant term. So for the quadratic epression 6 7 the coefficient of is 6, the coefficient of is 7 and the constant term is To factorise 6 7 : multiply the coefficient of (6) by the constant term () which gives find two numbers whose product is and whose sum is the coefficient of (7). The two numbers are 4 and 3 split the term using these two numbers then factorise by grouping So (3 ) (3 ) (3 )( ) So 6 7 when factorised gives (3 )( ) The order of 4 and 3 can be reversed ( ) ( ) ( )(3 ) To check this answer epand (3 )( ) as in Section 8.3 to get 6 7 Eample Factorise Solution Multiply the coefficient of by the constant term. Find two numbers whose product is 4 and whose sum is. 8 3 ( 4) 3( 4) ( 4)( 3) ( 4)( 3) Write as 8 3 or as 3 8 Factorise by grouping. The bracketed term must be the same. ( 4) is a common factor. This answer could also be written as ( 3)( 4) 456

2 8. Further factorising CHAPTER 8 Eample Factorise Solution Multiply the coefficient of by the constant term. Find two numbers whose product is 8 and whose sum is ( 3) ( 3) ( 3)(3 ) ( 3)(3 ) Write 7 as 9 Factorise by grouping. The bracketed term must be the same. ( 3) is a common factor. This answer could also be written as (3 )( 3) Epressions of the form p qy ry where p, q and r are numbers can be factorised in a similar way. Eample 3 Factorise 5y 3y Solution 3 5y 3y 5y 3y 5y 3y Multiply the coefficient of by the coefficient of y Find two numbers whose product is 6 and whose sum is 5 5y 3y y 3y 3y ( y) 3y( y) 5y 3y ( y)( 3y) Write 5y as y 3y. Factorise by grouping. Note that 3y is the same as 3y. The bracketed term must be the same. ( y) is a common factor. Check by epanding the brackets ( y)( 3y) 3y y 3y 5y 3y 457

3 CHAPTER 8 Further factorising, simplifying, completing the square and algebraic proof Eample 4 Factorise completely Solution ( ) Consider Take out the common factor 3 Factorise Multiply the coefficient of by the constant term. Find two numbers whose product is and whose sum is ( ) ( ) ( )( ) Write as. Factorise by grouping. The bracketed term must be the same. ( ) is a common factor ( ) 3[( )( )] ( )( ) Eercise 8A In questions 3 factorise the epressions y 7y 5 0 4y 5y 6 4y 9y 6 5y y 3 5 9y y In questions 4 6 factorise the epressions completely y 6y Simplifying rational epressions Algebraic epressions in the form of a fraction are called rational epressions. Each of these rational epressions can be simplified by factorising the numerator and denominator and then cancelling any epression which is common. The methods of factorising used in Sections 8., 8.5, 8.6 and 8. are required in this section. 6 ( 3)( 3)

4 8. Simplifying rational epressions CHAPTER 8 Eample 5 Simplify fully 4 4 Solution 5 4 ( ) 4 ( )( ) 4 ( ) 4 ( )( ) 4 4 ( ) ( )( ) Factorise the numerator by taking out the common factor. 4 is the difference of two squares. Factorise it by using A B (A B)(A B). Write 4 4 in a fully factorised form. Cancel the common factor ( ). is usually written as ( ) It is not possible to simplify further. Eample Simplify fully 3 Solution ( ) 3 ( )( ) 3 3 3( ) 3 ( ) ( ) 3( ) ( ) ( ) Factorise the numerator by taking out the common factor. Factorise the denominator. Find two numbers whose product is and whose sum is 3 The numbers are and Write in a fully factorised form. Cancel the common factor ( ). It is not possible to simplify 3 further. Eample 7 Simplify fully

5 CHAPTER 8 Further factorising, simplifying, completing the square and algebraic proof Solution ( 3) ( 3) ( 3)( ) (3 ) Factorise the numerator. Multiply the coefficient of by the constant term to get 6 Find two numbers whose product is 6 and whose sum is 5 The numbers are 6 and Write 5 as 6. Factorise the denominator by taking out the common factors ( 3) ( ) 4 3 ( 3 ) (3 ) ( ) 3 (3 ) (3 ) ( ) 3 (3 ) ( ) 4 3 Write ( 3) (3 ). in a fully factorised form. Cancel the common factor (3 ). Eercise 8B Simplify a 4 k3 k b (y ) c (p ) (p ) (y ) p(p ) ( ) d e f ( 3 ) ( ) ( )( ) ( 3) ( ) In Questions 5 simplify the epressions fully ( 3) y 4y y 7 3y 7y 3y y 460

6 8.3 Adding and subtracting rational epressions CHAPTER Adding and subtracting rational epressions Numerical fractions were added and subtracted in Section 4. Similar methods are used to add and subtract algebraic fractions. To add fractions with the same denominator, add the numerators but do not change the denominator. For eample To subtract fractions with the same denominator, subtract the numerators but do not change the denominator. For eample To add or subtract fractions with different denominators, each fraction must be written with a common denominator. For eample The LCM of 5 and 6 is To find firstly write each fraction with a common denominator of so and In general, to add or subtract algebraic fractions with different denominators factorise the denominators if possible write each fraction as a fraction with a common denominator add or subtract the fractions and factorise the numerator if possible simplify the algebraic fraction if possible as in Section 8. Eample 8 Write as a single fraction in its simplest form. Solution 8 Factorise the denominator ( ) Common denominator is ( ) Since and ( ) divide eactly into ( ). ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) Write as a fraction with denominator ( ). The other fraction has denominator ( ) so does not ( ) change. Since the denominators are the same just subtract the numerators. Cancel the common factor. The single fraction cannot be simplified further. 46

7 CHAPTER 8 Further factorising, simplifying, completing the square and algebraic proof Eample 9 Simplify Solution ( ) 5( ) 5( ) Common denominator is 5( ) 5 3 Write each fraction as a fraction with denominator 5( ). 3( ) 5( ) 5( ) 5( ) 5 3 3( ) 5( ) 5( ) 5( ) 5( ) 5( ) 5 3 5( ) 6 5( ) Factorise the denominators. All denominators divide eactly into 5( ). Since the denominators are the same just combine the numerators. (6 5 ) 5 Divide the numerator and the denominator by ( ) 46 Eample Write as a single fraction. 3 Solution The denominators do not factorise. 3 Common denominator is ( )( 3) Since the denominators have no common factor. 5 5( 3) 3 3( ) Write each fraction as a fraction ( ) ( )( 3) ( 3) ( )( 3) with denominator ( )( 3) ( 3) 3( ) 3 ( )( 3) ( )( 3) 5( 3) 3( ) Since the denominators are the ( )( 3) same just subtract the numerators Simplify the numerator. ( )( 3) 4 ( )( 3) 4( 3) Factorise the numerator. ( ) ( 3) 5 3 4( 3) 4( 3) The single fraction cannot be simplified further. 3 ( ) ( 3) ( )( 3)

8 8.3 Adding and subtracting rational epressions CHAPTER 8 Eercise 8C 3 Write 5 5 as a single fraction in its simplest form. 3 3 Write as a single fraction Write 5( ) ( ) as a single fraction. 7 a Factorise 5 6 b Write as a single fraction in its simplest form. 4 9 Write as a single fraction in its simplest form. Simplify 3 Simplify Epress as a single fraction. 3 3 Write as a single fraction Write as a single fraction a Factorise 3 b Simplify Write as a single fraction 9 3 in its simplest form Write as a single fraction. ( ) 4 Simplify 3 4 Epress as a single fraction Epress as a single fraction. 4 7 a Factorise b Write as a single fraction in its simplest form a Factorise b Write as a single fraction in its simplest form a Factorise i 5 6 ii 7 b Write as a single fraction in its simplest form a Factorise b Write as a single fraction in its simplest form

9 CHAPTER 8 Further factorising, simplifying, completing the square and algebraic proof 8.4 Completing the square The diagram shows 6. The 6 has been split in half To complete the square 9 (3 3) needs to be added on The diagram now shows 6 9 ( 3) Removing the 9 shows 6 ( 3) 9 Writing 6 in the form ( 3) 9 is called completing the square for 6. In general writing the quadratic epression a b c in the form a( p) q is called completing the square. In Eercise 8C these epansions of perfect squares were found ( 7) 4 49 ( 5) 0 5 In each case the final number in the bracket is half the coefficient of. In general ( A) A A so A ( A) A To complete the square for the epression A Step : Write ( ) Leave enough space inside the bracket for writing in a constant term Step : Write ( A) Since A is half the coefficient of ( of A A) Step 3: Write ( A) A Subtract the square of half the coefficient of (subtract A ) Eample The epression 8 can be written in the form ( p) q for all values of. Find the value of p and the value of q. Solution 8 Complete the square for 8. To complete the square for A find half the coefficient of then square it to give A Then write A as A A A ( A) A of ( 4) ( 4) 8 ( 4) 6 ( p) q is the same as ( 4) 6 when p 4 and q 6 Halve the coefficient of. Write ( ) with half the coefficient of before the end bracket then subtract the square of half the coefficient of. (4) 6 Compare ( 4) 6 with ( p) q to read off the value of p and the value of q. 464

10 8.4 Completing the square CHAPTER 8 In Section 3. quadratic graphs were drawn and maimum and minimum points were often estimated by reading from the graphs. The method of completing the square can be used to find the eact values for maimum and minimum points on quadratic graphs. For eample the graph of y 8 has a minimum point. The coordinates of this minimum point can be found by using the result of eample Completing the square for 8 gives 8 ( 4) 6 so the equation of the graph can be written as y ( 4) 6 Squaring any value always gives an answer which is positive or zero so for any value of, the smallest value of ( 4) is 0 When ( 4) 0, 4 and y The minimum point of the graph y 8 has coordinates (4, 6) The minimum value of 8 is 6 Eample The graph of the curve with equation y 6 has a minimum point. a Write the epression 6 in the form ( p) q. See Section 3. Eample 4 b Hence find the coordinates of the minimum point. Solution a ( ) 6 of ( ) 6 [( ) ( ) ] 6 6 ( ) ( ) 6.5 Separate the constant term and complete the square for. Halve the coefficient of. Write ( ) with half the coefficient of before the end bracket then subtract the square of half the coefficient of. Write the answer in the required form ( p) q so p and q 6.5 b y ( ) 6.5 The minimum value of y is 6.5 and occurs when ( ) 0 so The minimum point is (0.5, 6.5) Note: The minimum value of 6 is 6.5 Write y 6 in the completed square form using the answer to part a. The least value of ( ) is 0 Eample 3 The epression 7 0 can be written in the form a( p) q. a Find the value of a, the value of p and the value of q. b Use the answers to part a i to find the maimum value of 7 0 ii to find the value of for which 7 0 has its maimum value. 465

11 CHAPTER 8 Further factorising, simplifying, completing the square and algebraic proof Solution 3 Method a 7 0 ( 0) 7 of 0 5 ( 0) 7 [( 5) (5) ] 7 [( 5) 5] 7 ( 5) 50 7 Take out the coefficient of and separate the constant then complete the square for 0. Halve the coefficient of. 7 0 ( 5) 67 a( p) q is the same as ( 5) 67 when a, p 5 and q 67 Method a a( p) q 7 0 a( p p ) q 7 0 a ap ap q 7 0 a so a ap 0 so ap 0, 4p 0, p 5 ap q 7 so (5) q 7, 50 q 7 a, p 5 and q 67 b 7 0 ( 5) 67 i The maimum value of ( 5) 67 is 0 67 The maimum value of 7 0 is 67 ii The maimum value of 7 0 occurs when 5 Compare ( 5) 67 with a( p) q to read off the value of a, the value of p and the value of q. Equate the two epressions. Epand brackets using ( A) A A Using part a. For the two epressions to be the same for all values of the terms must be identical, the terms must be identical and the constant terms must be identical. ( 5) 0 so 67 ( 5) is maimum when least value is subtracted from 67 Least value of ( 5) is 0 The maimum value occurs when ( 5) 0 Eercise 8D Write in the form ( p) q. a b 4 c d 0 e 4 f 6 g 8 h 4 i 8 j Write in the form ( p) q. a 6 0 b 8 0 c 4 0 d 6 e 4 8 f 6 g 4 8 h 0 7 i 40 0 j Write in the form ( p) q. a b 3 c 5 d 3 e Write in the form a( p) q. a 4 b 8 c d e Write in the form p ( q). a 6 b 9 4 c 0 6 d 4 e 3 466

12 8.5 Algebraic proof CHAPTER 8 6 For all values of, 0 3 ( p) q. a Find the value of p and the value of q. b Write down the minimum value of The curve with equation y 3 has a minimum point. a Write the epression 3 in the form ( p) q. b Hence find the coordinates of the minimum point. 8 The curve with equation y 6 has a maimum point. a Write the epression 6 in the form p ( q). b Hence find the coordinates of the maimum point. 9 The curve with equation y has a maimum point. a Write the epression in the form a( p) q. b Hence find the coordinates of the maimum point. 0 Show that the minimum value of 4 3 is the same as the maimum value of Algebraic proof Jane finds this puzzle. Write down an odd integer. Write down the net integer (which will be an even number). Square your odd number and double your even number and add your two results. The answer is always more than the square of your even number. Jane chooses 7 as her odd integer. She writes the numbers 7 and 8 (7 and 8 are consecutive integers.) Jane squares 7 to get 49 and doubles 8 to get 6 She then adds her two results to get 65 ( 49 6). Jane says that she has proved the puzzle because 65 is more than the square of 8 (her even number). Jane is wrong because she has only shown that the puzzle works for the odd number 7 She has not proved that it works for all odd numbers. Here are eamples of some important facts including algebraic epressions for odd numbers, which will help when writing algebraic proofs three consecutive integers can be written in the form n, n, n where n is an integer. In some questions it is more useful to write three consecutive integers in the form n, n, n so that the middle term is the simplest any even number can be written in the form n where n is an integer the net even number after n is n so n and n are consecutive even integers any odd number can be written in the form n where n is an integer n and n are consecutive odd integers. The net eample shows an algebraic proof of Jane s puzzle. 467

13 CHAPTER 8 Further factorising, simplifying, completing the square and algebraic proof Eample 4 Prove algebraically that for any odd Understand the problem by first writing some numerical eamples number and the even number after it, e.g. 3, 4 gives the square of the odd number added to twice the even number is always one more than the square of the even number. Solution 4 For any integer n Odd number n Net (even) number n (odd number) twice the even number (n ) n 4n 4n 4n Set up the problem by using algebraic epressions. Remember to state what n stands for. The number after n is n (n ) (n )(n ) 4n (n) (the even number) 4n 4n 0 Interpret the result by linking the epression 4n to the problem. 4n n n The even number was n So the square of the odd number added to twice the even number is always one more than the square of the even number. Eample 5 a Factorise fully A B A B. b Prove that the difference between the squares of any two integers added to the difference between the two integers is an even number. Solution 5 a A B A B (A B)(A B) (A B) Factorise A B as the difference of two squares and write A B as (A B). (A B)( A B ) b Let m and n be any two integers with m n. The difference between the integers is m n m n m n (m n)(m n ) If m and n are both even or both odd, m n is even and m n is odd so (m n)( m n ) is even. (A B) is a common factor. Note that the question refers to any two integers so two different letters are used. Add the difference between the squares of the integers to the difference between the integers. Use part a even even even, odd odd even even even odd, odd odd odd even odd even If one of m and n is even and the other odd, then m n is odd and m n is even so again (m n)(m n ) is even. For any integers, m and n, m n m n is even. So the difference between the squares of any two integers added to the difference between the two integers is an even number. 468

14 Chapter summary CHAPTER 8 Eercise 8E In questions 7 prove the result algebraically. The sum of any three consecutive integers is a multiple of 3 The sum of any two consecutive odd numbers is a multiple of 4 3 The sum of any two odd numbers is an even number. 4 The sum of any four consecutive odd numbers is a multiple of 8 5 The sum of any three consecutive odd numbers is never a multiple of 6 6 The difference between the squares of any two consecutive even numbers is twice the sum of the two even numbers. 7 The sum of the squares of any three consecutive integers is never a multiple of 3 8 Prove algebraically that for any even number and the odd number after it, the square of the even number added to four times the odd number is always a square number. 9 Prove that the difference between the squares of any two integers added to the sum of the two integers is an even number. 0 The epression (n 3) (n 4) is the nth term of the sequence of numbers 0, 5,, 8, a Write down an epression in terms of n for the (n )th term of the sequence 0, 5,, 8, b By finding an epression for the sum of the nth term and the (n )th term of the sequence, prove that the sum of any two consecutive terms in the sequence is a square number. Chapter summary You should now know: how to factorise quadratic epressions of the form a b c and a by cy where a, b and c are numbers with a that in the epression a b c, a is the coefficient of, b is the coefficient of and c is the constant term how to simplify rational epressions by factorising both the numerator and denominator and cancelling any common factors how to add or subtract algebraic fractions with different denominators by applying these steps factorise the denominators if possible write each fraction as a fraction with a common denominator add or subtract the fractions and factorise the numerator if possible simplify the algebraic fraction if possible as in Section 8. how to complete the square by writing the quadratic epression a b c in the form a( p) q 469

15 CHAPTER 8 Further factorising, simplifying, completing the square and algebraic proof that the minimum value of ( p) q is q and occurs when p 0 that the maimum value of q ( p) is q and occurs when p 0 how to prove algebraically a given result. Chapter 8 review questions Simplify fully i m 4 m 5 ii p 6 p iii q 3 q Simplify fully a (3 4) 3(4 5) b (y 3 ) 5 c n n n 3 Write as a single fraction in its simplest form Simplify fully 5 5 q 6 iv 4( k 8) k 8 (387 June 003) (385 November 00) (387 November 005) 5 Simplify fully 4 a ( 3 y) 5 b 6 8 (5540 June 005) 6 The epression 8 can be written in the form p ( q) for all values of. a Find the value of p and the value of q. b The epression 8 has a maimum value. i Find the maimum value of 8. ii State the value of for which this maimum value occurs. 7 Given that 4 a ( b) for all values of,find the value of a and the value of b. (388 November 005) 8 a Simplify i (3 y) 3 ii (t 3 ) b Show that 4 5 can be written as ( p) q for all values of. State the values of p and q. (387 November 005) 9 Simplify (5 ) 5 0 a Factorise 9 6 b Simplify a Solve (388 March 004) 470 b Simplify fully Simplify 3 (387 June 004) (385 November 998)

16 Chapter 8 review questions CHAPTER 8 3 a Factorise b Write as a single fraction in its simplest form 3 4 Simplify a Factorise completely ( 5) 3( 5) b Simplify 3 (y 4) (y 4) (385 June 000) (388 January 005) 6 Show that 3 ( 3 ) ( ) 7 a Factorise 5 6y y b Hence, write as a product of two integers both of which are greater than 5 8 The epression 8 8 can be written in the form ( p) q for all values of. a Find the value of i p ii q. Here is a sketch of the graph of y 8 8 The minimum point on the curve is M. y b Write down the coordinates of M. The line with equation y k has two points of intersection with the graph of M y 8 8 when k a. c Write down the least possible value of a. O 9 a Factorise 4 3 b Simplify ( ) ( ) y 8 8 c Make f the subject of the formula u v (384 November 997) f 0 Prove algebraically that the sum of the squares of any two consecutive even integers is never a multiple of 8 Prove algebraically that the sum of the squares of any two odd numbers leaves a remainder of when divided by 4 (387 November 005) a Show that (a ) (b ) 4(a b)(a b ) b Prove that the difference between the squares of any two odd numbers is a multiple of 8 (You may assume that any odd number can be written in the form r where r is an integer.) (387 June 003) 3 Humera wrote down three consecutive square numbers in order of size. She then added the smallest and the largest of these numbers together and finally subtracted the middle number of the three numbers. Prove algebraically that Humera s answer should always be two more than the middle number. 47

A-LEVEL MATHS Bridging Work 2017

A-LEVEL MATHS Bridging Work 2017 A-LEVEL MATHS Bridging Work 017 Name: Firstly, CONGRATULATIONS for choosing the best A-Level subject there is. A-Level Maths at Wales is not only interesting and enjoyable but is highly regarded by colleges,