Algebra 2. Mathematics Worksheet

Transcription

1 Algebra Mathematics Worksheet This is one of a series of worksheets designed to help you increase your confidence in handling Mathematics. This worksheet contains both theory and eercises which cover:-. Epanding brackets. Simple factorisation. Quadratic factorisation. Completing the square There are often different ways of doing things in Mathematics and the methods suggested in the worksheets may not be the ones you were taught. If you are successful and happy with the methods you use it may not be necessary for you to change them. If you have problems or need help in any part of the work then there are a number of ways you can get help. For students at the University of Hull Ask your lecturers. You can contact a Mathematics Tutor from the Skills Team on the shown below. Access more Maths Skills Guides and resources at the website below. Look at one of the many tetbooks in the library. Web: skills@hull.ac.uk

2 . Epanding Brackets Brackets are used for convenience in grouping terms together. When removing brackets each term within the bracket must be multiplied by the quantity outside the bracket. For eample y y. We can see this by using the idea from Algebra where and y were the lengths of two pieces of wood (in whatever units you like!. y - y y - y - y y ( - y -y We can see from the diagram that ( y is not the same as y. In the same way 7( y ( 7 ( 7 ( y 7 y When simplifying epressions containing brackets, first multiply out the brackets and then collect like terms. Worked Eamples y 5 y 6y (a Simplify y y y 5 6 y 6 5 y 0 y 6 0 With practice you may be able to do the process in one step by collecting terms as you multiply. For instance, in the above, collecting y -terms y 5y y and then the numbers 6 6 and finally the -term 0. (b Simplify ( y z (y z ( y z ( y z y z y z y z y z y 6z this is better written as 6z y otherwise the sign can get lost! Multiplying brackets together can be done in a number of ways, here are four:-. ( y( a b ( a b y( a b multiply first bracket out then remaining brackets a b ya yb (remember to be careful if you have a miture of signs page

3 . The eye-brows method Collecting terms gives ( + y (a + b ( y( a b a ya b yb. We can use a diagram and think in terms of area y From the diagram we see that the total a a ya area is given by ( y( a b b b yb This is equivalent to adding all the bits together giving ( y( ab a ya b yb. Multiply, using a table and ignoring lengths y a a ya ( y( ab a ya b yb b b yb this method is especially useful when some of the terms are negative or the brackets contain more than terms. (see eample d below From the above, putting a and b y we have y y y y y ( y( y y y y y y y y y Similarly y y y y y - y These very important epressions occur often and should be learnt! Eamples With practice you will be able to cut out some intermediate stages. Epand and simplify the following (a ( ( 5 page

4 Method Method ( ( 5 ( 5 ( In both cases ( ( 5 5 (b ( y( y Method Method ( y( y ( y y( y y y y y y y y In both cases ( y( y (c ( Comparing this with the epression for ( y we have 9 y y y This could also have been done using either of the methods above. (d y y 5 using method (much easier than method in most cases y 6 y 9 y y 8y 6 y y -5 y y 5 = 6 8y 6y y 5 e ( using we have y y y and hence 6 8 page

5 Eercise Epand and simplify where possible. y a y ( 7( y y y ( y y z y z. Simple Factorisation Factorisation is the reverse process to epanding brackets Epanding ( y gives y so factorising y gives ( y. To do this we need to find common factors in the given terms. Eamples (a Factorise y Both terms can be divided by (i.e. they have a factor of, so y = ( y. (b Factorise 7 y Both terms can be divided by 7 (i.e. they have a factor of 7, so 7 y = 7( y. (c Factorise mn n p m p There is no factor common to all three terms hence the epression cannot be factorised. 5 (d Factorise mn 5n m n To factorise this epression we need to find a factor common to all terms, 5 and have a factor of 7 5 mn, n and m n have a factor of n, 5 so mn 5n m n = 7n (mn 5 m n 5 Eercise - Factorise fully (where possible. a b. a ab. ab ac 5ad ut 5t 6d d 8. a bc a b c 0. 5a b 0a c b c. 5 6ab 9a b b c gt ht ab 6ab a r rh r rh 8r page

6 . Quadratic Factorisation As ( ( 6 so factorising 6gives ( ( We can find clues to help in factorisation by considering the patterns of signs: Epanding ( ( 7 6 ( ( 7 6 Factorising 7 6 ( ( 7 6 ( ( The signs when factorising (.. +..( (....(.... If the constant term is positive then the brackets will both have the same sign. Epanding ( ( 6 ( ( 6 Factorising 6 ( ( 6 ( ( when factorising The signs or (....( or (.. +..(.... If the constant term is negative then the brackets will have different signs. So, when factorising the quadratic a b c (with a 0 then if b and c are both positive the brackets will both have plus signs in if b is negative and c is positive the brackets will both have minus signs in and if c is negative one bracket will have a plus and the other a minus sign. Eamples (The method used here is one of many. (a Factorise A B C To get the term both A and C must be D From the signs both B and D are positive so to get + we need B = and D = page 5

7 This leads to: This gives and in the other two boes a total -term of as required. Solution Factors of are ( + ( +. Or you could have started with A B C D coefficients. The solution is always the same. and equated (b Factorise 7 A B C To get the term A = and C = D From the signs both B and D are negative so to get +, B = - and D = - or vice-versa This leads to EITHER OR ( -term 5 incorrect ( -term 6 7 correct Solution Factors of 7 are ( (. (again with practice this will become easier (c Factorise 5 A B C To get we need A = & C = D - To get - we need B =, & D = - or vice-versa or B = & D = - or vice-versa This gives possibilities ( -term total ( -term total ( -term total 6 5 ( -term total From we have- Solution factors of 5 are ( ( In practice, once the correct pair have been found it is not necessary to work out any more ecept, perhaps as a check. page 6

8 (d Factorise 9 A B C To get we need A = & C = or vice-versa or A = & C = D -9 To get -9: B = & D = -9 or B = & D = - or vice-versa in both cases This gives a lot of possibilities but symmetry suggests we try 6 This gives -term total which is what we want Solution factors of 9 are ( ( This is an eample of the difference of two squares. Earlier we had, similarly, 9 ( y( y y (e Factorise p 5p q 5q Pairing the terms up to put the two p terms and the two q terms together gives p 5p q 5q p( 5 q( 5 This last epression is similar to pa qa which can be factorised as ( p q A giving p 5p q 5q ( p q( 5 Solution factors of p 5p q 5q are ( p q( 5 or ( 5( p q (f Factorise 6 7 A B C 6 To get 6 : A = & C = 6 or A = & B = D - To get -: B = & D = - or B = - & D = or vice-versa in both cases This gives 8 possibilities but you cannot have a 6 and a in the same bracket, nor a and a (why? - answer later so this reduces the possibilities to : ( -term total 8 7 ( -term total ( -term total 9 7 ( -term total 9 7 From we have Solution factors of 6 7 are ( - ( + page 7

9 To answer the earlier question consider ( (. This epands to give 6 ( which has a factor of and ( also has a factor of From this we can see that, unless the quadratic has a numerical factor, the terms in the brackets cannot have a common numerical factor. This is why we could ignore ( 6, ( as possible factors of 6 7. Eercise Factorise as far as possible y ( ( a 6a. ( mn mn d d. 9. Completing the Square Completing the square is the technique which is the basis for obtaining the formula for the solution of a quadratic equation. It can also be used in solving quadratic equations and finding maimum and minimum values of quadratics. It involves finding what needs to be added to epressions such as 8 and y 0 y to get them into the form ( a and ( y b respectively. (a Consider 8 we want to find a number a so that ( a 8... this gives ( a a a 8... where means equal for all values of. This means that the coefficient of on both sides must be the same, giving a 8 a and a 6. We need to add 6 to complete the square. 8 6 ( or 8 ( 6 Once this process is understood it becomes easy to complete the square as long as the coefficient of the squared term is. (b Consider y 0 y: Coefficient of y is -0 halve it giving This suggests that we need to add ( 5 5 giving ( - 5 y -0y + 5 y and y 0y y 5 5 page 8

10 c Consider p 9 p : We need to complete the square on p 9 p - coefficient of p is 9 halve it giving. This suggests we need to add giving p p + 9 p + 0 and p 9 p p 0 p 0 to p 9 p d Consider + 5 = ( : In the bracket the coefficient of is 6 halve it giving. This suggests that we add 9 (in the bracket giving ( , and + 5 = ( + 8 ( + 5 e Consider (the nasty p p p p : In the bracket the coefficient of p is 6 halve it giving. This suggests that we 7 7 add 6 (in the bracket p p p p p giving p p 6 6 and p p p Eercise In each of the following add the term which will complete the square for the given epressions. Then write the result down as the square of an epression. c k 9. a c 7k a b m y b m y.. y n p 5y n p. 8.. d k a 6d k 7a Complete the square in the following. p p. y y p 7 p 6. 5a a page 9

11 ANSWERS Eercise. 6y y y 6 0 8y a y y 6y. y z 8. 6 Eercise. a b. a b c 5d d d 0. no simpler form. a t u 5 8. abc a b t b. b a a b b c a b r r h r r t g h 6b h r Eercise.. y y aa mn mn d d Eercise ( c p ( k ( a ( b 7 ( m ( y y.. 5 ( y ( n 5 ( p p ( d ( k ( a 7 6 a We would appreciate your comments on this worksheet, especially if you ve found any errors, so that we can improve it for future use. Please contact the Maths tutor by at skills@hull.ac.uk Updated nd June 0 The information in this leaflet can be made available in an alternative format on request. Telephone page 0