Algebra Toolbox Part 2

8/13/2019 Algebra Toolbox Part 2

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The AlgebraToolbox

Part TwoQuadratics

Algebraic Fractions


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The Algebra Toolbox

Contents

Section Topics Page1. Factorising QuadraticExpressions

a. Basic Definition 3

b. What Makes Quadratics? 4c. Factorising Type 1 Quadratics 7d. Factorising Type 2 Quadratics 9e. Factorising Type 3 Quadratics 13

f. Factorising Harder Type 3 Quadratics 172. Solving QuadraticEquations

Quadratic Equations 21

3. Algebraic fractions &dividing with algebra

Algebraic Fractions 25


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1. QUADRATIC EXPRESSIONS - factorising

(a) BASIC DEFINITIONA Quadratic Expression is one that

must contain an x2 (or a 2, y 2 etc, depending on what letter youre using) may contain an x-term (such as 3x, 2x, 7x, 5a, 3t etc) may contain a plain number without a letter (such as 4, 5, 9 etc)must not contain any other kind of term other than these

The general form of a quadratic is

a x 2 + b x + c

where the x is called the variable, and a, b and c are all constants(numbers). a, b and c are sometimes negative, in which case you will see

negative signs in the expression.

The following are examples of quadratic expressions:

3 x2 + 8 x + 5 here, a = 3, b = 8 and c = 5

3 x2 8 x 5 here, a = 3, b = 8 and c = 5

x2 7 here, a = 1, b = 0 (theres no b-term) and c = 7

4a a 2 here, a = 1, b = 4 and c = 0 (do you agree??) Its a good idea to cutnpaste terms first, so insteadwrite it as a 2 + 4a

5t + 3 + t 2 Cutnpaste to rewrite it as t 2 + 5t + 3 so a = 1, b = 5 and c = 3

y2 here, a = 1, but b and c both = 0

u2

here, a = , and b = 0, c = 0 a(a 3) this is written in disguised form, because if you expand

it youll get a 2 3a, which is indeed a quadratic!a = 1, b = 3 , c = 0

Examples of non-quadratics are 2x 3 5x; x3 + 5x ; 2x + 3

Can you see which terms make them non-quadratics? (2x 3, 3/x and 2x)

The Algebra Toolbox 2003 R. Bowman. All rights reserved.3


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(b) WHAT MAKES QUADRATICS?

Try making quadratics by multiplying these pairs of linear terms

together

(i) (a + 3) (a + 5)The word linear refers to anyexpression of the formata + 5; 3x + 2; 2x 5; 7 x;2a + 1; 3b; 5w; etcThey all have just a letter, maybea number multiplying it, andmaybe another number added orsubtracted.

(ii) (x 2) (2x 4)

(iii) (3x 5) 4x

(iv) (y 2) (y + 2)

(v) (7 b) (2b + 1)

(vi) x (3 x + 1)

(vii) (3y + 5)(3y 5)Section 3(d) on Page 39 of Toolbox Part 1 shows you how to do these.

Here are the answers..

(i) (a + 3) (a + 5) = a2 + 8a + 15

(ii) ( x 2) (2 x 4) = 2 x2 8 x + 8

(iii) 4 x (3 x 5) = 12 x2 20 x Note no c term in the answer! Can yousee that this is because the 1 st term (4x)consists of 1 term only?

(iv) (y 2) (y + 2) = y2 4 Note no b term in the answer becauseeach bracketed term was identical &brackets have opposite signs

VITALINFO!!(v) (7 b) (2b + 1) = 2b 2 + 13b + 7

(vi) x (3 x + 1) = 3 x2 + x Same reasoning as for (iii)

(vii) (3y + 5)(3y 5) = 9y2 25 same reasoning as for (iv)



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Having done these, there are now some important observations youshould make:

usually , when you multiply two linear terms together, you

will get something like ax2

+ bx + c, with 3 terms. Thishappens in (i), (ii) and (v) above

There are, however, two important exceptions:

(1) when we have two linear expressions the same exceptfor their sign (e.g. (a + 6)(a 6) ) , the result is missing theb term. This happens in (iv) and (vii).

(2) when one of the linear expressions has an x-term only(e.g. 3x(x + 4)) , the result is missing the c term. Thishappens in (iii) and (vi) above.

Now here are some names to learn:

We will label expressions like

3 x 2 + x , 12 x 2 20x, 7y y 2 as Type 1 quadratics 9y 2 25, y 2 4, x 2 25 etc as Type 2 quadratics x2 3x + 5, 2x 2 + x 7 as Type 3 quadratics

Type 1 quadratics are missing their c term & look like ax 2 + bxType 2 quadratics are missing their b term & look like ax 2 + cType 3 quadratics are missing no terms (they have all 3 !)

Important note The type 2quadratics we deal with allhave a minus sign betweenthe 2 terms, so this will lookmore like ax 2 c



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Practice Exercises 1

Write down the values of a, b and c, and state whether thequadratic is Type 1, 2 or 3. Question 1 is done for you!

Quadratic Values of a, b and c? Type 1, 2 or3?

1 2a 2 + 3a + 5 a = 2, b = 3, c = 5 Type 32 5y 2 8y3 4 x2 11 x 4 y2 165 25 + y +3y 2 6 t 2 327 3b 2 + b + 78 3t2 9 7b + 8 3b 2 10 4t

CLICK HERE FOR ANSWERS!


Expand each of these, and categorise your answer as Type 1, 2 or 3.

1 (3a + 4)(2a + 5) 6 (5y 7)y2 (7a 2)(7a + 2) 7 (3 2p)(3 + 2p)3 3 x(2 x 5) 8 (a 3)(a 3)4 (b + 2)(b + 2) 9 (2b + 5)(b 5)5 (3a 5)(3a + 5) 10 (3y + 4)(3y 4)

CLICK HERE FOR ANSWERS !

When youve completed Exercise 2, make sure you know how to(a) categorise your answers (the quadratics) as Type 1, 2 or 3(b) recognise from the question what Type your answer will be.

In other words you should know that3a(2a + 5) will give a Type 1 answer because the one of the two factors (the 3a) hasonly 1 term (ans: 6a 2 + 15a) (2a 3)(2a + 3) will give a Type 2 answer because each bracket has the samethings, but signs are opposite (ans: 4a 2 9) (2a + 5)(3a 4) will give a Type 3 because its not Type 1 or Type 2 !!(ans: 6a 2+ 7a 20)



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(c) FACTORISING TYPE 1 QUADRATICS

The good news is..This has already been done in Toolbox Part 1, (page36) although not thoroughly.

Type 1 factorisations involve taking out a commonfactor!

Example 1

Factorise x2 + 8 x

Noting the common factor is x , place this at the front of the brackets

x( + )

Now fill in the blanks. THINK: x multiplies what to give x 2. ANS x x multiplies what to give 8 x . ANS 8

So place the x and the 8 in the brackets:

x( x + 8). ANSWER!!

Example 2Factorise 3 x2 9 x

Here there are TWO common factors the x and the 3. Take them both out thefront as 3 x .

3 x(. .)

THINK.. 3x multiplieswhat

to give 3x2

? ANS x.3x multiplies what to give 9x? ANS 3 .

So place the x and the 3 in the brackets

3 x( x 3) ANSWER!!



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Example 3

Factorise 18a 12a 2

Can you tell what the common factors will be?Certainly theres an a, but what about the numbers?

18 and 12 can both be divided by several numbers, including 2, 3 and 6.

Pick the biggest! So well go with 6 .

The common factors well use then are Key point!! Takenote!!6 and a

As for previous examples, put these in front of the brackets as 6a

6a(. .)THINK 6a multiplies what to give 18a ? ANS 3

6a multiplies what to give 12a 2 ? ANS 2a

Place the 3 and the 2a into the brackets :

= 6a(3 2a) ANSWER!!



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(d) FACTORISING TYPE 2 QUADRATICS

These quadratics are of the form

x2

94x 2 16y2 25a 2 49

etc.

Can you tell what they have in common?

If you said.

a minus sign in the middletwo termsboth terms are squares of numbers or letters (25 is 5 2, 49 is72, y 2 is y squared, 4x 2 is (2x) 2 etc

then youd be right!!

Because Type 2 Quadratics contain a minus sign and two terms

which are squares, they are known as

DIFFERENCE OF TWO SQUARES

The general rule can be written as

a 2 b 2 = (a b)(a + b)

The minus and plus signs can be swapped around.This is the same as a 2 b 2 = (a + b)(a b)



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Before we get into it, try these expansions . (See Toolbox Part 1, P39)

(i) (a 3)(a + 3) = .?

(ii) (5 y)(5 + y) = .?

(iii) (3 x + 2)(3 x 2) = ?

How did you go? Did you get these answers?(i) a 2 9 (ii) 25 y 2 (iii) 9x 2 4

What do the 3 questionshave in common?

What do the 3 answershave in common?

So if I was asked to write down the expansion of (3y 4)(3y + 4) hereswhat I should be thinking:

(i) The 2 terms in eachpair of brackets are thesame.

(ii) There is a minussign in one bracket, anda plus in the otherbracket.

(i) The 1 st term is thesquare of the 1 st term ineach bracket, and the2nd term is the square ofthe 2 nd term in eachbracket.

(ii) There is always aminus in the answer!

(i) Does each bracket have the same terms, andopposite signs? YES! This means our rulecan work! (a + b)(a b) = a 2 b 2

(ii) What is the square of 3y? 9y 2 (3y 3y = 9y 2) (iii) What is the square of 4? 16

(iv) Answer must then be 9y 2 16 !!



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Now lets move into Factorising Type 2 quadratics.

Example 1

Factorise x2 a 2 .

Remember FACTORISE is the reverse of EXPAND.So it is helpful if we can think backwards - what couldhave been expanded to get x 2 a 2 ?

Because this is a Type 2, we know somethinglike ( + )( ) must have been expanded to get x 2 a 2.

= x2

= a 2 so must be x& must be a

Remember Always try to take out a common

factor first - before you do anything!

THINK

= ( x a )( x + a) ANSWER!! (or you could write (x + a)(x a)

Example 2THINK

Factorise 16 t 2 .

What was expanded to get 16 t 2 ?

We know ( + )( ) = 16 t 2.

= (4 t )(4 + t) ANSWER!!

= 16 = t 2

so must be 4& must be t

Remember Always try to take out a common factor first - before you do anything!



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Example 3

Factorise 9 x2 25b 2 .

What was expanded to get 9x 2 25b 2 ?

We know ( + )( ) = 9x 2 25b 2.The brackets must each contain a 3x and an 5b.

= (3 x 5b )(3 x + 5b) ANSWER!!

THINK

= 9x 2 = 25b 2

so must be 3x& must be 5b

Example 4

Factorise 2a 2 50.

Remember to take out any common factor first! 2 is a common factor!

2a2 50= 2(a 2 25)

Now work on the bracketed term a 2 25 as a Type 2This becomes (a 5)(a + 5)

= 2(a 5)(a + 5) ANSWER!!




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(e) FACTORISING - TYPE 3 QUADRATICS

Type 3 quadratics are the hardest to factorise.

Try following this example of expanding which leads to a Type 3. Payspecial attention to tracking the numbers throughout each step:

Example 1

Expand & simplify (a + 3)(a + 5)

Remember to break up the first bracketed term (a + 3) and repeat the second (a + 5)See examples in Toolbox 1 Pages 39-40

Can you track the progress of the twonumbers (3 and 5) from the first step to thelast?Can you see that

the 8 in the answer has come fromADDING the 3 and 5the 15 in the answer has come fromMULTIPLYING the 3 and 5?

(a + 3)(a + 5)= a(a + 5) + 3(a + 5)= a 2 + 5a + 3a + 15= a 2 + 8a + 15 ANS!!

3 5 = 15

3 + 5 = 8

Example 2

Expand and simplify (y 5)(y + 9)

Can you do this the quick way using our discovery above?This time the two numbers are 5 and 9.

ADD THEM: 5 + 9 = +4MULTIPLY THEM: 5 9 = 45

We can now write down the answer straight away:

(y 5)(y + 9) = y 2 +4y 45 ANSWER!! You should check this bydoing it the long way!



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Example 3

Expand and Simplify (a 5) 2

This of course really means (a 5)(a 5)

So doing it the quick way again, the numbers are 5 and 5.

ADD THEM: 5 + 5 = 10 MULTIPLY THEM: 5 5 = + 25

ANSWER: a 2 10 a + 25

FACTORISING TYPE 3 QUADRATICS

Now we move into FACTORISING Type 3 quadratics. This is theREVERSE of EXPANDING (what we did in examples 1-3 above)

Example 4 (this is Example 1 backwards)

Factorise a2 + 8a + 15.

Answer will look like (a + )(a + )

(a + 3)(a + 5) ANS!!

THINK. The 8 is the result of ADDING twonumbers, and the 15 is the result ofMULTIPLYING them! What arethey? + = 8 = 15

is 3 is 5




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Example 5 (this is Example 2 backwards)

Factorise y2 + 4y 45

Answer will look like (y)(y.)But because the end number is a NEGATIVE, the 2 numbers were looking for must be aPOSITIVE and a NEGATIVE.

The smartest way to go about this is tofirst find all pairs of numbers multiplying togive 45, then pick the pair which add to 4.

TRY 9 AND 5 ?? NO they add to 49 AND 5 ?? YES !!

ANSWER (y + 9) (y 5)

The minus makesthis a bit harder!Ouch!!

THINK. The +4 is the result of ADDING

two numbers, and the 45 is theresult of MULTIPLYING them! Whatare they? + = 4 = - 45

is -5 is +9

Example 6

Factorise x 2 7x + 12

Begin with the brackets ( x.)( x)

Look for 2 numbers which MULTIPLY TO MAKE +12 = +12ADD TO MAKE 7 .. + = 7

Numbers must be 3 and 4 !!Check 3 4 = +12 and

3 + 4 = 7 . YEP were right!!

Mult. to make +12 . They must beeither both positive or bothnegative!

Add to make 7. This means theymust be both negative !

(x 3)(x 4) ANSWER !!The Algebra Toolbox 2003 R. Bowman. All rights reserved.15


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Example 7

Factorise 2a2 10a 28

Aha!! Theres a common factor of 2. So take it out!!

2a2 10a 28= 2(a 2 5a 14)

Now just work on whats in the brackets, as a Type 3. Ignore the 2 for the timebeing.

Look for 2 numbers which MULTIPLY to make 14ADD to make 5

Numbers are 7 and +2 !!

= 2(a 7)(a + 2) ANSWER!!

Example 8

Factorise 15a 3a 2 18.

The first thing we should do is cutnpaste to get the terms in the right order (soit looks like ax 2 + bx + c format):

3a 2 + 15a 18.

Now we see theres a common factor of 3, but as theres a negative sign to beginwith, make the common factor 3, and take it out the front:

= 3 (a 2 5a + 6) Remembering about the sign changes inside the brackets(because of the negative term in front).

Taking 3 out as the common factor rather than 3 is a good ideabecause it leaves a normal-looking quadratic inside the brackets.

Now work on a 2 5a + 6 as a regular Type 3.Look for 2 numbers which multiply to give +6, add to give 5.Numbers are 3 and 2 !!

= 3 (a 3)(a 2)



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(f) FACTORISING - HARDER TYPE 3QUADRATICS

Up to now, the only quadratics weve been dealing with have been oneswhere the a value (in front of the x 2 ) is equal to 1.

In Examples 7 and 8 on the previous pages, the a value was not 1, but wewere able to take out a common factor which then led to a quadratic wherea was equal to 1.

Now we meet quadratics where a is something other than 1. Examplesinclude

(i) 2x2

+ 9x 5(ii) 3a 2 5a 2(iii) 3b 2 11b 20(iv) 6t 2 + 17t + 12

Note that in these, there are no common factors that can be taken out.

If the number in front of the x2 term is something other than 1, thenthis needs to be done by a different method.

There are a number of different methods of doing these, but Ill showyou what I believe to be the quickest and easiest. Lets do the onesabove.



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Example 1

Factorise 2x 2 + 9x 5

STEP 1: Set up brackets like this.

=2

......)2......)(2( x x

Had the leading term been 3x 2 rather than 2x 2, then all the 2swould be 3s !

STEP 2 Multiply a by c. 2 5 = 10. ac = 10b = 9.

Look for two numbers whichMULTIPLY to give 10 = 10

+ = 9

ADD to give 9

Numbers are+10 and 1

STEP 3 Insert these numbers into the brackets in the fraction above

=2

)12)(102( + x x

STEP 4 Cancel the 2 in the bottom into whichever bracket it will

divide fully into (in this case the first the 2 divides fully the 2xand the 10, whereas in the 2 nd bracket it will divide the 2x but notthe 1.)

= 2

)12)(102( + x x x + 5

STEP 5 Write the answer!

= ( x + 5)(2 x 1).. ANSWER!!!

STEP 6 CHECK by expanding this answer to see if you get the original.



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Example 2

Factorise 3a 2 5a 2

STEP 1: Set up brackets like this.

=3

......)3......)(3( aa

Had the leading term been 3x 2 rather than 2x 2, then all the 2swould be 3s !

STEP 2 Multiply a by c. 3 2 = 6. ac = 6b = 5 .

Look for two numbers whichMULTIPLY to give 6 = 6

+ = 5

ADD to give 5

Numbers are+1 and 6

STEP 3 Insert these numbers into the brackets in the fraction above

=3

)63)(13( + aa

STEP 4 Cancel the 3 in the bottom into whichever bracket it will

divide fully into (in this case the second the 3 divides fully the3x and the 6 , whereas in the 1 st bracket it will divide the 3xbut not the + 1.)

= 3

)63)(13( + aa

a - 2

STEP 5 Write the answer!

= (3a + 1)(a 2).. ANSWER!!!

STEP 6 CHECK by expanding this answer to see if you get the original.



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First decide whether Type 1 , Type 2 or Type 3 , then factorise!Remember in every question, to always look for a common factor first

1 a2 5a 11 3a2 752 t2 9 12 t2 14 5t3 16 25p 2 13 a + a 2

4 x2 5 x + 6 14 2t + t 2 155 z2 1 15 3a 2 12a + 366 4 49a 2 16 2y2 5y 37 5 x2 + 15 x 17 3a2 + 4a + 18 p2 + 4p + 3 18 p2 p9 p2 4p + 3 19 3b2 11b 20

10 y2 + y 6 20 6t2 + 13t + 6

CLICK HERE FOR ANSWERS



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2. QUADRATIC EQUATIONS - solving

The good news is. if you can understand factorising (Section 1 above)

then you can solve quadratic equations!!

In the previous section we learnt how to

FACTORISE QUADRATIC EXPRESSIONS

Now we learn how to

SOLVE QUADRATIC EQUATIONS

Heres an example see if you can spot what the differences are betweenthis example, and the examples in the previous section.

Example 1

Solve a2 + 7a + 10 = 0

STEP 1 Factorise the expression on the left side (Its a Type 3 - usethe skills learnt in the last section!) Check the example ifyouve forgotten!

(a + 2) (a + 5) = 0

STEP 2 We now have two things being multiplied ( ) to give zero. This means either one of them must be zero

Either a + 2 = 0So a = 2

Or a + 5 = 0So a = 5

a = 2 or 5 ANSWERS!!


easy!!!


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Example 2

Solve 3a2 12a = 0

STEP 1 Factorise the expression on the left side (Its a Type 1 - usethe skills learnt in the last section!) Check the example ifyouve forgotten!

3a (a 4) = 0

STEP 2 We now have two things being multiplied ( ) to give zero. This means either one of them must be zero

Note that if 3a = 0, this means3 times a is equal to 0.

The only thing a can be forthis to happen is zero!Either a 4 = 0So a = 4 or 3a = 0So a = 0

a = 0 or 4 ANSWERS!!

too easy!!!

NOTE. If you wanted to instead, you could have begun with thefirst line 3a 2 12a = 0, and then divided all 3 terms by 3. This wouldthen give you

3a 2 3 = a 2 12a 3 = 4a0 3 = 0

a 2 4a = 0

Proceeding as a Type 1 would then give

a (a 4) = 0

So either a = 0, or a 4 = 0

In other words, a = 0 or a = 4 (same answers as above!!)



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Example 3 (hard)

Solve 10t = 8 + 3t 2

STEP 1 Tidy it up. Cut n paste all terms to the one side, so ZERO ISON THE OTHER SIDE. Which side is best to move terms to?

I think its better to get the squared term (3t 2) so theres nonegative sign in front of it. Can you see then that we shouldaim to get the terms to the RIGHT? Well only need to movethe 10b.

10t = 8 + 3t 2 0 = 3t 2 10t + 8 Note weve slotted the 10t into its proper position (remember

the ax 2 + bx + c format?) & placed the 3t 2 at the front, 8 at theend

3t2 10t + 8 = 0 Turning everything around (makes it look more familiar when 0 ison the right side)

STEP 2 FACTORISE THE LEFT (Its a harder Type 3), using yourwonderful knowledge from the previous section! Check theexample if youve forgotten!

03

.......)3.......)(3( =t t Look for 2 numbers which MULTIPLY to give 24, ADD to 10

Theyre 6 and 4 . Insert these into brackets

03

)43)(63( = t t Cancelling gives

(t 2) (3t 4) = 0

STEP 3 Either/or.

Either t 2 = 0 or 3t 4 = 0t = 2 or 3t = 4

t = 4/3

t = 2 or t = 4/3.ANSWERS!!



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Solve these quadratics. Be sure to categorise them as Type 1, 2 or 3 first.Remember to look for a common factor before you do anything!

1 x2 5 x = 0 6 3a2 48 = 02 p2 16 = 0 7 x2 = x + 123 a2 + 5a + 6 = 0 8 8d + 2d 2 = 04 2b2 6b 56 = 0 9 2y2 + 13y 7 = 05 9y = y 2 10 3a2 = 11a + 4

CLICK HERE FOR ANSWERS !



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3. ALGEBRAIC FRACTIONS

This is a continuing source of stress for students from Year 9 up.

The key thing to remember here is there are times when you can cancelthe same terms in the top and bottom, and times you cant:

RIGHT WRONG

108 =

2524

=54 right! (cancel the 2s)

7353

108

++

= =75 (cant cancel the 3s)

108 is not the same as

75 !

573345

= 74

right! (cancel 3s & 5s) 573345

++++

is not the same as 74

!(cant cancel the 3s and 5s)

=>

> right! (cancel the > ) =

>

> wrong!

=>

> wrong!

cb

acab = right! (cancel the as)

cb

caba =

++

wrong!

(cant cancel the as)

c xy

abcabxy

22= right! (cancel the as & bs)

cb

caba =

wrong!

(cant cancel the as)

)()()(

y x y xba

+

= (a + b) right!

(cancel the ( x y)s)a

d acab ++ cannot be simplified!

Nothing can be cancelled!

x xy x +3

can be written as x y x )3( +

and then simplified by cancelling the

xs to give 3 + y.ANS!!

x x 23 +

cannot be simplified!Nothing can be cancelled!

cant cancelthe >



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Important note: The expressions in the left column are the ones wherewe can cancel things . Do you notice anything in common that theseexpressions in the left column have?

Every time we have cancelled something (in the left column), it isconnected to other terms by a multiplication. Cancelling cannot be donewhen the terms are connected by addition or subtraction , like the onesin the right column!

The general rule is:

If you have an algebraic fraction, and want to cancel the same termfrom top and bottom to simplify it, this can only be done provided all

terms are connected by multiplication! In other words, if you have any

terms connected by addition or subtraction signs, then cancelling cantbe done. The exception is when + or appears within a set of brackets .Read on

Sometimes the top or bottom may require some manipulation (e.g.factorising) before cancelling can be done.

Example 1

Simplify x

x x 32 +

x x x 32 + Cant cancel yet as theres an addition sign connecting 2 of the terms

= x

x x )3( + Factorising the top (check example ). This has now created 3 terms (x,

x+3 and x) and the only operation connecting any of them ismultiplication. The top really is x ( x+3 ). The addition sign, because itsin the brackets, is considered now to be part of the term (x + 3) and wontaffect the cancelling operation. This is what Im talking about in the box above (in blue)

= x + 3 Cancelling the x s. This is legal here because the 2 terms in the top (thex and the (x + 3)) are connected by multiplication and there are no otheraddition or subtraction operations connecting terms anywhere else.



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Example 2

Simplify9

652

2

+

x x x

9 6522

+

x x x

=)3)(3()3)(2(

+

x x x x

factorising top & bottom separately. Check examples .

Now there are four terms (x-2, x-3, x+3, x-3)The only operations connecting the 4 terms now aremultiplications, so we can cancel similar terms (x 3 ) appearing in top& bottom. Remember the and + signs appearing inside thebrackets are of no concern because they are not situated betweenterms.

= 3+ x2 x

This is as far as we can go. There are 4 terms present here (x, 2, x, 3)

and some of them are connected by addition and subtraction signs.Further factorising is impossible.Cancelling is therefore not allowed.

Example 3

Simplify y x

y x42

3

2

In this example, there are no additions or subtractions, so

Phew! This is hardyakka to explain (letalone understand!) Ihope youre getting it!

we can cancel pairs of similar terms, as long as one is in the topand one is in the bottom! (either fraction!)Begin by cancelling the ys (you can cancel diagonally)

=42

3

2

x

x

Now cancel the 2 and 4 (you can cancel vertically)

2 into 2 goes 12 into 4 goes 2

=23

2

x1 x

Now cancel the x and x 2 (again cancelling vertically) x into x goes 1x into x 2 goes x(like 3 into 3 2 goes 3)

Remember x 2 is x times x !

= 231

x

Nothing left to cancel (we can only cancel diagonally and vertically). Somultiply

along the top ( x times 1 is x ) andalong the bottom (3 times 2y is 6y)

=6 x ANSWER !!



28/30


Determine which of these expressions can be simplified. Simplify those thatcan be done.

1ad ab 6

2)2)(2(

++

x x x

2 xy

xyz 5

7 x

x x3

52

3 x

y x2

8

242

+

x x

43

)3(

aaa 9

1823

2 +

x x

5a

aa )1( + 1065

122

2

++

x x x x

CLICK HERE FOR ANSWERS



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ANSWERS Exercise 1

Quadratic Values of a, b and c? Type 1, 2 or3?

1 2a2 + 3a + 5 a = 2, b = 3, c = 5 Type 3

2 5y 2 8y a = 5, b = 8, c = 0 Type 13 4 x2 11 x a = 4, b = 11 Type 14 y2 16 a = 1, b = 0, c = 16 Type 25 25 + y +3y 2 a = 3, b = 1, c = 25 Type 36 t 2 32 a = , b = 0, c = 32 Type 27 3b 2 + b + 7 a = 3, b = 1 , c = 7 Type 38 3t2 a = 3, b = 0, c = 0 Type 1 or 29 7b + 8 3b 2 a = 3, b= 7, c = 8 Type 310 4t a = 0, b = 4, c = 0 Not a quadraticNote Q8 can be categorised as either Type 1 or 2Q10 is not a quadratic because the squared term is missing. To be a quadratic,there must be a squared term!!

ANSWERS Exercise 2

(1) 6a 2 + 23a + 20 Type 3; (2) 49a 2 4 Type 2; (3) 6x 2 15x Type 1; (4) b 2 + 4b + 4 Type 3;(5) 9a 2 25 Type 2; (6) 5y 2 7y Type 1; (7) 9 4p 2 Type 2; (8) a 2 6a + 9 Type 3;(9) 2b 2 5b 25 Type 3; (10) 9y 2 16 Type 2

ANSWERS Exercise 3

(1) a(a 5) Type 1(2) (t 3)(t + 3) Type 2(3) (4 5p)(4 + 5p) Type 2(4) (x 3)(x 2) Type 3(5) (z 1)(z + 1) Type 2(6) (2 7a)(2 + 7a) Type 2(7) 5x(x + 3) Type 1(8) (p + 1)(p + 3) Type 3(9) (p 1)(p 3) Type 3(10) (y + 3)(y 2) Type 3(11) 3(a 5)(a + 5) Common factor out first then Type 2(12) (t 7)(t + 2) Cut n paste first to get t 2 5t 14 then Type 3(13) a(1 + a) Type 1(14) (t + 5)(t 3) Cut n paste first to get t 2 + 2t 15 then Type 3

(15) 3 (a + 6)(a 2) Take out common factor ( 3) then Type 3(16) (2y + 1)(y 3) Type 3 harder(17) (3a + 1)(a + 1) Type 3 harder(18) p(p 1) Type 1(19) (3b + 4 )(b 5 ) Type 3 harder(20) (3t + 2)(2t + 3) Type 3 harder



30/30

ANSWERS Exercise 4

(1) x = 0 or 5 (2) p = 4 or 4 (3) a = 2 or 3 (4) b = 7 or 4 (5) y = 0 or 9(6) a = 4 or 4 (7) x = 4 or 3 (8) d = 0 or 4 (9) y = or 7 (10) a = 4 or 1/3

ANSWERS Exercise 5

(1)d b

(2)5 z

(3) cant be done (4) a (5) a+1 (6) x 2 (7)3

5 x

(8) x 2 (9))3(2

1 x

(10)24

+

x x

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Algebra Toolbox Part 2