Factorising - Numeracy Workshop...workshopExpressions and Expansion). Topics include extracting common factors, factorising quadratic expressions and polynomials. Workshop resources:These

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FactorisingNumeracy Workshop

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[email protected] Factorising 2 / 43

IntroductionThese slides extend on a basic knowledge of algebra (such as the previous Algebraworkshop Expressions and Expansion). Topics include extracting common factors,factorising quadratic expressions and polynomials.

Workshop resources: These slides are available online:

www.studysmarter.uwa.edu.au → Numeracy and Maths → Online Resources

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Email: [email protected]@uwa.edu.au Factorising 3 / 43

Factors of numbers

A factor of a number is a number that divides into it evenly.

Example: 4 is a factor of 12 since 3× 4 = 12.

Of course, 3 is also a factor of 12. (The others are 1, 2, 6 and 12.)


Factors of numbers





Factors of numbers



Of course, 3 is also a factor of 12.

(The others are 1, 2, 6 and 12.)


Factors of numbers



Of course, 3 is also a factor of 12. (The others are

1, 2, 6 and 12.)


Factors of numbers





Factors of terms

A factor of a term is a number, variable or combination that divides into it evenly.

Example: The term 12x has a factor of 4 since 3x × 4 = 12x .

Important point: the variable x is also a factor of 12x since 12× x = 12x . (Thisis true even though x could be any number, including non-whole numbers.)

Factors of 12x can be made up of combinations of other factors, such as 2, 6x, 1,3x, 12, etc.


Factors of terms






Factors of terms






Factors of terms




Factors of 12x can be made up of combinations of other factors, such as

2, 6x, 1,3x, 12, etc.


Factors of terms






Factors of expressions

In the previous algebra workshop we looked at the expression

4(2x + 3)

This has two clear factors, 4 and 2x + 3 because

4× (2x + 3) = 4(2x + 3)

When we expanded the brackets, we got

4(2x + 3) = 8x + 12

Our task now is to reverse this process. That is, take an expression like8x + 12 and extract the factors which are common to both terms (8x and 12).

This process is called factorisation.




4(2x + 3)

This has two clear factors,

4 and 2x + 3 because

4× (2x + 3) = 4(2x + 3)


4(2x + 3) = 8x + 12






4(2x + 3)

This has two clear factors, 4 and

2x + 3 because

4× (2x + 3) = 4(2x + 3)


4(2x + 3) = 8x + 12






4(2x + 3)


4× (2x + 3) = 4(2x + 3)


4(2x + 3) = 8x + 12






4(2x + 3)


4× (2x + 3) = 4(2x + 3)


4(2x + 3) = 8x + 12






4(2x + 3)


4× (2x + 3) = 4(2x + 3)


4(2x + 3) = 8x + 12

Our task now is to reverse this process.

That is, take an expression like8x + 12 and extract the factors which are common to both terms (8x and 12).





4(2x + 3)


4× (2x + 3) = 4(2x + 3)


4(2x + 3) = 8x + 12






4(2x + 3)


4× (2x + 3) = 4(2x + 3)


4(2x + 3) = 8x + 12




Factorisation example

Example: Factorise 28x + 98x2.

What factors do the terms 28x and 98x2 have in common?

It might help to imagine some multiplication signs:

28× x and 98× x × x

So, x is a factor common to both terms. Both numbers are even:

2× 14× x and 2× 49× x × x

So, 2 is also a common factor. In fact, this step reveals another one.

2× 2× 7× x and 2× 7× 7× x × x

7 is also a common factor.






28× x and 98× x × x


2× 14× x and 2× 49× x × x


2× 2× 7× x and 2× 7× 7× x × x







28× x and 98× x × x


2× 14× x and 2× 49× x × x


2× 2× 7× x and 2× 7× 7× x × x







28× x and 98× x × x

So, x is a factor common to both terms.

Both numbers are even:

2× 14× x and 2× 49× x × x


2× 2× 7× x and 2× 7× 7× x × x







28× x and 98× x × x


2× 14× x and 2× 49× x × x


2× 2× 7× x and 2× 7× 7× x × x







28× x and 98× x × x


2× 14× x and 2× 49× x × x


2× 2× 7× x and 2× 7× 7× x × x







28× x and 98× x × x


2× 14× x and 2× 49× x × x

So, 2 is also a common factor.

In fact, this step reveals another one.

2× 2× 7× x and 2× 7× 7× x × x







28× x and 98× x × x


2× 14× x and 2× 49× x × x


2× 2× 7× x and 2× 7× 7× x × x







28× x and 98× x × x


2× 14× x and 2× 49× x × x


2× 2× 7× x and 2× 7× 7× x × x







28× x and 98× x × x


2× 14× x and 2× 49× x × x


2× 2× 7× x and 2× 7× 7× x × x





2× 2× 7× x and 2× 7× 7× x × x

The combination of common factors is 2× 7× x = 14x .

We call this the highest common factor and write it outside some brackets:

28x + 98x2 = 14x(

2 + 7x

)

The remaining bits of both terms go inside the brackets.

This is the fully factorised form of the original expression.

Note: You can check your answer by expanding the factorised form.




2× 2× 7× x and 2× 7× 7× x × x



28x + 98x2 = 14x(

2 + 7x

)







2× 2× 7× x and 2× 7× 7× x × x



28x + 98x2 = 14x(

2 + 7x

)







2× 2× 7× x and 2× 7× 7× x × x



28x + 98x2 = 14x(

2 + 7x

)







2× 2× 7× x and 2× 7× 7× x × x



28x + 98x2 = 14x(

2 + 7x

)







2× 2× 7× x and 2× 7× 7× x × x



28x + 98x2 = 14x(2 + 7x)







2× 2× 7× x and 2× 7× 7× x × x



28x + 98x2 = 14x(2 + 7x)







2× 2× 7× x and 2× 7× 7× x × x



28x + 98x2 = 14x(2 + 7x)





Factorisation exercises

6x − 3x2 =

3x(2− x)

8y + 16y2 = 8y(1 + 2y)

(Note the use of the “hidden” factor of 1 in the first term.)

2x2y + 4x3 = 2x2(y + 2x)

2xy − 4x + 3x2 = x(2y − 4 + 3x)

(Factors must be common to all terms.)



6x − 3x2 = 3x(

2− x

)

8y + 16y2 = 8y(1 + 2y)


2x2y + 4x3 = 2x2(y + 2x)

2xy − 4x + 3x2 = x(2y − 4 + 3x)




6x − 3x2 = 3x(2− x)

8y + 16y2 = 8y(1 + 2y)


2x2y + 4x3 = 2x2(y + 2x)

2xy − 4x + 3x2 = x(2y − 4 + 3x)




6x − 3x2 = 3x(2− x)

8y + 16y2 =

8y(1 + 2y)


2x2y + 4x3 = 2x2(y + 2x)

2xy − 4x + 3x2 = x(2y − 4 + 3x)




6x − 3x2 = 3x(2− x)

8y + 16y2 = 8y(

1 + 2y

)


2x2y + 4x3 = 2x2(y + 2x)

2xy − 4x + 3x2 = x(2y − 4 + 3x)




6x − 3x2 = 3x(2− x)

8y + 16y2 = 8y(1 + 2y)


2x2y + 4x3 = 2x2(y + 2x)

2xy − 4x + 3x2 = x(2y − 4 + 3x)




6x − 3x2 = 3x(2− x)

8y + 16y2 = 8y(1 + 2y)


2x2y + 4x3 = 2x2(y + 2x)

2xy − 4x + 3x2 = x(2y − 4 + 3x)




6x − 3x2 = 3x(2− x)

8y + 16y2 = 8y(1 + 2y)


2x2y + 4x3 =

2x2(y + 2x)

2xy − 4x + 3x2 = x(2y − 4 + 3x)




6x − 3x2 = 3x(2− x)

8y + 16y2 = 8y(1 + 2y)


2x2y + 4x3 = 2x2(

y + 2x

)

2xy − 4x + 3x2 = x(2y − 4 + 3x)




6x − 3x2 = 3x(2− x)

8y + 16y2 = 8y(1 + 2y)


2x2y + 4x3 = 2x2(y + 2x)

2xy − 4x + 3x2 = x(2y − 4 + 3x)




6x − 3x2 = 3x(2− x)

8y + 16y2 = 8y(1 + 2y)


2x2y + 4x3 = 2x2(y + 2x)

2xy − 4x + 3x2 =

x(2y − 4 + 3x)




6x − 3x2 = 3x(2− x)

8y + 16y2 = 8y(1 + 2y)


2x2y + 4x3 = 2x2(y + 2x)

2xy − 4x + 3x2 = x(

2y − 4 + 3x

)




6x − 3x2 = 3x(2− x)

8y + 16y2 = 8y(1 + 2y)


2x2y + 4x3 = 2x2(y + 2x)

2xy − 4x + 3x2 = x(2y − 4 + 3x)




6x − 3x2 = 3x(2− x)

8y + 16y2 = 8y(1 + 2y)


2x2y + 4x3 = 2x2(y + 2x)

2xy − 4x + 3x2 = x(2y − 4 + 3x)



Double bracket expressions

Here is a common type of double bracket expression.

Let’s expand the brackets tosee whether there are any useful patterns we can use to reverse the process:

(x + 2)(x + 5) =

x2 + 5x+ 2x + 2× 5

= x2 + (5 + 2)x + 2× 5

= x2 + 7x + 10

The pattern is here: 5 + 2 = 7 and 2× 5 = 10.



Here is a common type of double bracket expression. Let’s expand the brackets tosee whether there are any useful patterns we can use to reverse the process:

(x + 2)(x + 5) =

x2 + 5x+ 2x + 2× 5

= x2 + (5 + 2)x + 2× 5

= x2 + 7x + 10





(x + 2)(x + 5) =

x2 + 5x+ 2x + 2× 5

= x2 + (5 + 2)x + 2× 5

= x2 + 7x + 10





(x + 2)(x + 5) = x2

+ 5x+ 2x + 2× 5

= x2 + (5 + 2)x + 2× 5

= x2 + 7x + 10





(x + 2)(x + 5) = x2

+ 5x+ 2x + 2× 5

= x2 + (5 + 2)x + 2× 5

= x2 + 7x + 10





(x + 2)(x + 5) = x2 + 5x

+ 2x + 2× 5

= x2 + (5 + 2)x + 2× 5

= x2 + 7x + 10





(x + 2)(x + 5) = x2 + 5x

+ 2x + 2× 5

= x2 + (5 + 2)x + 2× 5

= x2 + 7x + 10





(x + 2)(x + 5) = x2 + 5x+ 2x

+ 2× 5

= x2 + (5 + 2)x + 2× 5

= x2 + 7x + 10





(x + 2)(x + 5) = x2 + 5x+ 2x

+ 2× 5

= x2 + (5 + 2)x + 2× 5

= x2 + 7x + 10





(x + 2)(x + 5) = x2 + 5x+ 2x + 2× 5

= x2 + (5 + 2)x + 2× 5

= x2 + 7x + 10





(x + 2)(x + 5) = x2 + 5x+ 2x + 2× 5

= x2 + (5 + 2)x + 2× 5

= x2 + 7x + 10





(x + 2)(x + 5) = x2 + 5x+ 2x + 2× 5

= x2 + (5 + 2)x + 2× 5

= x2 + 7x + 10





(x + 2)(x + 5) = x2 + 5x+ 2x + 2× 5

= x2 + (5 + 2)x + 2× 5

= x2 + 7x + 10



A Rule

So, in general, whenever we are asked to expand an expression of the form:

(x + a)(x + b)

we always end up with

x2 + (a + b)x + ab

That is, you always add the two numbers a and b together to get the numbermultiplying x , and you multiply them to get the constant term.


A Rule

So, in general, whenever we are asked to expand an expression of the form:

(x + a)(x + b)

we always end up with

x2 + (a + b)x + ab

That is, you always add the two numbers a and b together to get the numbermultiplying x , and you multiply them to get the constant term.


Factorisation

If we are asked to factorise:

x2 + 5x + 6

then we are being asked to write the above in the form

(x + a)(x + b)

where a + b = 5 and a× b = 6. Can you find two numbers which do this?

The answer is 2 and 3 (2 + 3 = 5, 2× 3 = 6).

So we can factorise the expression as follows:

x2 + 5x + 6 = (x + 2)(x + 3)


Factorisation


x2 + 5x + 6


(x + a)(x + b)


The answer is 2 and 3 (2 + 3 = 5, 2× 3 = 6).


x2 + 5x + 6 = (x + 2)(x + 3)


Factorisation


x2 + 5x + 6


(x + a)(x + b)


The answer is 2 and 3 (2 + 3 = 5, 2× 3 = 6).


x2 + 5x + 6 = (x + 2)(x + 3)


Factorisation


x2 + 5x + 6


(x + a)(x + b)


The answer is 2 and 3 (2 + 3 = 5, 2× 3 = 6).


x2 + 5x + 6 = (x + 2)(x + 3)


Factorisation

Factorise x2 − 3x − 10

So x2 − 3x − 10 = (x + a)(x + b)

where a + b = −3 and a× b = −10. Which two numbers do this?

The answer is −5 and 2 (−5 + 2 = −3, −5× 2 = −10).


x2 − 3x − 10 = (x − 5)(x + 2)


Factorisation


So x2 − 3x − 10 = (x + a)(x + b)


The answer is −5 and 2 (−5 + 2 = −3, −5× 2 = −10).


x2 − 3x − 10 = (x − 5)(x + 2)


Factorisation


So x2 − 3x − 10 = (x + a)(x + b)


The answer is −5 and 2 (−5 + 2 = −3, −5× 2 = −10).


x2 − 3x − 10 = (x − 5)(x + 2)


Factorisation


So x2 − 3x − 10 = (x + a)(x + b)


The answer is −5 and 2 (−5 + 2 = −3, −5× 2 = −10).


x2 − 3x − 10 = (x − 5)(x + 2)


Removing Factors

We have seen a method which usually works when the multiplier of x2 isequal to 1.

What if we were asked to factorise:

3x2 + 9x + 6

We notice that the multipliers of all three terms in the above expression aredivisible by 3. Hence, we can factor out this common factor as follows:

3x2 + 9x + 6 = 3(x2 + 3x + 2)

Now, the expression in brackets is just like we saw on the previous slides, we lookfor two numbers which add to 3 and multiply to 2. This gives us:

3x2 + 9x + 6 = 3(x + 1)(x + 2)


Removing Factors



3x2 + 9x + 6


3x2 + 9x + 6 = 3(x2 + 3x + 2)


3x2 + 9x + 6 = 3(x + 1)(x + 2)


Removing Factors



3x2 + 9x + 6


3x2 + 9x + 6 = 3(x2 + 3x + 2)


3x2 + 9x + 6 = 3

(x + 1)(x + 2)


Removing Factors



3x2 + 9x + 6


3x2 + 9x + 6 = 3(x2 + 3x + 2)


3x2 + 9x + 6 = 3(x + 1)(x + 2)


Removing Factors

Factorise the following expression.

5x2 + 40x + 60

Here we see that the multiplier of x2 is 5. We also notice that all multipliers inthe above expression are divisible by 5. Hence, we can factor out this common

factor as follows:

5x2 + 40x + 60 = 5(x2 + 8x + 12)

Now we look for two numbers which add to 8 and multiply to 12. This gives us:

5x2 + 40x + 60 = 5(x + 2)(x + 6)


Removing Factors


5x2 + 40x + 60


factor as follows:

5x2 + 40x + 60 = 5(x2 + 8x + 12)


5x2 + 40x + 60 = 5(x + 2)(x + 6)


Removing Factors


5x2 + 40x + 60


factor as follows:

5x2 + 40x + 60 = 5(x2 + 8x + 12)


5x2 + 40x + 60 = 5

(x + 2)(x + 6)


Removing Factors


5x2 + 40x + 60


factor as follows:

5x2 + 40x + 60 = 5(x2 + 8x + 12)


5x2 + 40x + 60 = 5(x + 2)(x + 6)


Factorisation


x2 − 9


(x + a)(x + b)

It looks different to the expressions we have been factorising because it appears tohave no x term. However, we can make it the same if we include an x term with a

multiplier of 0:

x2+0x − 9

So we know that we need two numbers which add up to 0 and multiply to give−9. Can you find two numbers which do this?

The answer is 3 and −3 (3 +−3 = 0, 3×−3 = −9).


x2 − 9 = (x + 3)(x − 3)


Factorisation


x2 − 9


(x + a)(x + b)

It looks different to the expressions we have been factorising because it appears tohave no x term.

However, we can make it the same if we include an x term with amultiplier of 0:

x2+0x − 9


The answer is 3 and −3 (3 +−3 = 0, 3×−3 = −9).


x2 − 9 = (x + 3)(x − 3)


Factorisation


x2 − 9


(x + a)(x + b)


multiplier of

0:

x2+0x − 9


The answer is 3 and −3 (3 +−3 = 0, 3×−3 = −9).


x2 − 9 = (x + 3)(x − 3)


Factorisation


x2 − 9


(x + a)(x + b)


multiplier of 0:

x2+0x − 9


The answer is 3 and −3 (3 +−3 = 0, 3×−3 = −9).


x2 − 9 = (x + 3)(x − 3)


Factorisation


x2 − 9


(x + a)(x + b)


multiplier of 0:

x2+0x − 9


The answer is 3 and −3 (3 +−3 = 0, 3×−3 = −9).


x2 − 9 = (x + 3)(x − 3)


Factorisation


x2 − 9


(x + a)(x + b)


multiplier of 0:

x2+0x − 9


The answer is 3 and −3 (3 +−3 = 0, 3×−3 = −9).


x2 − 9 = (x + 3)(x − 3)


Factorisation


x2 − 9


(x + a)(x + b)


multiplier of 0:

x2+0x − 9


The answer is 3 and −3 (3 +−3 = 0, 3×−3 = −9).


x2 − 9 = (x + 3)(x − 3)


The Difference of Two Squares

If you are asked to factorise x2 − 16 we get:

(x + 4)(x − 4)


(x + 6)(x − 6)

This leads us to a general formula called the difference of two squares:

x2 − a2 = (x + a)(x − a)

Tip: Try to avoid memorising too many formulas. It’s handy to remember thatthis is just a simple special case of a more general process.




(x + 4)(x − 4)


(x + 6)(x − 6)


x2 − a2 = (x + a)(x − a)





(x + 4)(x − 4)


(x + 6)(x − 6)


x2 − a2 = (x + a)(x − a)





(x + 4)(x − 4)


(x + 6)(x − 6)


x2 − a2 = (x + a)(x − a)





(x + 4)(x − 4)


(x + 6)(x − 6)


x2 − a2 = (x + a)(x − a)





(x + 4)(x − 4)


(x + 6)(x − 6)


x2 − a2 = (x + a)(x − a)



The Difference of Two Squares: Examples

x2 − 25 =

(x − 5)(x + 5)

x2 − 49 = (x + 7)(x − 7)

4x2 − 25 = (2x)2 − 52

= (2x − 5)(2x + 5)

9x4 − 64 =(3x2

)2 − 82

=(3x2 − 8

) (3x2 + 8

)



x2 − 25 = (x − 5)(x + 5)

x2 − 49 = (x + 7)(x − 7)

4x2 − 25 = (2x)2 − 52

= (2x − 5)(2x + 5)

9x4 − 64 =(3x2

)2 − 82

=(3x2 − 8

) (3x2 + 8

)



x2 − 25 = (x − 5)(x + 5)

x2 − 49 =

(x + 7)(x − 7)

4x2 − 25 = (2x)2 − 52

= (2x − 5)(2x + 5)

9x4 − 64 =(3x2

)2 − 82

=(3x2 − 8

) (3x2 + 8

)



x2 − 25 = (x − 5)(x + 5)

x2 − 49 = (x + 7)(x − 7)

4x2 − 25 = (2x)2 − 52

= (2x − 5)(2x + 5)

9x4 − 64 =(3x2

)2 − 82

=(3x2 − 8

) (3x2 + 8

)



x2 − 25 = (x − 5)(x + 5)

x2 − 49 = (x + 7)(x − 7)

4x2 − 25 =

(2x)2 − 52

= (2x − 5)(2x + 5)

9x4 − 64 =(3x2

)2 − 82

=(3x2 − 8

) (3x2 + 8

)



x2 − 25 = (x − 5)(x + 5)

x2 − 49 = (x + 7)(x − 7)

4x2 − 25 = (2x)2 − 52

=

(2x − 5)(2x + 5)

9x4 − 64 =(3x2

)2 − 82

=(3x2 − 8

) (3x2 + 8

)



x2 − 25 = (x − 5)(x + 5)

x2 − 49 = (x + 7)(x − 7)

4x2 − 25 = (2x)2 − 52

= (2x − 5)(2x + 5)

9x4 − 64 =(3x2

)2 − 82

=(3x2 − 8

) (3x2 + 8

)



x2 − 25 = (x − 5)(x + 5)

x2 − 49 = (x + 7)(x − 7)

4x2 − 25 = (2x)2 − 52

= (2x − 5)(2x + 5)

9x4 − 64 =

(3x2

)2 − 82

=(3x2 − 8

) (3x2 + 8

)



x2 − 25 = (x − 5)(x + 5)

x2 − 49 = (x + 7)(x − 7)

4x2 − 25 = (2x)2 − 52

= (2x − 5)(2x + 5)

9x4 − 64 =(3x2

)2 − 82

=

(3x2 − 8

) (3x2 + 8

)



x2 − 25 = (x − 5)(x + 5)

x2 − 49 = (x + 7)(x − 7)

4x2 − 25 = (2x)2 − 52

= (2x − 5)(2x + 5)

9x4 − 64 =(3x2

)2 − 82

=(3x2 − 8

) (3x2 + 8

)



Sometimes we need to factor out the highest common factor:

3x2 − 75 =

3(x2 − 25)

= 3(x + 5)(x − 5)

2x2 − 8 = 2(x2 − 4)

= 2(x + 2)(x − 2)

x3 − 25x = x(x2 − 25)

= x(x − 5)(x + 5)

18x3 − 32x = 2x(9x2 − 16)

= 2x(3x + 4)(3x − 4)




3x2 − 75 = 3(x2 − 25)

= 3

(x + 5)(x − 5)

2x2 − 8 = 2(x2 − 4)

= 2(x + 2)(x − 2)

x3 − 25x = x(x2 − 25)

= x(x − 5)(x + 5)

18x3 − 32x = 2x(9x2 − 16)

= 2x(3x + 4)(3x − 4)




3x2 − 75 = 3(x2 − 25)

= 3(x + 5)(x − 5)

2x2 − 8 = 2(x2 − 4)

= 2(x + 2)(x − 2)

x3 − 25x = x(x2 − 25)

= x(x − 5)(x + 5)

18x3 − 32x = 2x(9x2 − 16)

= 2x(3x + 4)(3x − 4)




3x2 − 75 = 3(x2 − 25)

= 3(x + 5)(x − 5)

2x2 − 8 =

2(x2 − 4)

= 2(x + 2)(x − 2)

x3 − 25x = x(x2 − 25)

= x(x − 5)(x + 5)

18x3 − 32x = 2x(9x2 − 16)

= 2x(3x + 4)(3x − 4)




3x2 − 75 = 3(x2 − 25)

= 3(x + 5)(x − 5)

2x2 − 8 = 2(x2 − 4)

= 2

(x + 2)(x − 2)

x3 − 25x = x(x2 − 25)

= x(x − 5)(x + 5)

18x3 − 32x = 2x(9x2 − 16)

= 2x(3x + 4)(3x − 4)




3x2 − 75 = 3(x2 − 25)

= 3(x + 5)(x − 5)

2x2 − 8 = 2(x2 − 4)

= 2(x + 2)(x − 2)

x3 − 25x = x(x2 − 25)

= x(x − 5)(x + 5)

18x3 − 32x = 2x(9x2 − 16)

= 2x(3x + 4)(3x − 4)




3x2 − 75 = 3(x2 − 25)

= 3(x + 5)(x − 5)

2x2 − 8 = 2(x2 − 4)

= 2(x + 2)(x − 2)

x3 − 25x =

x(x2 − 25)

= x(x − 5)(x + 5)

18x3 − 32x = 2x(9x2 − 16)

= 2x(3x + 4)(3x − 4)




3x2 − 75 = 3(x2 − 25)

= 3(x + 5)(x − 5)

2x2 − 8 = 2(x2 − 4)

= 2(x + 2)(x − 2)

x3 − 25x = x(x2 − 25)

= x

(x − 5)(x + 5)

18x3 − 32x = 2x(9x2 − 16)

= 2x(3x + 4)(3x − 4)




3x2 − 75 = 3(x2 − 25)

= 3(x + 5)(x − 5)

2x2 − 8 = 2(x2 − 4)

= 2(x + 2)(x − 2)

x3 − 25x = x(x2 − 25)

= x(x − 5)(x + 5)

18x3 − 32x = 2x(9x2 − 16)

= 2x(3x + 4)(3x − 4)




3x2 − 75 = 3(x2 − 25)

= 3(x + 5)(x − 5)

2x2 − 8 = 2(x2 − 4)

= 2(x + 2)(x − 2)

x3 − 25x = x(x2 − 25)

= x(x − 5)(x + 5)

18x3 − 32x =

2x(9x2 − 16)

= 2x(3x + 4)(3x − 4)




3x2 − 75 = 3(x2 − 25)

= 3(x + 5)(x − 5)

2x2 − 8 = 2(x2 − 4)

= 2(x + 2)(x − 2)

x3 − 25x = x(x2 − 25)

= x(x − 5)(x + 5)

18x3 − 32x = 2x(9x2 − 16)

= 2x

(3x + 4)(3x − 4)




3x2 − 75 = 3(x2 − 25)

= 3(x + 5)(x − 5)

2x2 − 8 = 2(x2 − 4)

= 2(x + 2)(x − 2)

x3 − 25x = x(x2 − 25)

= x(x − 5)(x + 5)

18x3 − 32x = 2x(9x2 − 16)

= 2x(3x + 4)(3x − 4)


Harder Factorisation

What if we can’t easily factor out the multiplier of x2?

First, consider expanding

(7x + 4)(3x + 5)

If we do this we get

21x2 + 35x + 12x + 20

which then equals

21x2 + 47x + 20

Expansion is easy, but doing this problem backwards is tricky if we don’t knowwhere we started. Our previous methods don’t work here.





(7x + 4)(3x + 5)


21x2 + 35x + 12x + 20

which then equals

21x2 + 47x + 20






(7x + 4)(3x + 5)


21x2 + 35x + 12x + 20

which then equals

21x2 + 47x + 20






(7x + 4)(3x + 5)


21x2 + 35x + 12x + 20

which then equals

21x2 + 47x + 20






(7x + 4)(3x + 5)


21x2 + 35x + 12x + 20

which then equals

21x2 + 47x + 20




In general, when we see an expression of the form

Ax2 + Bx + C

we want to factorise it by writing it in the following form:

(ax + b)(cx + d)

Note that a and c multiply to ptoduce the multiplier of x2 (A), and that b and dmultiply to produce the constant on the end (C).

Then we need to play around with it a bit.




Ax2 + Bx + C


(ax + b)(cx + d)






Ax2 + Bx + C


(ax + b)(cx + d)




Harder Factorisation: Example

Factorise 2x2 + 3x + 1.

We need to write this in the form (ax + b)(cx + d).

The numbers a and c must multiply up to 2, and so one of them must be 1 andthe other must be 2. (It doesn’t matter which is which because multiplication is

commutative.)

(2x + b)(x + d)

The numbers b and d must multiply up to 1, and so one of them must be 1 andthe other must be 1.

(2x + 1)(x + 1)






commutative.)

(2x + b)(x + d)


(2x + 1)(x + 1)





The numbers a and c must multiply up to 2, and so one of them must be 1 andthe other must be 2.

(It doesn’t matter which is which because multiplication iscommutative.)

(2x + b)(x + d)


(2x + 1)(x + 1)






commutative.)

(2x + b)(x + d)


(2x + 1)(x + 1)






commutative.)

(2x + b)(x + d)


(2x + 1)(x + 1)





The numbers a and c must multiply to 7, and so one of them must be 1 and theother must be 7. It doesn’t matter which is which.

(7x + b)(x + d)

The numbers b and d must multiply to 2, so one of them must be 1 and the othermust be 2. The question is, which one is which? There are two possibilities:

(7x + 2)(x + 1) (7x + 1)(x + 2)

To decide which one is correct, expand them both:

7x2 + 9x + 2 7x2 + 15x + 2






(7x + b)(x + d)


(7x + 2)(x + 1) (7x + 1)(x + 2)


7x2 + 9x + 2 7x2 + 15x + 2






(7x + b)(x + d)


(7x + 2)(x + 1) (7x + 1)(x + 2)


7x2 + 9x + 2 7x2 + 15x + 2






(7x + b)(x + d)


(7x + 2)(x + 1) (7x + 1)(x + 2)


7x2 + 9x + 2 7x2 + 15x + 2






(7x + b)(x + d)


(7x + 2)(x + 1) (7x + 1)(x + 2)


7x2 + 9x + 2 7x2 + 15x + 2






(7x + b)(x + d)


(7x + 2)(x + 1) (7x + 1)(x + 2)


7x2 + 9x + 2

7x2 + 15x + 2






(7x + b)(x + d)


(7x + 2)(x + 1) (7x + 1)(x + 2)


7x2 + 9x + 2 7x2 + 15x + 2






(7x + b)(x + d)


(7x + 2)(x + 1) (7x + 1)(x + 2)


7x2 + 9x + 2 7x2 + 15x + 2



Factorise 6x2 − 11x − 10.


The problem now is that both 6 and 10 have multiple possible factorisations!There are in fact 16 potential answers to test.

The best way to navigate through these options is with a combination of educatedguessing and trial-and-error. Start by writing out the potential factorisations of 6

and 10 as follows:

3

2

5

2

10

1

-

-

��

��*HHHHHHj-

-

In this case, the number term (−10) is negative so we need to get a pair whosedifference is 11.

6=







and 10 as follows:

3

2

5

2

10

1

-

-

��

��*HHHHHHj-

-


6=







and 10 as follows:

3

2

5

2

10

1

-

-

��

��*HHHHHHj-

-


6=






The best way to navigate through these options is with a combination of educatedguessing and trial-and-error.

Start by writing out the potential factorisations of 6and 10 as follows:

3

2

5

2

10

1

-

-

��

��*HHHHHHj-

-


6=







and 10 as follows:

3

2

5

2

10

1

-

-

��

��*HHHHHHj-

-


6=







and 10 as follows:

3

2

6

1

5

2

10

1

-

-

��

��*HHHHHHj-

-








and 10 as follows:

3

2

6

1

5

2

10

1

-

-

��

��*HHHHHHj-

-








and 10 as follows:

3

2

6

1

5

2

10

1

-

-

��

��*HHHHHHj-

-


1× 2 = 2 and 6× 5 = 30. Difference 6= 11. Try [email protected] Factorising 25 / 43






and 10 as follows:

3

2

6

1

5

2

10

1

-

-

��

��*HHHHHHj

-

-


1× 5 = 5 and 6× 2 = 12. Difference 6= 11. Try [email protected] Factorising 25 / 43






and 10 as follows:

3

2

6

1

5

2

10

1

-

-

��

��*HHHHHHj

-

-


2× 2 = 4 and 3× 5 = 15. Difference = 11. We have a [email protected] Factorising 25 / 43




We have found that 2× 2 = 4 and 3× 5 = 15.

(2x5)(3x2)

All we need to do now is place the “+” and “−” signs in the appropriate brackets.

So our answer is

6x2 − 11x − 10 = (2x − 5)(3x + 2).






(2x 5)(3x 2)


So our answer is

6x2 − 11x − 10 = (2x − 5)(3x + 2).






(2x 5)(3x 2)


So our answer is

6x2 − 11x − 10 = (2x − 5)(3x + 2).






(2x − 5)(3x + 2)


So our answer is

6x2 − 11x − 10 = (2x − 5)(3x + 2).






(2x − 5)(3x + 2)


So our answer is

6x2 − 11x − 10 = (2x − 5)(3x + 2).


A final note on double bracket factorisations

Note: Not all expressions of the form ax2 + bx + c can be factorised into twobrackets easily.

Some can’t be facorised at all. For example

x2 + 2x + 2

looks easy but cannot be factorised.


A final note on double bracket factorisations

Note: Not all expressions of the form ax2 + bx + c can be factorised into twobrackets easily. Some can’t be facorised at all. For example

x2 + 2x + 2

looks easy but cannot be factorised.


Why is factorising useful?

Factorising is handy for simplifying expressions and equations, which makesformulas more efficient to use and problems easier to solve.

Solve for x : x2 = 6x .

x2 − 6x = 0 (get x terms together)

x(x − 6) = 0 (factorise)

We know that 0× a = 0, whatever a is, so only one of the two factors aboveneeds to be 0 to solve the equation:

Either x = 0 or x − 6 = 0.

Hence, the solutions are x = 0 or 6.








Either x = 0 or x − 6 = 0.









Either x = 0 or x − 6 = 0.









Either x = 0 or x − 6 = 0.









Either x = 0 or x − 6 = 0.









Either x = 0 or x − 6 = 0.









Either x = 0 or x − 6 = 0.




Factorising can also simplify algebraic fractions:

4x + 6

2=

2(2x + 3)

2(factorise)

=1�2(2x + 3)

�21(cancel common factors)

= 2x + 3




4x + 6

2=

2(2x + 3)

2(factorise)

=1�2(2x + 3)


= 2x + 3




4x + 6

2=

2(2x + 3)

2(factorise)

=1�2(2x + 3)


= 2x + 3




4x + 6

2=

2(2x + 3)

2(factorise)

=1�2(2x + 3)


= 2x + 3




4x + 6

2=

2(2x + 3)

2(factorise)

=1�2(2x + 3)


= 2x + 3




x2 + 3x + 2

x + 1=

(x + 1)(x + 2)

x + 1(factorise)

=1��

��(x + 1)(x + 2)

��x + 11(cancel common factors)

= x + 2

Note: Watch out when you cancel terms involving variables. The original fractionmakes it clear that there is a problem when x = −1 because the fraction becomes

00 , which is an indeterminate quantity. This problem is no longer obvious in the

simplified version. Usually, we would write the answer as

x + 2, x 6= −1




x2 + 3x + 2

x + 1=

(x + 1)(x + 2)

x + 1(factorise)

=1��

��(x + 1)(x + 2)


= x + 2




x + 2, x 6= −1




x2 + 3x + 2

x + 1=

(x + 1)(x + 2)

x + 1(factorise)

=1��

��(x + 1)(x + 2)


= x + 2




x + 2, x 6= −1




x2 + 3x + 2

x + 1=

(x + 1)(x + 2)

x + 1(factorise)

=1��

��(x + 1)(x + 2)


= x + 2




x + 2, x 6= −1




x2 + 3x + 2

x + 1=

(x + 1)(x + 2)

x + 1(factorise)

=1��

��(x + 1)(x + 2)


= x + 2




x + 2, x 6= −1




x2 + 3x + 2

x + 1=

(x + 1)(x + 2)

x + 1(factorise)

=1��

��(x + 1)(x + 2)


= x + 2




x + 2, x 6= −1


Polynomials

This final section is about polynomials, a topic which may not be on your mathssyllabus.


Polynomials

A polynomial is a bunch of terms involving whole number powers (ie. positiveintegers) of a variable added/subtracted together.

We might also have a constant term (a single number) as well.

3x2 − 6x + 7x5 + 2

Look at the above polynomial. Every term in it is either a positive integer powerof the variable x or a constant term.

Term Power of x

3x2 27x5 5−6x 1

2 constant term or “2x0”(so the power of x is 0)


Polynomials



3x2 − 6x + 7x5 + 2


Term Power of x

3x2 27x5 5−6x 1



Polynomials



3x2 − 6x + 7x5 + 2


Term Power of x

3x2 27x5 5−6x 1



Polynomials



3x2 − 6x + 7x5 + 2


Term Power of x

3x2 27x5 5−6x 1



Polynomials



3x2 − 6x + 7x5 + 2


Term Power of x

3x2 27x5 5−6x 1



Polynomials



3x2 − 6x + 7x5 + 2


Term Power of x

3x2

27x5 5−6x 1



Polynomials



3x2 − 6x + 7x5 + 2


Term Power of x

3x2 2

7x5 5−6x 1



Polynomials



3x2 − 6x + 7x5 + 2


Term Power of x

3x2 27x5

5−6x 1



Polynomials



3x2 − 6x + 7x5 + 2


Term Power of x

3x2 27x5 5

−6x 12 constant term or “2x0”

(so the power of x is 0)


Polynomials



3x2 − 6x + 7x5 + 2


Term Power of x

3x2 27x5 5−6x

12 constant term or “2x0”



Polynomials



3x2 − 6x + 7x5 + 2


Term Power of x

3x2 27x5 5−6x 1



Polynomials



3x2 − 6x + 7x5 + 2


Term Power of x

3x2 27x5 5−6x 1

2

constant term or “2x0”(so the power of x is 0)


Polynomials



3x2 − 6x + 7x5 + 2


Term Power of x

3x2 27x5 5−6x 1

2 constant term

or “2x0”(so the power of x is 0)


Polynomials



3x2 − 6x + 7x5 + 2


Term Power of x

3x2 27x5 5−6x 1

2 constant term or “2x0”



Polynomials



3x2 − 6x + 7x5 + 2


Term Power of x

3x2 27x5 5−6x 1



Polynomials: Examples

7p8 − 62p2 + 5− p3

7y2 − 3y4

3 + x

23q

5 + q4 + 3q7 − 2q + 1

(Note that the numbers multiplying the variable parts do not have to be wholenumbers.)



7p8 − 62p2 + 5− p3

7y2 − 3y4

3 + x

23q

5 + q4 + 3q7 − 2q + 1




7p8 − 62p2 + 5− p3

7y2 − 3y4

3 + x

23q

5 + q4 + 3q7 − 2q + 1




7p8 − 62p2 + 5− p3

7y2 − 3y4

3 + x

23q

5 + q4 + 3q7 − 2q + 1



Polynomials: A Convention

Polynomials are usually written with their powers in descending order.

7p8 − 62p2 + 5− p3

−→ 7p8 − p3 − 62p2 + 5

7y2 − 3y4 −→ −3y4 + 7y2

3 + x −→ x + 3

23q

5 + q4 + 3q7 − 2q + 1 −→ 3q7 + 23q

5 + q4 − 2q + 1




7p8 − 62p2 + 5− p3 −→ 7p8 − p3 − 62p2 + 5

7y2 − 3y4 −→ −3y4 + 7y2

3 + x −→ x + 3

23q

5 + q4 + 3q7 − 2q + 1 −→ 3q7 + 23q

5 + q4 − 2q + 1




7p8 − 62p2 + 5− p3 −→ 7p8 − p3 − 62p2 + 5

7y2 − 3y4

−→ −3y4 + 7y2

3 + x −→ x + 3

23q

5 + q4 + 3q7 − 2q + 1 −→ 3q7 + 23q

5 + q4 − 2q + 1




7p8 − 62p2 + 5− p3 −→ 7p8 − p3 − 62p2 + 5

7y2 − 3y4 −→ −3y4 + 7y2

3 + x −→ x + 3

23q

5 + q4 + 3q7 − 2q + 1 −→ 3q7 + 23q

5 + q4 − 2q + 1




7p8 − 62p2 + 5− p3 −→ 7p8 − p3 − 62p2 + 5

7y2 − 3y4 −→ −3y4 + 7y2

3 + x

−→ x + 3

23q

5 + q4 + 3q7 − 2q + 1 −→ 3q7 + 23q

5 + q4 − 2q + 1




7p8 − 62p2 + 5− p3 −→ 7p8 − p3 − 62p2 + 5

7y2 − 3y4 −→ −3y4 + 7y2

3 + x −→ x + 3

23q

5 + q4 + 3q7 − 2q + 1 −→ 3q7 + 23q

5 + q4 − 2q + 1




7p8 − 62p2 + 5− p3 −→ 7p8 − p3 − 62p2 + 5

7y2 − 3y4 −→ −3y4 + 7y2

3 + x −→ x + 3

23q

5 + q4 + 3q7 − 2q + 1

−→ 3q7 + 23q

5 + q4 − 2q + 1




7p8 − 62p2 + 5− p3 −→ 7p8 − p3 − 62p2 + 5

7y2 − 3y4 −→ −3y4 + 7y2

3 + x −→ x + 3

23q

5 + q4 + 3q7 − 2q + 1 −→ 3q7 + 23q

5 + q4 − 2q + 1


Polynomials: Terminology

Consider the following polynomial.

3x7 − 4x5 + 37x

4 − 2x + 9

We say it is a “polynomial in the variable x”.

The order of a polynomial is the highest power of x present. In this case, theorder is 7.

The coefficient of x7 is 3.(The number multiplying the x part.)

The coefficient of x5 is −4.

The coefficient of x4 is 37 .

The coefficient of x is −2.

The coefficient of x0 is 9 (constant term).

The coefficient of x2 is 0 (because it isn’t there!).




3x7 − 4x5 + 37x

4 − 2x + 9












3x7 − 4x5 + 37x

4 − 2x + 9


The order of a polynomial is the highest power of x present. In this case, theorder is

7.










3x7 − 4x5 + 37x

4 − 2x + 9












3x7 − 4x5 + 37x

4 − 2x + 9












3x7 − 4x5 + 37x

4 − 2x + 9




The coefficient of x5 is

−4.








3x7 − 4x5 + 37x

4 − 2x + 9












3x7 − 4x5 + 37x

4 − 2x + 9






37 .







3x7 − 4x5 + 37x

4 − 2x + 9












3x7 − 4x5 + 37x

4 − 2x + 9






The coefficient of x is

−2.






3x7 − 4x5 + 37x

4 − 2x + 9












3x7 − 4x5 + 37x

4 − 2x + 9








9 (constant term).





3x7 − 4x5 + 37x

4 − 2x + 9












3x7 − 4x5 + 37x

4 − 2x + 9









0 (because it isn’t there!).




3x7 − 4x5 + 37x

4 − 2x + 9












4x3 − 5x − 9

What is the order of this polynomial?

3

What is the coefficient of x3? 4


What is the coefficient of x? −5

What is the coeficient of x0? −9




4x3 − 5x − 9

What is the order of this polynomial? 3








4x3 − 5x − 9


What is the coefficient of x3?

4







4x3 − 5x − 9









4x3 − 5x − 9



What is the coefficient of x2?

0






4x3 − 5x − 9









4x3 − 5x − 9




What is the coefficient of x?

−5





4x3 − 5x − 9









4x3 − 5x − 9





What is the coeficient of x0?

−9




4x3 − 5x − 9







Importance

Why are polynomial expressions important?

They are easier to work with than other expressions (hard to know this unless youhave worked with other expressions).

Most other expressions can be very closely approximated by polynomialexpressions.

In a sense, polynomial expressions are mathematical “building blocks”.


Importance






Importance






Importance






Adding and Subtracting Polynomials

When we add or subtract polynomials we get a new polynomial (just like we dowith numbers!). This can be done by adding and subtracting their like terms.

(4x2 + 3x + 7) + (2x2 + 5x + 2)

We may drop the brackets in this case (why?).

4x2 + 3x + 7 + 2x2 + 5x + 2

If you need to, shift the signed terms around so that like terms are next to eachother.

4x2 + 2x2 + 3x + 5x + 7 + 2

We now add like terms together.

6x2 + 8x + 9




(4x2 + 3x + 7) + (2x2 + 5x + 2)


4x2 + 3x + 7 + 2x2 + 5x + 2


4x2 + 2x2 + 3x + 5x + 7 + 2


6x2 + 8x + 9




(4x2 + 3x + 7) + (2x2 + 5x + 2)


4x2 + 3x + 7 + 2x2 + 5x + 2


4x2 + 2x2 + 3x + 5x + 7 + 2


6x2 + 8x + 9




(4x2 + 3x + 7) + (2x2 + 5x + 2)


4x2 + 3x + 7 + 2x2 + 5x + 2


4x2 + 2x2 + 3x + 5x + 7 + 2


6x2 + 8x + 9



Simplify the following:

(3x3 − 4x2 + 5)− (x3 + 3x − 4)

There is a negative sign to the left of the second set of brackets. This iseffectively a multiplier of −1. From the previous workshop, we know that this −1

gets distributed to each term in the bracket.

3x3 − 4x2 + 5− (x3 + 3x − 4)

= 3x3 − 4x2 + 5− x3− 3x + 4

(In short, every sign in the 2nd bracket changes.)


3x3 − x3 − 4x2 − 3x + 5 + 4


2x3 − 4x2 − 3x + 9




(3x3 − 4x2 + 5)− (x3 + 3x − 4)



3x3 − 4x2 + 5− (x3 + 3x − 4)

= 3x3 − 4x2 + 5− x3− 3x + 4



3x3 − x3 − 4x2 − 3x + 5 + 4


2x3 − 4x2 − 3x + 9




(3x3 − 4x2 + 5)− (x3 + 3x − 4)



3x3 − 4x2 + 5− (x3 + 3x − 4)

= 3x3 − 4x2 + 5

− x3− 3x + 4



3x3 − x3 − 4x2 − 3x + 5 + 4


2x3 − 4x2 − 3x + 9




(3x3 − 4x2 + 5)− (x3 + 3x − 4)



3x3 − 4x2 + 5− (x3 + 3x − 4)

= 3x3 − 4x2 + 5

− x3− 3x + 4



3x3 − x3 − 4x2 − 3x + 5 + 4


2x3 − 4x2 − 3x + 9




(3x3 − 4x2 + 5)− (x3 + 3x − 4)



3x3 − 4x2 + 5− (x3 + 3x − 4)

= 3x3 − 4x2 + 5− x3

− 3x + 4



3x3 − x3 − 4x2 − 3x + 5 + 4


2x3 − 4x2 − 3x + 9




(3x3 − 4x2 + 5)− (x3 + 3x − 4)



3x3 − 4x2 + 5− (x3 + 3x − 4)

= 3x3 − 4x2 + 5− x3

− 3x + 4



3x3 − x3 − 4x2 − 3x + 5 + 4


2x3 − 4x2 − 3x + 9




(3x3 − 4x2 + 5)− (x3 + 3x − 4)



3x3 − 4x2 + 5− (x3 + 3x − 4)

= 3x3 − 4x2 + 5− x3− 3x

+ 4



3x3 − x3 − 4x2 − 3x + 5 + 4


2x3 − 4x2 − 3x + 9




(3x3 − 4x2 + 5)− (x3 + 3x − 4)



3x3 − 4x2 + 5− (x3 + 3x − 4)

= 3x3 − 4x2 + 5− x3− 3x

+ 4



3x3 − x3 − 4x2 − 3x + 5 + 4


2x3 − 4x2 − 3x + 9




(3x3 − 4x2 + 5)− (x3 + 3x − 4)



3x3 − 4x2 + 5− (x3 + 3x − 4)

= 3x3 − 4x2 + 5− x3− 3x + 4



3x3 − x3 − 4x2 − 3x + 5 + 4


2x3 − 4x2 − 3x + 9




(3x3 − 4x2 + 5)− (x3 + 3x − 4)



3x3 − 4x2 + 5− (x3 + 3x − 4)

= 3x3 − 4x2 + 5− x3− 3x + 4



3x3 − x3 − 4x2 − 3x + 5 + 4


2x3 − 4x2 − 3x + 9




(3x3 − 4x2 + 5)− (x3 + 3x − 4)



3x3 − 4x2 + 5− (x3 + 3x − 4)

= 3x3 − 4x2 + 5− x3− 3x + 4



3x3 − x3 − 4x2 − 3x + 5 + 4


2x3 − 4x2 − 3x + 9




(3x3 − 4x2 + 5)− (x3 + 3x − 4)



3x3 − 4x2 + 5− (x3 + 3x − 4)

= 3x3 − 4x2 + 5− x3− 3x + 4



3x3 − x3 − 4x2 − 3x + 5 + 4


2x3 − 4x2 − 3x + 9




(3x3 − 4x2 + 5)− (x3 + 3x − 4)



3x3 − 4x2 + 5− (x3 + 3x − 4)

= 3x3 − 4x2 + 5− x3− 3x + 4



3x3 − x3 − 4x2 − 3x + 5 + 4


2x3 − 4x2 − 3x + 9




(3x3 − 4x2 + 5)− (x3 + 3x − 4)



3x3 − 4x2 + 5− (x3 + 3x − 4)

= 3x3 − 4x2 + 5− x3− 3x + 4



3x3 − x3 − 4x2 − 3x + 5 + 4


2x3 − 4x2 − 3x + 9




−2(6x4 − 7x2 − 3) + 5(−8x6 + 3x4 − 4)

Once again, the number out the front of each brackets gets distributed to each ofthe terms.

−2(6x4 − 7x2 − 3) + 5(−8x6 + 3x4 − 4)

= −12x4 + 14x2 + 6− 40x6 + 15x4− 20


−12x4 + 15x4 + 14x2 + 6− 20− 40x6

We now add like terms together and write with decreasing powers:

−40x6 + 3x4 + 14x2 − 14




−2(6x4 − 7x2 − 3) + 5(−8x6 + 3x4 − 4)


−2(6x4 − 7x2 − 3) + 5(−8x6 + 3x4 − 4)

= −12x4 + 14x2 + 6− 40x6 + 15x4− 20


−12x4 + 15x4 + 14x2 + 6− 20− 40x6


−40x6 + 3x4 + 14x2 − 14




−2(6x4 − 7x2 − 3) + 5(−8x6 + 3x4 − 4)


−2(6x4 − 7x2 − 3) + 5(−8x6 + 3x4 − 4)

= −12x4

+ 14x2 + 6− 40x6 + 15x4− 20


−12x4 + 15x4 + 14x2 + 6− 20− 40x6


−40x6 + 3x4 + 14x2 − 14




−2(6x4 − 7x2 − 3) + 5(−8x6 + 3x4 − 4)


−2(6x4 − 7x2 − 3) + 5(−8x6 + 3x4 − 4)

= −12x4 + 14x2

+ 6− 40x6 + 15x4− 20


−12x4 + 15x4 + 14x2 + 6− 20− 40x6


−40x6 + 3x4 + 14x2 − 14




−2(6x4 − 7x2 − 3) + 5(−8x6 + 3x4 − 4)


−2(6x4 − 7x2 − 3) + 5(−8x6 + 3x4 − 4)

= −12x4 + 14x2 + 6

− 40x6 + 15x4− 20


−12x4 + 15x4 + 14x2 + 6− 20− 40x6


−40x6 + 3x4 + 14x2 − 14




−2(6x4 − 7x2 − 3) + 5(−8x6 + 3x4 − 4)


−2(6x4 − 7x2 − 3) + 5(−8x6 + 3x4 − 4)

= −12x4 + 14x2 + 6− 40x6

+ 15x4− 20


−12x4 + 15x4 + 14x2 + 6− 20− 40x6


−40x6 + 3x4 + 14x2 − 14




−2(6x4 − 7x2 − 3) + 5(−8x6 + 3x4 − 4)


−2(6x4 − 7x2 − 3) + 5(−8x6 + 3x4 − 4)

= −12x4 + 14x2 + 6− 40x6 + 15x4

− 20


−12x4 + 15x4 + 14x2 + 6− 20− 40x6


−40x6 + 3x4 + 14x2 − 14




−2(6x4 − 7x2 − 3) + 5(−8x6 + 3x4 − 4)


−2(6x4 − 7x2 − 3) + 5(−8x6 + 3x4 − 4)

= −12x4 + 14x2 + 6− 40x6 + 15x4− 20


−12x4 + 15x4 + 14x2 + 6− 20− 40x6


−40x6 + 3x4 + 14x2 − 14




−2(6x4 − 7x2 − 3) + 5(−8x6 + 3x4 − 4)


−2(6x4 − 7x2 − 3) + 5(−8x6 + 3x4 − 4)

= −12x4 + 14x2 + 6− 40x6 + 15x4− 20


−12x4 + 15x4 + 14x2 + 6− 20− 40x6


−40x6 + 3x4 + 14x2 − 14




−2(6x4 − 7x2 − 3) + 5(−8x6 + 3x4 − 4)


−2(6x4 − 7x2 − 3) + 5(−8x6 + 3x4 − 4)

= −12x4 + 14x2 + 6− 40x6 + 15x4− 20


−12x4 + 15x4 + 14x2 + 6− 20− 40x6


−40x6 + 3x4 + 14x2 − 14


Multiplying Polynomials

We can use the distributive law to multiply two polynomials together

(3x2 + 2x)(4x5 + 3x) =

12x7 + 9x3+ 8x6+ 6x2

Every term in the first bracket meets every term in the second bracket.(We saw this in the previous workshop.)

Sometimes, we work with “longer” polynomials. The rule is still the same. Everyterm in the first bracket must meet every term in the second bracket.




(3x2 + 2x)(4x5 + 3x) =

12x7 + 9x3+ 8x6+ 6x2






(3x2 + 2x)(4x5 + 3x) = 12x7

+ 9x3+ 8x6+ 6x2






(3x2 + 2x)(4x5 + 3x) = 12x7

+ 9x3+ 8x6+ 6x2






(3x2 + 2x)(4x5 + 3x) = 12x7 + 9x3

+ 8x6+ 6x2






(3x2 + 2x)(4x5 + 3x) = 12x7 + 9x3

+ 8x6+ 6x2






(3x2 + 2x)(4x5 + 3x) = 12x7 + 9x3+ 8x6

+ 6x2






(3x2 + 2x)(4x5 + 3x) = 12x7 + 9x3+ 8x6

+ 6x2






(3x2 + 2x)(4x5 + 3x) = 12x7 + 9x3+ 8x6+ 6x2






(3x2 + 2x)(4x5 + 3x) = 12x7 + 9x3+ 8x6+ 6x2






(3x2 + 2x)(4x5 + 3x) = 12x7 + 9x3+ 8x6+ 6x2




Expansion

In general

(polynomial) × (polynomial)

expands out to

polynomial

Note: We saw earlier that 2nd order polynomials can (sometimes) be factorised.In general, polynomials can also (sometimes) be be factorised.

However, that’s a topic for another time . . .


Expansion

In general


expands out to

polynomial




Expansion

In general


expands out to

polynomial

Note: We saw earlier that 2nd order polynomials can (sometimes) be factorised.

In general, polynomials can also (sometimes) be be factorised.



Expansion

In general


expands out to

polynomial




Expansion

In general


expands out to

polynomial




Using STUDYSmarter Resources

This resource was developed for UWA students by the STUDYSmarter team forthe numeracy program. When using our resources, please retain them in their

original form with both the STUDYSmarter heading and the UWA crest.


Documents

Factorising - Numeracy Workshop...workshopExpressions and Expansion). Topics include extracting common factors, factorising quadratic expressions and polynomials. Workshop resources:These