REFLECTION

From this two chapters, I have learnt that not only do expressions come in linear forms but also in quadratic forms with a highest power of two. I have learnt more ways to expand my expressions such as using special identities and also common factor. No matter how confusing sometimes the expressions may be, with a strong grasp of the foundation of expansion of expression, we would all be able to solve our expression easily and effortlessly.


I have also learnt from factorisation of quadratic expressions very useful and interesting. Not only do they have more variety of ways to factorise our expressions, we also can apply them into real world context which I find really helpful since our nowadays living uses a lot of mathematical formats and we are using them in our everyday lifes with or without knowing. From all six ways that I have learnt to factorise, I find special identity the easiest to be applied into real world context questions. However, that does not mean that the rest are not applicable to the real world context.


To me, both this chapters are very interesting and very useful for real world. I would really like having more of such chapters coming because not only it is open to a wide variation of ways to solve questions, but also I have such a fun time brainstorming to factorise and expand the expressions.

FACTORISATION USING GROUPING

Lastly, sometimes an expression can have more than four terms in it. For such cases, they would have to be factorised using the grouping method. Grouping is when you take the two similar expression and group them together. Note that this method could only be used for certain expressions. One example to illustrate this would be:

ax-5a+4x-20

= (ax-5a)+(4x-20)

= a(x-5)+4(x-5)

= (x-5)(a+4)

One final example for grouping would be:

(x+y)(a+b)-(y+z)(a+b)

= (a+b)((x+y)-(y+z))

= (a+b)(x+y-y-z)

= (a+b)(x-z)

Lets watch a short video explaining a little bit more about factorising using grouping. 

FACTORISING USING SPECIAL IDENTITY

Special Identities

Special identities have three different kind with a fixed format that has to be applied all the time when solving using this method. The formats would be: <br>

PERFECT SQUARES

à   (a+b) ² = a²+2ab+b^{2  <br>}

DIFFERENCE OF 2 SQUARES

à   (a-b) ² = a²-2ab+b^{2 <br>} 

à   (a+b)(a-b) = a²-b^{2  <br>} 

For special identity, only certain expressions that meet the criteria would be able to be solved using these methods.<Br>

The criteria are: <br>

à The first term has to be a perfect square<br>

àThe last term has to be a constant and also a perfect square <br>

Only then will you be able to apply this method on. <br> 

An example would be: <br>

<br>

x² +12x+36 <br>

Since that x² is a perfect square and 36 is a perfect square and a constant, we can thus apply our special identity to this expression. <br>

x² +12x+36 is the same as a²+2ab+b². Thus, we know that the factorized form would be (a+b)^{2 <br>} 

a = x        b = 6 <br>

x² +12x+36 in its factorized form would then be (x+6)^2<br> 

The next example would be using the second perfect squares.

y²-6yz+49

Since y is a perfect square and 49 is also a perfect square, we can apply the second special identity.

y²-6yz+49 is the same as a²-2ab+b²

Thus, a = y           b = 7

The factorised form would then be (y-7)

The third one would be:

(11y+10)(11y-10)

The expression above is the same as the difference of squares so thus we can apply the special identity.

a = 11y     b = 10

The factorised form of (11y+10)(11y-10) would be 121y²-100.

Some of the more complex questions that could be solved using

Special identity would be:

 Evaluate the following without the use of calculator: 659²-341²

= (660-1)²-(340+1)²

= (660²-2(660)(1)+1²)-(340²+2(340)(1)+1²)

= (435600-1320+1)-(115600+680+1)

= 434281-116281

= 318000

FACTORISATION USING CROSS-METHOD

Cross-method.

 Sometimes a quadratic expression can have more than 2 terms in it. Thus, we would have to use another method to factorise it. To be able to apply this method into the expression, the expression must first have 3 terms and must be a quadratic expression.
Some examples to illustrate this would be:
 c2-4cd-21d2
To do this, we need a table to put in all our terms. Remember that the terms have to be put in this order:

 First term in the first column
 Last term in the second column
 Middle term in the final column

àDraw the way that is done in the picture. àPut in the terms following the position that is given. àThe terms written before the line will have their numbers    multiplying together from top to bottom. àThe term that is written after the vertical line would have       the answers adding together to get the answer.     è After finding out what values are needed to be multiplied to get the answers, cross multiply the answers and write them after the vertical line.   è Add them together and ensure that the answer is what is written.

In the context of the question given, we would have to do and ensure that the answer is the same as the terms written underneath the horizontal line. Do remember to take extra notice of the signs of the terms.Thus, the factorized form for c²-4cd-21d² is (c+3d)(c-7d). 



Let's take a look at another example using the cross-method.

2x²-7xy+6y²

Doing exactly the same thing like we did earlier, we would come to an answer of (x-2y)(2x-3y).

FACTORISATION USING COMMON FACTOR

Common Factor
Factorising the common factor out from any expressions would always be the first method we would all try to use. It is simple and efficient but can only be used for expressions with not more than two terms in the expression. An example would be:

44b2+33ab
For the above expression, we can see that there is a common factor of b and 11. So, we would have to take it out of the expression.
44b2+33ab
= 11b(4b+3a)
Thus, we can see that the factorized form of 44b2+33ab is 11b(4b+3a).
Another example would be:
21ab-28ac
From the above expression, we know that there is a common factor of 7 and a. We would have to factor those two out and thus giving us an expression as such:
21ab-28ac
=7a(3b-4c)
Thus, we can see that the factorized form of 21ab-28ac is 7a(3b-4c).

TYPES OF FACTORISATION FOR QUADRATIC EXPRESSION

From the video, we have seen that there are six methods of factorization that could be used for quadratic expressions. The six are as following:
- Common Factor
- Cross-method
- Special Identities
Difference of two squares
Perfect squares (-)
Perfect squares (+)
- Grouping

FACTORISATION OF QUADRATIC EXPRESSION

Factorisation is the process of expressing an algebraic expression as a product of two or more algebraic expression. By doing this, we would know what needs to be multiplied to be able to get the expression.

In secondary 1, we have learnt about the factorization of linear expressions. The first way of factorizing the expression would be by looking for the common factor and taking it out. An example would be:

44b2+33ab


For the above expression, we can see that there is a common factor of b and 11. So, we would have to take it out of the expression.

44b2+33ab

= 11b(4b+3a)

Thus, we can see that the factorized form of 44b2+33ab is 11b(4b+3a).


Another example would be

-3a(2+b)+18a(b-1)

(taken from Shinglee Math TB)

For the example above, we would first have to expand and simplify the expression before factorizing the common factor out. The solution would be:

-3a(2+b)+18a(b-1)

= -6a-3ab+18ab-18a

= -24a-3ab+18ab

= 18ab-24a-3ab (expanded and simplified answer)

18ab-24a-3ab

The common factor here is 3 and a. We would have to factor that out.

18ab-24a-3ab

= 3a(6b-8-b)

= 3a(5b-8)

Thus, the factorized form of 18ab-24a-3ab is 3a(5b-8).

Now, after learning quadratic expressions, we have learnt new ways to factorize our expressions. Let us watch this video which explains of the different factorizing methods that could be used for quadratic expression.