Difference of Two Squares

Today we looked at factorising using the Difference of Two Squares method. The official mathematical definition of this is as follows:

What this formula is basically saying is, if we have a number (i.e. a) squared and we subtract another number (i.e. b) which has also been squared, we can easily write this in a factorised form (a+b)(a-b). 

This is a little confusing, so i will put a few examples on here, which hopefully may clear it up a little.

Example: Factorise the following: 

a² – 49

This can be written as: a² - 7² 

This is because 49 is actually 7²

We now have "something squared" minus "something else squared", which means we can simply factorise this to be:

(a + 7)(a – 7)  which is our final answer. 

Example: Factorise the following: 

t² – 64

Again, we can see that 64 can be written as 8² 

^{So we now have:} t² - 8²

^{Similarly to the previous example, we have "something squared" minus "something else squared". This means we can factorise this to be:} 

(t + 8)(t – 8) which is our final answer. 

Example: Factorise the following:

(6m)² – (5n)²

Here again, we have "something squared" minus "something else squared". It does look slightly different, but it is still in this same format so we can still factorise using this difference of two squares method. This means it factorises to be: 

(6m + 5n)(6m – 5n)

The following website gives a fantastic description and examples of factorising using the difference of two squares method: http://www.purplemath.com/modules/specfact.htm

I have also posted a worksheet to practice this difference of two squares method under the "Worksheets" tab. 

As always if you do have any questions of concerns feel free to comment on this post or any of the other posts. 

Factorising Quadratics

So today we are looking at factorising quadratics. Now a quadratic is an algebraic expression that looks like this:

x² + 5x + 6

Now the first thing we have to remember is that factorising is the opposite of expanding. So when we were doing expanding and we had something like (x + 3)(x + 2), we would expand it using FOIL and get:

x² + 2x + 3x + 6

We would then collect like terms and get a final answer of:

x² + 5x + 6

This is a quadratic.

Since factorising is the opposite of expanding, we want to start with an expression like x² + 5x + 6 and manipulate it so we then get (x + 3)(x + 2). Basically all we want to find out is the numbers that go in the brackets. We know there will always be the x’s [i.e. (x + ….)(x + ….)], we need to find out the numbers that go after the x’s.

If we look at the example above, we can see that these numbers in the brackets are 2 and 3. If we look carefully we can see that the numbers 2 and 3 in our brackets add together to give 5, which is the number out the front of our second term from our expression. We also can see that 2 and 3 multiply together to give 6, which is our third (last) term in our expression. This tells us that the numbers in the brackets have to:

• ·         Add together to give the second term
• ·         Multiply together to give the third term.

So looking at an example:

Factorise x² + 11x + 24

First look at pair of numbers that multiply together to give our third term, 24 (i.e. factors of 24). These are:

·         1 and 24

·         2 and 12

·         3 and 8

·         4 and 6

Now we need to pick one of these pairs that add up to give our second term, 11. This can only be 3 and 8. We now know that the numbers in our brackets are 3 and 8. So our answer, fully factorised, is:

(x + 8)(x + 3)

So that is how we factorise quadratics. A key point to remember is that factors of our third term may also be negative. The following website shows examples of these: http://www.mathcentre.ac.uk/resources/workbooks/mathcentre/web-factorisingquadratics.pdf -> (only look up to and including section 3). 

Remember if you do have any question on this, feel free to comment on this post or email me. There is a worksheet on this topic in the "Worksheets" tab. 

Factorising by Grouping (Part 2)

This shorter Friday afternoon lesson was spent re-enforcing the ideas from the previous lesson on factorising by grouping. We then continued on with the work sheet on factorising.

Next lesson we will be looking more into factorising more complicated problems, so be sure to have that factorising work sheet finished off by Monday. 

Factorising by Grouping

Today we looked at factorising by grouping. This involves grouping the terms of an expression into pairs and factorising each pair individually. For example:

2ax + 8ay + cx + 4cy 

We can group the 2ax and the 8ay together, as well as the cx and the 4cy.
If we look at the first group (2ax and 8ay), we see a common factor of 2a. If we take this out to factorise we get: 2a(x + 4y)
If we look at the second group (cx and 4cy) there is a common factor of c. If we take this out to factorise we get: c(x + 4y)

Now that we have factorised each group, we can go and put them back into the expression in their factorised form. We then get:
2a(x + 4y) + c(x + 4y)
We can now see a common factor of (x + 4y) in this expression. If we take this out and factorise, we get:
(x + 4y)(2a + c) which is our fully factorised final answer.

If you are unsure as to how we factorise, see my previous post below.
If you want more information on this, have a look here: http://www.khanacademy.org/math/algebra/multiplying-factoring-expression/factoring-by-grouping/v/factoring-trinomials-by-grouping-1 -> really good video on this topic!
or here: http://www.compuhigh.com/demo/factoringinpairs.htm

Welcome

Welcome to your year 10 maths page. I hope this is a useful resource for you to use. The idea of this blog is for you to be able to keep up, even if you do miss a class or two.
We are currently working on a unit of algebra.
We have already looked at expanding brackets. This involves multiplying either a number through a set of brackets, or multiplying a full bracket trough using the FOIL technique. FOIL stands for Firsts, Outers, Inners, Lasts, and it helps us to remember the order for multiplying two brackets with one another. For example:

For more information on this, see: http://www.mathsisfun.com/algebra/expanding.html

                                                        http://www.mathsisfun.com/algebra/polynomials-multiplying.html

We also have to remember to collect like terms. This means adding or subtracting the terms which are the same as other terms. For terms to be "like terms" they must have exactly the same combination of pronumerals (letters). For example, 2x and 2xy are not like terms, because one has an x, however the other has an x and a y. 4a and 26a are like terms, because they both only have an a and no other pronumerals. 

For more information on this, see: http://www.mathsisfun.com/algebra/like-terms.html

We are now looking at factorising. This involves taking out the highest common factor from the terms in an expression, and then writing the rest of the expression in brackets. A factor of a number is another number that goes into (or divides) this number. For example, 8 has the factors of 1, 8, 2, and 4.
Common factors are factors that two numbers share. For example, 8 has the factors 1, 8, 2, and 4, and 12 has the factors 1, 12, 2, 6, 3, and 4. This means that the common factors of 12 and 8 are 1, 2, and 4.
The highest common factor is the common factor that contains the most "stuff". This means it has the highest number, as well as the most pronumerals (letters). If we look at the factors of 8 and 12 again, the highest common factor is 4.
Finally, if we are asked to factorise an expression, the first step is to find the highest common factor.

For example: Factorise 4x + 8

The factors of 4x are: 1, 4x, 4, x, 2, 2x
The factors of 8 are: 1, 8, 2, 4
So the highest common factor is 4.

We now have to take this out the front of the brackets, and put the rest of the stuff in the brackets.
So we have: 4(? + #)

To work out what ? is: "What do we multiply 4 (our highest common factor out the front) by to get 4x?" The answer to this is x, so our ? is x.
To work out what # is: "What do we multiply 4 (our highest common factor out the front) by to get 8?" The answer to this is 2, so our # is 2.

So finally, we get: 4(x + 2) as our final answer when we are asked to factorise 4x + 8

For more examples, see https://www.mathsisfun.com/algebra/factoring.html