Notes

Answers with working can be found at the bottom of this page. The first few questions are relatively straight forward. The thing to remember with algebra is that you don't always get a clean answer. when presented with an equation like $(x+2)(x+1)=$ you're unlikely to get to a simple $x=??$ position.
So try not to get too hung up on looking for that.

What I found when I was learning algebra was that sometimes I would forget the basic rules. For examples sometimes I would do the following:
$$(x+y{)}^2=$$
$$(x+y)(x+y)$$
$$=x^2+xy+yx+y^2$$
But that goes against one of the primary rules, namely that we keep to alphabetical order. The correct answer is $x^2+xy+xy+y^2$ which can then be simplified to $x^2+2xy+y^2$. Messing it up made me think that $xy$ and $yx$ were different numbers, when they're actually the same number.

Substitution

mathematicians are lazy. They like to shorten everything they can in an equation. for example $\frac{3}{4}$ is used as a shorthand for $3÷4$, and when you see two mathematical objects next to each other without a symbol between them then multiplication is implied.
For example, $3(x)$ is really just $3\times x$ and $3abc$ is really $3\times (a\times b\times c)$.
So for a substitution equation like $3a+2b-c=T$ then what they're actually asking is $(3\times a)+(2\times b)-c=T$.
The substitution stuff appears to be more or less to improve your ability to use formula's, which is a large part of advanced mathematics.

Algebra is really just a way to represent complex relationships in mathematics. If you're having trouble understanding a question below, try to relate it back to something in the real world (a lot of your stuff will be based on geometric shapes. Senior maths generally uses graphs for Algebra).

So with regards to these questions, we're expressing values in terms of there relationships. So for question a) we can see for the sides representing the otherside of the shape there are two values, $y$ and 8. We can represent there relationship with the $\ast$ side with the following:
$$\ast =y+8$$.

For b), we're doing the same. writing an expression that represents the relationship of these values to one another, and in relation to the perimeter of the object (where the $perimeter=2(x)+2(y)$ or put another way $P=2(x+y)$).

$$2((y+2)+6)+2(y+8)$$
would give us the full rectangle. However, there's a bit missing from the object, which we need to subtract from our rectangle in order to accurately represent the object. The bit that's missing has a perimeter that can be described with the following expression (because it's a rectangle too):
$$2(y+6)$$
You'll note that not the whole perimeter of the missing bit is missing, only the bottom half of it. Thus we only need to deduct half of it's perimeter. Giving the following:

$$2((y+2)+6)+2(y+8)-(y+6)$$
$$2(y+8)+2(y+8)-(y+6)$$
$$2y+16+2y+16-y-6$$
$$4y+32-y-6$$
$$3y+26$$

For c), we plug the new value into our final equation from b) so that we have $3y+26=44$ giving us an equation we can solve.
$$3y+26=44$$
$$3y=44-26$$
$$3y=18$$
$$3y÷3=18÷3$$
$$y=6$$

Factorizing

Factorizing is the inverse of expanding in normal algebra. Where normally we're given an equation like $12y(3-x)$ and asked to expand it, factorising asks the opposite of us.
It gives us $36y-12xy$ and asks us to re-arrange it. In this case, we can see that 36 and 12 are both divisible by a few numbers, 3, 4, 6, and 12. For the greatest simplification we choose the largest number that is a factor of both numbers (it, when multiplied by two other numbers fits into both of the numbers in our equation). so the first step is $12(3y-xy)$. But we can see further to this that both items in the parentheses are being multiplied by y so we can remove it too. Thus giving us $12y(3-x)$.

In general, the point of factorising is to re-arrange the equation. We're not trying to solve it, so we shouldn't be disappearing numbers (just re-arranging them).

The Hard Part

It get's a bit more complex if you're doing quadratic polynomials. I suspect that's a bit beyond what's expected of you, so feel free to skip this section and questions 10 & 11 if you're pressed for time.
With a quadratic you're given an equation that will look like the following $a{x}^2+bx+c=0$ which you'll hopefully recognize as being the end result of expanding an equation like $(x+i)(x+j)$.

(often the = 0 might be skipped, that's OK, we're just showing it's there so that we know there's nothing on the otherside of the equation).

Of particular importance in the above equation is the relationship between the numbers b and c.
Specifically, they're made up of two smaller numbers, we'll call them i and j. The relationship between b and c and there factors i and j can be expressed as $b=i+j$ and $c=i\times j$.
The hard part is working out what i and j are. I'm not currently aware of any easy formula for working this out. You just have to pick numbers and check that they work.
For example, if we have a quadratic to factorize that looks like $x^2+17x+72$ then we have $b=17$ and $c=72$. If we cycle threw numbers looking for values for i and j that add up to 17 but multiply out to 72 we should eventually end up with 8 and 9 ( $8\times 9=72$ and $8+9=17$.
Thus we can factorize out the equation as follows:
$$x^2+17x+72$$
$$x^2+17x+72=x^2+x(8+9)+72$$
$$x^2+x(8+9)+8(9)$$
$$x(x)+8x+9x+8(9)$$
$$x(x+9)+8(x+9)$$
Note in the above line we've ended up with $(x+9)$ twice. We can rewrite the equation so that it only shows once by group the two things it's multiplying against, giving us:
$$(x+9)(x+8)$$
Q.E.D.:
$$x^2+17x+72=(x+8)(x+9)$$

So to recap:
We're doing the inverse of what we normally do. Instead of expanding the equation out, we're contracting it by looking for common base numbers throughout the equation.
It also helps that when you're multiplying numbers it doesn't matter which way round they go $4\times 5$ is the same as $5\times 4$. This means the order when your factoring the equations doesn't matter.
What does matter though, is the sign. $-4\times 5$ is not the same as $4\times 5$.
Also, keep in mind the shorthand we're using. $8(9)$ really just means $8\times 9$ and $x(8+9)$ really means $x\times (8+9)$.

Formulas

Perimeter of a rectangle:
$$P=2x+2y$$
Area of a Circle:
$$A=\pi r^2$$
$$\pi =3.14$$

Questions

1. $$(x+4)(x-8)=0$$

2. $$3(x+y)=0$$

3. $$(5-x{)}^2=0$$

4. $$(x-y{)}^2=0$$

5. Tommy is measuring a $20 bill that he has in his possession. He notes that the top and bottom sides are three times as long as the left and right sides of the bill. Write an equation that expresses this relationship with regards to the perimeter of the $20 bill.
   To answer this question You'll need the perimeter formula, and it might be helpful to draw the object being described.

6. $$64-4x=48-5x$$

7. $$20-5x=48-6x$$

8. $$x^2-40=24$$

9. $$x^2(x+y)=0$$

10. (HARD) Factorize the following: $$x^2-10x+25=0$$

11. (HARD) Factorize the following: $$x^2+32x+48=0$$

12. Factorize the following: $$12x-12=0$$

13. Factorize the following: $$36x^2-12x=0$$

14. Factorize the following: $$11x^3+15x^2=0$$

15. Given a circle with a radius of 4cm's, find the area of the circle (show your working). You can find the radius formula for substitution in the Formula section above.

Answers

Example Answers: $$(x+2)(x+1)=0$$
$$(x+2)(x+1)=x^2+x+2x+2$$
$$(x+2x=3x)$$
$$(x+2)(x+1)=x^2+3x+2$$

Answers to the questions begin here:

$$(x+4)(x-10)=0$$
$$x^2-8x+4x-32$$
$$(-8x+4x=-4x)$$
$$x^2-4x-32=0$$

$$3(x+y)=0$$
$$3x+3y=0$$

$$(5-x{)}^2=0$$
$$(5-x)(5-x)$$
$$x^2-5x-5x+25$$
$$(-5x-5x=-10x)$$
$$x^2-10x+25=0$$

$$(x-y{)}^2=0$$
$$(x-y)(x-y)$$
$$x^2+y^2-xy-xy$$
$$(-xy-xy=-2xy)$$
$$x^2+y^2-2xy=0$$

The formula for the perimeter of a rectangle is $perimeter=2x+2y$ where $x$ is the top/bottom and $y$ is the left/right side. As per the question, we know that $x=3(y)$ (because the top/bottom is 3 times as long as left/right side is high). So our equation should look something like the following:
$$perimeter=2(y)+2(3y)$$
$$perimeter=2y+6y$$
$$perimeter=8y$$

$$64-4x=48-5x$$
$$5x-4x=48-64$$
$$x=-16$$

$$20-5x=48-6x$$
$$6x-5x=48-20$$
$$x=28$$

$$x^2-40=24$$
$$x^2=24+40$$
$$x^2=64$$
$$\sqrt{x}=\sqrt{6}4$$
$$x=8$$

$$x^2(x+y)=0$$
$$x^2(x+y)=x^3+x^{2}y$$

$$x^2-10x+25=0$$
$$x^2-x(5+5)+5(5)$$
$$x^2-5x-5x+5(5)$$
$$x^2-5(x)-x(5)+5(5)$$
$$x(x)-5(x)-x(5)+5(5)$$
N.b $5\times 5$ equals the same as $-5\times -5$ (as a double negative is a positive). So we can swap the sign of $5(5)$ to match the rest of the equation at this point.
$$x(x)-5(x)-x(5)-5(-5)$$
$$x(x-5)-5(x-5)$$
$$(x-5)(x-5)$$
$$(x-5{)}^2=0$$

$$x^2+19x+48=0$$
$$x^2+x(19)+48$$
$$x^2+x(3+16)+3(16)$$
$$x(x)+3x+16x+3(16)$$
$$x(x)+3(x)+x(16)+3(16)$$
$$x(x+16)+3(x+16)$$
$$(x+3)(x+16)=0$$

$$12x-12=0$$
$$12x-12=12(x-1)$$

$$36x^2-12x=0$$
$$36x^2-12x=12x(3x-1)$$

$$11x^3+15x^2=0$$
$$11x^3+15x^2=x^2(11x+15)$$
Keen in mind that $11x^3$ just means $11\times x\times x\times x$, which is how we can factor it out without issue.

$$A=\pi r^2$$
$$\pi \times 4^2$$
$$3.14\times 4^2$$
$$3.14\times 16$$
$$=50.24cm$$
This is a simple substitution given a formula which you can find under the formula section at the top of the page.