It's important for you to know the general rules for manipulating numbers and variables (algebra) so we will do a little refresher on these concepts to make sure you feel comfortable on that front. We will also review some important algebra tricks like factoring and completing the square which are useful when solving equations.

When an expression contains multiple things added together, we call those things terms. Furthermore, terms are usually composed of many things multiplied together. If we can write a number $x$ as $x=abc$, we say that $x$ factors into $a$, $b$ and $c$. We call $a$, $b$ and $c$ the factors of $x$.

Given any four numbers $a,b,c$ and $d$, we can use the following algebra properties:

1. Associative property: $a+b+c=(a+b)+c=a+(b+c)$ and $abc=(ab)c=a(bc)$.
2. Commutative property: $a+b=b+a$ and $ab=ba$.
3. Distributive property: $a(b+c)=ab+ac$.

We use the distributive property every time we expand a bracket. For example $a(b+c+d)=ab + ac + ad$. The opposite operation of expanding is called factoring and consists of taking out the common parts of an expression to the front of a bracket: $ac+ac = a(b+c)$. We will discuss both of these operations in this section and illustrate what they are used for.

The distributive property is useful when you are dealing with polynomials: \[ (x+3)(x+2)=x(x+2) + 3(x+2)= x^2 +x2 +3x + 6. \] We can now use the commutative property on the second term $x2=2x$, and then combine the two $x$ terms into a single one to obtain \[ (x+3)(x+2)= x^2 + 5x + 6. \]

This calculation shown above happens so often that it is good idea to see it in more abstract form: \[ (x+a)(x+b) = x(x+b) + a(x+b) = x^2 + (a+b)x + ab. \] The product of two linear terms (expressions of the form $x+?$) is equal to a quadratic expression. Furthermore, observe that the middle term on the right-hand side contains the sum of the two constants on the left-hand side while the third term contains the their product.

It is a very common for people to get this wrong and write down false equations like $(x+a)(x+b)=x^2+ab$ or $(x+a)(x+b)=x^2+a+b$ or some variation of the above. You will never make such a mistake if you keep in mind the distributive property and expand the expression using a step-by-step approach. As a second example, consider the slightly more complicated algebraic expression and its expansion: \[ \begin{align*} (x+a)(bx^2+cx+d) &= x(bx^2+cx+d) + a(bx^2+cx+d) \nl &= bx^3+cx^2+dx + abx^2 +acx +ad \nl &= bx^3+ (c+ab)x^2+(d+ac)x +ad. \end{align*} \] Note how we grouped together all the terms which contain $x^2$ in one term and all the terms which contain $x$ in a second term. This is a common pattern when dealing with expressions which contain different powers of $x$.

Suppose we are asked to solve for $t$ in the following equation \[ 7(3 + 4t) = 11(6t - 4). \] The unknown $t$ appears on both sides of the equation so it is not immediately obvious how to proceed.

To solve for $t$ in the above equation, we have to bring all the $t$ terms to one side and all the constant terms to the other side. The first step towards this goal is to expand the two brackets to obtain \[ 21 + 28t = 66t - 44. \] Now we move things around to get all the $t$s on the right-hand side and all the constants on the left-hand side \[ 21 + 44 = 66t - 28t. \] We see that $t$ is contained in both terms on the right-hand side so we can rewrite the equation as \[ 21 + 44 = (66 - 28)t. \] The answer is now obvious $t = \frac{21 + 44}{66 - 28} = \frac{65}{38}$.

Factoring means to take out some common part in a complicated expression so as to make it more compact. Suppose you are given the expression $6x^2y + 15x$ and you are asked to simplify it by “taking out” common factors. The expression has two terms and when we split each terms into it constituent factors we obtain: \[ 6x^2y + 15x = (3)(2)(x)(x)y + (5)(3)x. \] We see that the factors $x$ and $3$ appear in both terms. This means we can factor them out to the front like this: \[ 6x^2y + 15x = 3x(2xy+5). \] The expression on the right is easier to read than the expression on the right since it shows that the $3x$ part is common to both terms.

Here is another example of where factoring can help us simplify an expression: \[ 2x^2y + 2x + 4x = 2x(xy+1+2) = 2x(xy+3). \]

When dealing with a quadratic function, it is often useful to rewrite it as a product of two factors. Suppose you are given the quadratic function $f(x)=x^2-5x+6$ and asked to describe its properties. What are the roots of this function, i.e., for what values of $x$ is this function equal to zero? For which values of $x$ is the function positive and for which values is it negative?

When looking at the expression $f(x)=x^2-5x+6$, the properties of the function are not immediately apparent. However, if we factor the expression $x^2+5x+6$, we will be able to see its properties more clearly. To factor a quadratic expression is to express it as product of two factors: \[ f(x) = x^2-5x+6 = (x-2)(x-3). \] We can now see immediately that its solutions (roots) are at $x_1=2$ and $x_2=3$. You can also see that, for $x>3$, the function is positive since both factors will be positive. For $x<2$ both factors will be negative, but a negative times a negative gives positive, so the function will be positive overall. For values of $x$ such that $2<x<3$, the first factor will be positive, and the second negative so the overall function will be negative.

For some simple quadratics like the above one you can simply guess what the factors will be. For more complicated quadratic expressions, you need to use the quadratic formula. This will be the subject of the next section. For now let us continue with more algebra tricks.

Any quadratic expression $Ax^2+Bx+C$ can be written in the form $A(x-h)^2+k$. This is because all quadratic functions with the same quadratic coefficient are essentially shifted versions of each other. By completing the square we are making these shifts explicit. The value of $h$ is how much the function is shifted to the right and the value $k$ is the vertical shift.

Let's try to find the values $A,k,h$ for the quadratic expression discussed in the previous section: \[ x^2+5x+6 = A(x-h)^2+k = A(x^2-2hx + h^2) + k = Ax^2 - 2Ahx + Ah^2 + k. \]

By focussing on the quadratic terms on both sides of the equation we see that $A=1$, so we have \[ x^2+\underline{5x}+6 = x^2 \underline{-2hx} + h^2 + k. \] Next we look at the terms multiplying $x$ (underlined), and we see that $h=-2.5$, so we obtain \[ x^2+5x+\underline{6} = x^2 - 2(-2.5)x + \underline{(-2.5)^2 + k}. \] Finally, we pick a value of $k$ which would make the constant terms (underlined again) match \[ k = 6 - (-2.5)^2 = 6 - (2.5)^2 = 6 - \left(\frac{5}{2}\right)^2 = 6\times\frac{4}{4} - \frac{25}{4} = \frac{24 - 25}{4} = \frac{-1}{4}. \] This is how we complete the square, to obtain: \[ x^2+5x+6 = (x+2.5)^2 - \frac{1}{4}. \] The right-hand side in the above expression tells us that our function is equivalent to the basic function $x^2$, shifted $2.5$ units to the left, and $\frac{1}{4}$ units downwards. This would be really useful information if you ever had to draw this function, since it is easy to plot the basic graph of $x^2$ and then shift it appropriately.

It is important that you become comfortable with the procedure for completing the square outlined above. It is not very difficult, but it requires you to think carefully about the unknowns $h$ and $k$ and to choose their values appropriately. There is a simple rule you can remember for completing the square in an expression of the form $x^2+bx+c=(x-h)^2+k$: you have to use half of the coefficient of the $x$ term inside the bracket, i.e., $h=-\frac{b}{2}$. You can then work out both sides of the equation and choose $k$ so that the constant terms match. Take out a pen and a piece of paper now and verify that you can correctly complete the square in the following expressions $x^{2} - 6 x + 13=(x-3)^2 + 4$ and $x^{2} + 4 x + 1=(x + 2)^2 -3$.