Chapter 3: Algebra

Before we solve problems related to algebra, we must know what algebra is.
Algebra is the part of math where it uses constants (ex: 5, 3), variables (ex: a, xy), and more to form an equation called algebraic expression/equation. An example of this is: $$\mathrm{4a²\; +\; 2bc\; =\; d}$$ In this example, there are two terms. But first, what exactly is a term? An algebraic expression is formed with two or more terms. A term is either a single number or variable, or the product of several numbers or variables.

In one term, we have multiple parts called coefficients. A coefficient is a number by which another number or symbol is multiplied. We have three types of coefficients:

1. Numerical coefficient
   
   A numerical coefficient is the number being multiplied on. Let's take the first term of the example above, as an example. In $\mathrm{4a²}$, the numerical coefficient there is 4.
2. Literal coefficient
   
   A literal coefficient is the variable being multiplied on. In $\mathrm{4a²}$, $\mathrm{a²}$ is the literal coefficient.
3. Leading coefficient
   
   A leading coefficient is part of the literal coefficient. This is the exponent of the variable, and determines the order of the terms, hence the name. In $\mathrm{4a²}$, 2 is the literal coefficient due to it being the exponent. Let's take $\mathrm{2bc}$. The leading coefficient of this is 2, because $\mathrm{bc}$ can be reimagined as $\mathrm{b¹c¹}$, and 1+1 is 2.

With this, you should be good!

Operations on Algebraic Expressions

Adding

You can only combine the like terms. Like terms have the exact literal coefficient. $$\mathrm{5a\; +\; 2a\; +\; 6ab\; +\; 3ab}$$ $$\mathrm{ =\; 7a\; +\; 6ab\; +\; 3ab}$$ $$\mathrm{ =\; 7a\; +\; 9ab}$$

Subtracting

You can only subtract the like terms. $$\mathrm{5a\; -\; 2a\; -\; 6ab\; -\; 3ab}$$ $$\mathrm{ =\; 3a\; -\; 6ab\; -\; 3ab}$$ $$\mathrm{ =\; 3a\; -\; 3ab}$$

Multiplying

Example 1: $$\mathrm{5a(2a)\; +\; 6ab(3b)}$$ $$\mathrm{ =\; 10a²\; +\; 6ab(3b)}$$ $$\mathrm{ =\; 10a²\; +\; 18ab²}$$ Example 2: $$\mathrm{(5\; +\; b)(c\; +\; 9)}$$ For you to solve this, you must use the FOIL method. FOIL means First, Outside, Inside, Last. $$\mathrm{(a\; +\; b)(c\; +\; d)}\mathrm{ =\; ac\; +\; ad\; +\; bc\; +\; bd}$$ So let's solve! $$\mathrm{(5\; +\; b)(c\; +\; 9)}$$ $$\mathrm{ =\; 5c\; +\; 5(9)\; +\; bc\; +\; b(9)}$$ $$\mathrm{ =\; 5c\; +\; 45\; +\; bc\; +\; 9b}$$

Special Products

• $\mathrm{(x\; +\; y)²\; =\; x²\; +\; 2xy\; +\; y²}$
  because $$\mathrm{(x\; +\; y)²}$$ $$\mathrm{=\; (x\; +\; y)(x\; +\; y)}$$ $$\mathrm{=\; x(x)\; +\; xy\; +\; xy\; +\; y(y)}$$ $$\mathrm{=\; x²\; +\; 2xy\; +\; y²}$$
• $\mathrm{(x\; +\; y)(x\; -\; y)\; =\; x²\; -\; y²}$
  because $$\mathrm{(x\; +\; y)(x\; -\; y)}$$ $$\mathrm{=\; x(x)\; +\; x(-y)\; +\; xy\; +\; y(-y)}$$ $$\mathrm{=\; x²\; -\; xy\; +\; xy\; -\; y²}$$ $$\mathrm{=\; x²\; -\; y²}$$
• $\mathrm{(x\; +\; y)(x²\; -\; xy\; +\; y²)\; =\; x³\; +\; y³}$