Class 5 Notes I. Monomial Exponential Expressions A. Addition and Subtraction B. Multiplication and Division C. Exponentiation D. Practice Problems II. Polynomials A. Multiplying B. Dividing Polynomials by Monomials C. Dividing Polynomials by Polynomials D. Practice Problems I. Monomial Exponential Expressions This week we're going to start investigating a somewhat new type of algebraic expression. Thus far we've dealt with expressions involving the 4 basic arithmetic operations: addition, subtraction, multiplication, and division. For example, in the equation of a line we have these operations only. Thus far, we've left out the operation of exponentiation. That's what we're going to start thinking about this week. We're going to consider expressions that look like x^2, or x^5: expressions involving variables raised to powers. The first thing we're going to need is some basic rules for how to manipulate these expressions. A. Addition and Subtraction Consider the expression x^3. (on board) 2 x^3 We call this a monomial, which just means that it is a single number, variable, or product of numbers and variables. Suppose we try to add two monomials: (on board) x^3 + 2x^3 The x^3 and 2x^3 are like terms, so we can combine them to get (on board) 3x^3 Subtraction is just addition using negative numbers, so it works the same way: (on board) 5x^4 - x^4 What is this? (let class answer) (on board) 4x^4 So if we have monomials where the variables are the same, for example two monomials involving x^4, we can just add the number parts. This is true even if we have more complicated monomial expressions: (on board) 4 x^3 y^2 z^5 + 2 x^3 y^2 z^5 = 6 x^3 y^2 z^5 In contrast, if the variable parts of monomials are different, there is no way to combine terms. (on board) x^2 + x There's no way we can simplify this and combine the x^2 and x to get a single monomial. So the rule is: (on board) You can add/subtract the number parts of monomials that have the same variable parts. You cannot combine monomials with different variable parts. B. Multiplication and Division Now we know how to add and subtract, so let's move on and learn to multiply and divide. Consider (on board) x^3 . x^2 To multiply these, we have to remember what the powers mean. Recall that raising something to the 3rd power means multiplying that thing by itself 3 times, and similarly for squaring. (on board) x^3 = x . x . x x^2 = x . x Well then, we can write (on board) x^3 . x^2 = (x . x . x) . (x . x) This is just x muplitiplied by itself 5 times, so (on board) x^3 . x^2 = (x . x . x) . (x . x) = x^5 We can see the rule emerging from this quite naturally: when multiplying monomials, add the exponents. The same rule applies, variable by variable, when we have several variables: (on board) x^3 y^2 . x y^3 What is this? (let class answer) (on board) x^3 y^2 . x y^3 = x^4 y^5 Note that if there's no exponent written, there is an invisible 1. It's just like the coefficient in front of a variable. We can do division by a similar argument: (on board) x^5 / x^2 = (x . x . x . x . x) / (x . x) Here, we cancel two x's, and are left with (on board) x^5 / x^2 = (x . x . x . x . x) / (x . x) = x . x . x = x^3 So the rule with division is quite simple too: instead of adding, we subtract. We can write down these rules in symbolic form: (on board) x^m x^n = x^(m+n) x^m / x^n = x^(m-n) In other words, for addition add the exponents, for division subtract them. We'll see later this works whether m and n are positive, negative, or zero. C. Exponentiation The last basic operation we want to be able to do to monomial expressions is exponentiate them. In other words, we want to be able to simplify: (on board) (x^3)^2 Let's do exactly as we did with multiplication and division: let's write out what this means. The x^3 part means take x and multiply it by itself 3 times: (on board) (x^3)^2 = (x . x . x)^2 The squared part means take the thing inside the parenthesis and multiply it itself: (on board) (x^3)^2 = (x . x . x)^2 = (x . x . x) . (x . x . x) This is just x multiplied by itself 6 times, so (on board) (x^3)^2 = (x . x . x)^2 = (x . x . x) . (x . x . x) = x^6 Thus, we can see the rule for raising monomials to powers: just multiply the exponents. This again applies to multiple variables in a single monomial. For example, what is: (on board) (x^2 y^4)^3 We just take 2 times 3 for the exponent of x, and 4 times 3 for the exponent of y: (on board) (x^2 y^4)^3 = x^3 y^12 We can write out this rule symbolically as well: (on board) (x^m)^n = x^(m . n) In words, when raising a power to a power, just multiply the exponents. D. Group Work Practice Simplify: (-2x^2 y^3)(4 x y^2) Find the area of a rectangle whose sides have length 3x^3 y^2 and 2x^4 y^3 z^2 II. Polynomials Now we know a little something about monomials, so the natural thing to do is to consider expressions consisting of several of them. If we have monomials with different variable parts combined with addition or subtraction, we generally can't combine them to get a single monomial. This leads to expressions like (on board) x^3 + x + 2 These expressions, consisting of two or more monomials, are called polynomials. We want to learn how to manipulate expressions like these. A. Multiplication Adding polynomials is very simple, since it just consists of adding monomials. We won't bother talking about that expicitly. I'd like instead to focus on multiplication. We can multiply a monomial by a polynomial easily just using the distributive law. Here's what I mean. Consider: (on board) 3x^2 (x^3 + x + 2) We can do this just by distributing out the multiplication. The 3x^2 term is multiplying each of the monomials in the parentheses, so we can rewrite this as (on board) 3x^2 (x^3 + x + 2) = (3x^2)(x^3) + (3x^2)(x) + (3x^2)(2) This is just multiplying monomials, which we know how to do. How do we mutliply these? (let class answer) (on board) 3x^2 (x^3 + x + 2) = (3x^2)(x^3) + (3x^2)(x) + (3x^2)(2) = 3x^5 + 3x^4 + 6x^2 And that's it. We can use this distribution rule to multiply polynomials with arbitrary numbers of terms. One of the most common operations is multiplying two binomials -- a binomial is a polynomial with two terms. Consider: (on board) (2x - 1)(3x + 6) We can do this by noting that the 2x is multiplying everything in the second set of parentheses, and similarly for the -1. So this is (on board) (2x - 1)(3x + 6) = (2x)(3x + 6) + (-1)(3x + 6) Now we can distribute again. How do we do this? (let class answer) (on board) (2x - 1)(3x + 6) = (2x)(3x + 6) + (-1)(3x + 6) = (2x)(3x) + (2x)(6) + (-1)(3x) + (-1)(6) Now we've back to multiplying monomials. Before we go ahead and do the multiplication, I just want to point out a useful mnemonic for this called FOIL. FOIL stands for (on board) FOIL = first, outside, inside, last We got 4 terms when we multiplied the two binomials. The first one is (2x)(3x), which is just the product of the first terms of each binomial. So first means multiply the first two terms, in this case 2x and 3x. Next we have (2x)(6), which is just the product of the two outside terms, 2x and 6. After that, we have (-1)(3x), which is the product of the inside two terms, -1 and 3x. Finally, we have (-1)(6), which is the product of the two last terms. So FOIL means that to multiply two binomials, multiply first, then outside, then inside, then last, and add them all together. It's just a useful mnemonic. In any event, let's finish the problem. How do we simplify from here? (let class answer) (on board) (2x - 1)(3x + 6) = (2x)(3x + 6) + (-1)(3x + 6) = (2x)(3x) + (2x)(6) + (-1)(3x) + (-1)(6) = 6x^2 + 12x - 3x - 6 = 6x^2 - 9x - 6 Note that we can combine like terms in the middle here. That is often the case when we're multiplying binomials. B. Dividing Polynomials by Monomials We've already talked about how to divide monomials. It is also possible to divide polynomials, using a method called synthetic division. The easiest example of this is when we have a polynomial divided by a monomial. Here's an example: (8x^4 + 4x^3) / 2x^3 To handle this, we just break it up into two monomial divisions: (8x^4 + 4x^3) / 2x^3 = 8x^4 / 2x^3 + 4x^3 / 2x^3 These we know how to do: (8x^4 + 4x^3) / 2x^3 = 8x^4 / 2x^3 + 4x^3 / 2x^3 = 4 x + 2 It's also possible we will be able to divide some of monomials we get in this process but not others. For example: (4x^3 y^2 + 2x^4 y) / (x^3 y^2) = 4x^3 y^2 / (x^3 y^2) + 2x^4 y / (x^3 y^2) = 4 + 2x / y In this case we get something that's not a polynomial, but that's ok. C. Dividing Polynomials by Polynomials What if we want to divide by something that's not a monomial. For example: (x^2 + 12x - 8) / (x + 2) In this case, we set up the problem like a long division: _______________ x+2 ) x^2 + 12x - 8 We start by comparing the leading terms of the divisor and the dividend. In this case, the leading term of the divisor is x, and the leading term of the dividend is x^2. Since x^2 / x = x, we put x up top first, and multiply below: __________x____ x+2 ) x^2 + 12x - 8 -(x^2 + 2x) ----------- 10x - 8 Now we repeat with the leading term of the next line: 10x / x = 10, so we put 10 next: __________x_+_10 x+2 ) x^2 + 12x - 8 -(x^2 + 2x) ----------- 10x - 8 -(10x + 20) --------- - 28 We can't bring anything else down, so we have a remainder: -28 / (x+2) So the final answer is: (x^2 + 12x - 8) / (x+2) = x + 10 + (-28) / (x+2) Let's try another: (3x^4 + x^3 - 2x^2 + 1) / (x^2 + 1) In this case, the most important thing is to line up the columns properly: _______________3x^2_+__x_-_5 x^2 + 1 ) 3x^4 + x^3 - 2x^2 + 0x + 1 -(3x^4 + 3x^2) ------------------- x^3 - 5x^2 + 0x + 1 -(x^3 + x) ----------------- - 5x^2 - x + 1 -(-5x^2 - 5) ---------------- - x + 6 The final answer is: 3x^2 + x - 5 + (-x+6)/(x^2 + 1) D. Practice Problems Divide: (10 x^5 - 5x^4 + 20 x^2 + 2) / (5x^3 - 2x + 1)