MAT 51 Wladis Project 3: Combining Like Terms and the Distributive Property Simplifying expressions with like terms When adding like terms, the “common part” ALWAYS STAYS THE SAME. This is because combining like terms is just like counting up how many instances of that “common part” are present in the expression. This is also just like combining exponential expressions with common bases: the base never changes. Expression w Common Simplifcation Exponential Common Simplification like terms part expression base with common base 2
3 2
3 ⋅
⋅ ⋅ 2 3 5 2
3 ⋅
2
3 0
0
2
1
Now you try! Expression with like terms 4
3
2 4
3
⋅
Exponential Common Simplification expression base with common base 1⋅1⋅
Common Simplifcation part ⋅
Like Terms When two terms have parts in common that are identical, we can call them like terms, and we can combine them by adding together the remaining part of each term that is separate from the common part. For simple cases, this means that we can add together the coefficients of two terms if the variable parts of each term are identical. But this idea of like terms can be generalized to a much wider set of cases. Which of these are like terms? (Sometimes there is more than one right answer, depending on how you look at the problem!): 5
3
5
3
8
12
8√5
5√5
2√5 2√3
3√2 5
4 3
1
2
These two terms are like terms because they both have 3 in them: 5
These two terms are NOT like terms, because they don’t have anything in common: one term has an and the other has a These two terms are like terms because they both have in them: 8
12
These three terms are like terms because all three terms have √5 in them: 8 √5 5 √5 2 √5 These are NOT like terms, because the part under the radical is NOT the same. √3 and √2 are NOT the same and cannot be combined. Usually we say that these are NOT like terms, because and are not the same. This is correct. 1 However, we could also be creative and look at the problem this way: These two terms ARE like terms, because both terms have in them: 5
4 Usually we say that these are NOT like terms, because 1 and 1 are not the same. This is correct. However, we could also be creative and look at the problem this way: These two terms ARE like terms, because both terms have 1 in them: 3
1
2
1 5
3
Usually we say that these are NOT like terms, because and are not the same. This is correct. However, we could also be creative and look at the problem this way: These two terms ARE like terms, if we rewrite the as the product of two x’s: 5
3 ⋅ Now you try! Which terms are like terms and why? For like terms, circle the common part. Use the previous chart as a guide. If like terms, circle the common part in each term: 2
3 2
3 2
3 4
5
3√7
4√3
√7 3√2
2
2 2
√3 3 1
3 2
1 If these are not like terms, explain why: Combining like terms: When two or more terms are like terms, they can be combined into a single term, by identifying the identical part, and adding together the remaining part, or the coefficients of each term. We note that when adding like terms, the “common part” ALWAYS STAYS THE SAME. This is because combining like terms is just like counting up how many instances of that “identical part” are present in the expression. This is also just like combining exponential expressions with common bases: the base never changes. Here are some examples: Original expression 2
3 5
Common part 7
2√3
3
3
4
√2 2
1 1 2
2
√3 √2 2
5
5√3 4√2
Remaining part Combining like terms (coefficient) 2 3
2
3
2 3
5 1
3
7
5
1
3
5
7
5 7
2
2√3 5√3 2 5 √3
7√3 4√2 √2 4 1 √2
2
2
5√2 3
2
3
1
2
2
3
3
1
1 Explanation If we are adding 2 ’s and we add them to 3 other ’s, then we have 5 ’s altogether. ’s and we add If we are adding 5 them to 7 other ’s, then we have 2 ’s altogether. If we are adding 2 √3’s and we add them to 5 other √3’s, then we have 7 √3’s altogether. If we are adding 4 √2’s and we add them to 1 other √2, then we have 5 √2’s altogether. If we are adding 2 ’s and we add them to 3 ‐many other ’s, then we have 2 3 ‐many ’s altogether. If we are adding 3 many 1 ’s and we add them to 2 other 1 ’s, then we have 3
2 ‐many 1 ’s altogether. Now you try! Use the problems above as a model to combine the following like terms, or to explain why they cannot be combined. Original expression 5
2
7√2
7
7
3√2
2 2
3 √2 3
1
Common part 3 2
1 Remaining part Combining like terms (coefficient) The Distributive Property (and multiplying polynomials): Usually we simplify whatever is inside a set of parentheses before we work on whatever is outside the parentheses. But this isn’t always possible. Consider the following expression: 2 3
The terms inside the parentheses are NOT like terms, so they can’t be combined. So what can we do in this case? Examples 2 3
3
3
3
3
6
2 2
4 2
4
⋯
2
2 ⋅
2
2
3
3
⋯
This is just 2 3
2
2
Now you try! Simplify the following expressions: 3
4
2
! 2 3
4 4
⋯
4 This is just 2
2
4 ! 4 4 4
4
2
3
⋯
3
4
5
3
⋯
3
4
⋯
4
2 ⋅3
⋅ 2 4
2⋅3
⋅
4⋅2
⋅
⋅
6
8
3 2
3
9 2
3
9
2
3
9
2
2
3
3
3
2
2
This is just 3 2
3 3
3 9 ! 6
9
27 This is just 2
3
2
4
! 2 3
3
9
9 9
9 4
5 So what is the pattern? Look at each simplification problem in the table above (the ones that were already worked out and the ones that you worked out on your own). For each of these problems, here are the patterns. Fill in the ones for the problems that you did: Pre‐worked examples: Examples that you completed: 2 3
2 3
2
3
4
2
4
2
4
2 3
2
2
5
2
3
2
4
3
2
4
3
3 2
3
9
3 2
3 3
3
9 2 3
4
5
Looking at each of the above problems, what is the overall pattern? Explanation: This makes sense because in the original expression we are adding
thing as adding
-many times, and then adding
-many times. This is the same
-many times to that. Now you try!: Write your explanation here: Question: Does it matter how many terms are in the parentheses? So, for example, what would happen if we had this? Fill in below: Generalizing the distributive property to multiplying polynomials: Simplify 1 2
1 2
3 : 3
1 ⋅ 2
1 ⋅
3 We can distribute the polynomial regular distributive property. 1 to each term in the polynomial 2
3 using just the ⋅ 2
1⋅ 2
⋅
3
1⋅
3 We can distribute the 2 to each term in the first instance of the polynomial 1 and then distribute the 3 to each term of the second instance of the polynomial 1 . 2
2
3
3 We always do multiplication before addition when there are no parentheses, so we multiply each term out until it is simplified. 2
3 We combine the two like terms 2 , 3 . These are the only like terms in this expression. And since the order and grouping of terms does not matter during addition, we can go ahead and add these two terms together. 2
3 Adding a negative is the same thing as subtraction. Now you try! For this one, we have given you some of the steps, and you just need to fill in the missing ones, using the problem above as a reference: Simplify 2
2
5 3
5 3
2 : 2
⋅ 3
⋅
⋅
2 We can distribute the polynomial to each term in the polynomial 3
2 using just the regular distributive property. We can distribute the 3 to each term in the first instance of the polynomial , distribute the to each term of the second instance of the polynomial and then distribute the 2 to each term of the third instance of the polynomial . We always do ___________ before ___________ when there are no parentheses, so we multiply each term out until it is simplified. There are two pairs of like terms: ____ , _____ and _____, _____. We combine each pair of like terms to get _____ and _____. These are the only like terms in this expression. And since the order and grouping of terms does not matter during _______, we can go ahead and add these two terms together. Adding a negative is the same thing as ________________. Now use these previous two problems to practice the distributive property, by using them as models to do some problems on your own: Simplify, and show your math work here: 3
1 2
7 4
3 Explanation of each math step to the left: 3 2
3
2 2
3 4
2
3
7 Another worked examples, and ones for you to try on your own: Mathematical work showing one way to simplify the expression: 2
3
1
3
2
5 3
1
1 3
2
5 2
1
2
1 5 2
3
1
1 3
2
3
1
3
2
5 3
2
1
5 3
2
1
1
4 4 3
5
3
2
5
2
5
2
3
2 Explanation of each math step to the left: First we have to rewrite the negative sign as a 1, so that we can distribute the 1 to each term in the second polynomial. Now we distribute the 1 to each term in the second polynomial. Now we simplify each place where we are multiplying by 1 by doing the multiplication. Now we group like terms together. Now we combine like terms. The answer in the previous row is fine, but we can also rewrite it this way.