Rules of Exponents
Lesson 28
In lesson 28, our warm up evaluates exponents and evaluated expressions. We are going to evaluate coefficients with our monomials. So it is important for your student to remember the rule for evaluating exponents. We are going to “S” our solve problem, we will first read the problem, the length of a rectangle is represented 14x2 and the width of 3x. What is the area of the rectangle in terms of x? Our first step in “S” or study the problem is to underline the question. Our question is, what is the area of the rectangle in terms of x? Our second step is to answer the question, what is this problem asking me to find? This problem is asking me to find the area of the rectangle in terms of x. You will begin lesson 28 by discussing a monomial. 2x3is a monomial; in the monomial, the two is the coefficient, the x is your base, and your three is your the three is your exponent. Our coefficient is multiplied by the base and our exponent tells us how many of the bases we have. In other words, x to the third means x times x times x. We have several monomials listed. A monomial is a product of numbers and variables. It is one term, that is why we use the word monomial, because mono means one. Below several non monomials are listed. The difference here is there is addition or subtract; there is more than one term. A monomial is one term and a non monomial have more than one term.
In lesson 28 we have our graphic organizer which is given in the book. Your students will use this graphic organizer to record all of the rules they find for monomials; they can use this as a cheat sheet as they go through other lessons. In our first table we have 2 to the 3 times 22. We are going to expand this problem: 23, actually means 2 times 2 times 2; and 22, means 2 times 2. Our base of 2 is repeated 3 times and then repeated twice. How many 2’s are there all together? 1-2-3-4-5; there are 5 two’s all together. If we want to find our answer, looking back here, we have 2x2x2, which is 8; times 2x2, which is 4; and 8x4 is 32. We said that our 23 times 22 was actually the same as 25; and 25 is also equaled to 32. Our next problem, we have 24 times 23; 24 is 2x2x2x2, four times; times 23; we have 1-2-3-4-5-6-7 two’s; 24 is 16 times 23, which is 8, so we have an answer of 128; so we see that 24 times 23, is the same as 27, or 128; if we have 22, that is two 2’s times 25, that’s 5 two’s, 1-2-3-4-5; we have a total of 1-2-3-4-5-6-7 two’s; 2x2=4; 2x2=4x2=8x2=16x2=32; and 4x32 also has an answer of 128; so we see here that 22 times 25, is the same as 27 or 128. Look at the questions below the table, what is being done to the exponents in the first column to get the answer in the last column. When we had 23 and 22, they became 25, the 3+2=5, so we are adding the exponents; 24 times 23, 4+3=7, or 27; 4+3=7, we added our exponents; and 22times 25, 2+5=7 again, so we added our exponents.
Looking at our second table, we have x1 times x2; x1 is just one x; times x2, which is x times x; how many x’s are there all together; 1-2-3, so our answer is x3. In our second problem, x2 times x3; x2 is x times x, and x3 is x times x times x; this gives us 5 x’s, or an answer of x5. If we have x3, 1-2-3 x’s times x5, 1-2-3-4-5, we have a total of 1-2-3-4-5-6-7-8 x’s, or an answer of x8. Once again, looking at our exponents, we have 1 and 2 and they end up as 3; 2 and 3, end up as 5, 3 and 5 end up as 8, so we are adding our exponents. 1+2=3; 2+3=5; 3+5 is x8.
If we look at our numerical expression, we know that we can write this in expanded form. We have our 6, times 22 which would be 2 x 2 and this is multiplied by 4 times 23 which is 2 x 2 x 2. Because of the commutative property of multiplication we can change the order in which we multiply. If we change and bring our 6 and our 4 to the front, and move all of our 2’s together, we will have 1-2 from the first term, 1-2-3 from the second term, so that we have 5 twos, or 6 x 4 x 25. You multiply the 6 and 4 and add the exponents of the like terms, to get 25. In our table, we look at our first problem, 6x22 times 4x2. In this case the 6 and the 4 are our coefficients and we treat them just as we did in the problem above, we can bring them to the front and multiply them; we also have 2 powers, x2 and x3 that has the same base, so we can also multiply those together. If we expand the problem, we have 6 times x times x times 4 times x times x times x. The exponents on the variables are 2 and 3, which gives us an x5. When we multiply our coefficients we have 6 times 4, which gives us a final answer of 24x5. Our second problem we will expand, -3 times y times y times y times y and 5 times y times y. Our variable is y and we have 4 and 2, so 4 + 2=6, or y6; our coefficients are -3 and 5, so -3 times 5 is -15y6. In our next example, we have 12 x times y and we have 2 times x times x times y; if we look at our variables, we have 1-2-3 x’s, or x3; 1-2 y’s or y2; our coefficients are 12 and 2, so we have 12 x 2; which leaves us with a final answer of 24x3y2. Our next example, if we write our problem in expanded form, we will have 6 times a times b times b times b; times 3 times a times a times b times b times b times b; we have three variables in this case, so we have 1-2-3 a’s, or a3; 1-2-3-4-5-6 b’s or b6; and 1 c. Our coefficients are 6 and 3, so we have 6 x 3; so our final answer is 18 a3b6c. In our last problem, our expanded form is going to be 2 times a times b; times 3 times a times a times c; times 4 times b; this gives us 1-2-3 a’s, 1-2 b’s; and 1 c; our coefficients in this case are 2 x 3 x 4, which is 24 a3b2c.
We look at the first table we see that we need to find the expanded form. Our first example has 23 divided by 21 . 23 is 2x2x2 divided by 21 which is just one two. If we look again at our expanded form, we see that if there’s a 2 in the numerator and a 2 in the denominator, they can cancel. This leaves us with an answer of 22 or 4. Our second example is 25 , or 2x2x2x2x2, divided by 22 , which is 2x2; if we want to cancel out the twos in the numerator and in the denominator, we have 2 pair that cancel, we can cancel these because 2 divided by 2 is 1, so it will not affect the product or the quotient, and we have 23 which equals 8. Notice what is happening, 23 divided by 21 is 22 , 3-1=2; 25 divided by 22 is 23 ; 5-2=3. We are subtracting the exponents when we divide monomials with the same base. Our second table, looking at the expanded form, x4 divided by x3 , we have 4 x’s on top and 3 on the bottom, if we want to cancel out because x divided by x is also 1, 1-2-3 cancel, leaving us with x1 which is x. We have 6 x’s divided by x3 , we can cancel 1-2-3, leaving us with x3 . And we can see that our rule works again, they have the same base of x, x4 divided by x3 , is x4-3 or x1 ; x6 divided by x3 , they have the same base and 6-3=3. Moving to the next table you can see that these are the same problems and the same tables that are found in the student book. We have x3 divided by x4 , the expanded form, x3 , we have 1-2-3 x’s on top, 1-2-3-4 x’s in the denominator, so if we want to cancel, in this case we will have an x left in the denominator; if we were to subtract our exponents we would have 3-4, and 3-4 is x-1 We know also by canceling, looking at our expanded form and canceling, that we are left with a coefficient of 1 in the numerator because all of these cancel, x divided by x is 1, and so we have 1 over x as our expanded form. So this tells us that x-1 is the same as 1 over x. Look at the next example, we have x3 divided by x6 , that means we have 3 x’s in the numerator and 6 x’s in the denominator; and then we cancel, once again everything cancels in the numerator, leaving us with our coefficient of 1 and we have 3 x’s in the denominator; 3-6= -3; so x-3 is the same as 1 over x3 . The negative exponent actually moves the variable from the numerator to the denominator.
If we are going to look at our next table. This table involves monomials being divided and coefficients. Just as we multiplied our coefficients when we multiplied the monomials, we will do the same with our coefficients in division. They are like a separate problem, we are going to divide our coefficients as if there were no variables involved, and then we will simplify the expression with the variables. The expanded form of this problem would be 5 times 1-2-3-4 x’s over 10 and 2 x’s, what are the exponents of the variables? If we notice it is possible to cancel out two of the x’s, leaving us with x2 in the numerator. If we have our coefficient of 5 divided by 10, we can simplify this fraction by reducing it to ½ . From here we have a final answer of x2 times ½ or x2 divided by 2. Looking at the next problem, we have -36 times 2x’s and 1 y and 6 times 5 x’s and 3 y’s. Our exponents subtract, and if you see 2-5, that will leave you with a -3, which makes sense because when we cancel we are left with 3 x’s in the denominator. We have 1 y and 3 y’s so when they subtract you have -2, which makes sense because you are left with 2 y’s in the denominator. So you have 1 over x3 y2 . Looking at your coefficients, -36 is divisible by 6, so that equals -6. And our final answer is -6 divided by x3 y2 . You treat your coefficients as if the variables are not there and divide them and then you subtract your exponents of variables that have the same base.
Looking at our next table, if we want to write 23 divided by 23 in expanded form, we will have 3 two’s on top and 3 two’s on the bottom. Our 3 two’s in the numerator and denominator will cancel out, because 2 divided by 2 is one, 2 divided by 2 is 1, and 2 divided by 2 is 1, leaving us with an answer of 1. Now if you look at your exponents, you had 23 divided by 23 . When we were dividing in our last table, we subtracting the exponents when they had the same base. 23-3 is 20 which equals 1. So any base, raised to the 0 power is equaled to 1. Looking at our next table with the variables of x, x3 divided by x3 , we have 1-2-3 x’s divided by three x’s, if we want to cancel out, x divided by x =1, x divided by x=1, x divided by x=1. So our answer is 1, we know that everything simplifies to 1. Make sure your students understand that this in 1 not 0, x divided by x is 1. From here if we look at our exponent, x is our base, and 3-3 is the same as x0 . So any variable raised to the 0 power is equal to 1. Here we have the quantity of x raised to the 3rd power. If we were going to expand this, we would have the quantity of x times the quantity of x times the quantity of x, which leaves us with 1-2-3 x’s or x3 . If we have x2 raised to the third power (x2 )3 that means we have the quantity of x2 times the quantity of x2 times the quantity of x2 . This in turn means that we have (x times x) times (x times x) times (x times x), or 6 x’s, leaving us with x6 . If we have (x5 )2 that means we have x5 times x5 In other words we have 5 x’s times 5 more x’s, or a total of 10 x’s or x10 . When you have a power raised to a power, (x2 )3 … you have 2,3 and you end up with 6; 5, 2 and you end up with 10…so you are going to multiply the power. 2 x 3 gives me an exponent of 6; 5 x 2 gives me an exponent of 10. In our next table, we have a product raised to a power, (xy)3 would be (x times y) times (x times y) times (x times y); the product was repeated 3 times and it leaves us with x3 y3 . If we have (x2 y)2 , that means we have (x2 y) times (x2 y) ; this gives us 2 x’s and 1 y and 2 more x’s and 1 y, leaving us with 4 x’s and 2 y’s; we repeated the product twice and are left with x4 y2 . When you have a power you actually distribute it to the product. When we had xy, we distributed the power so that we had x3 y3 ; when we had x2 y, we distributed the square but in doing that, you have to multiply exponents, so you have x to the 2 x 2 or 4 and then y2 .
In our last table we are going to expand our problem. We have x divided by y to the third power. That means we have x divided by y times x divided by y times x divided by y. we have repeated this quotient three times leaving us with an answer of x to the third divided by y to the third. If we have looking at our next problem, a to the second times b and we repeat it twice, then this is the same as a times as over b, a times an over, b we have repeated our quotient twice and we have four a’s, 1,2,3,4 over 2 b’s or a to the fourth b to the second. Looking at our last example x to the third y to the second times x to the third y to the second that means we have three x’s 1, 2, 3 divided by two y’s.
And three x’s 1,2,3, divided by 2 y’s for when we repeat the quotient twice we have 1,2,3,4,5,6, or x to the 6th over 1,2,3,4 y to the fourth, what we have done is when we have a quotient raised to a power we are distributing that power to the numerator and the denominator in the quotient. X divided by y raised to the third power, we distribute the three to the x and to the y. we distribute the 2 to the A squared which gives us a to the fourth and to the b which gives us b squared. We distribute the 2 to the x to the third and to the y squared which gives us x to the 6th and two times two is y to the 4th.
In our solve problem, we have already “S” the problem and we know this problem is asking us to find, the area of the rectangle in terms of x. In “O” organize the facts, we will first identify our facts. The length of a rectangle is represented by 14x2, this is a fact; and the width of 3x, another fact. We have to determine if these facts are necessary or unnecessary. Our first fact, the length of a rectangle is represented 14x2 . This is a necessary fact. The width is 3x; this also is a necessary fact. We list our necessary facts. In “L” we line up a plan. We have to choose our operation or operations. Because we are finding the area of a rectangle, our operation is going to be multiplication. We will now write in words what our plan of action will be. I will multiply the length and width since the figure is a rectangle. The area of a rectangle equals length times width. Keep in mind that when you are lining up a plan and you are writing your plan, it does not have to be exactly the same. All students are not going to write this exact plan but as long as they have the correct answer, it will give them the correct answer, and they are not using numbers, then their plan is good. In “V” verify your plan with action, we will first estimate. Because we see that we are multiplying two monomials, we know that our answer will also be a monomial. To carry out our plan, we know that the area of a rectangle is equal to the length times width (A=l x w). Our length is 14x2 and our width is 3x. When we multiply two monomials, we multiply the coefficients, 14 x 3= 42; x2 times x, that means we have 3 x’s or x3. In “E” examine your results, you will ask yourself several questions, Number 1, does your answer make sense? If we go back to our question, what is the area of a rectangle in terms of x…an answer of 42x3 does make sense. Is your answer reasonable? We said in our estimate that it would be a monomial, so our answer is a monomial. Is your answer accurate? Your students could rework the problem again to make sure they multiplied correctly. And then we write our answer as a complete thought, the area of the rectangle is 42x3. To close our lesson, we will review the essential questions. Number 1, what happens to the coefficients and exponents when we multiply monomials? The coefficients are multiplied and the exponents are added as long as they have the same base. Number 2, what happens to the coefficients and exponents when we divide monomials? The coefficients are divided or simplified and the exponents are subtracted as long as they have the same base. Negative exponents may because positive by moving to the denominator.