Hi. Chris and I wanted to say congratulations on finishing the first third of Visualizing Algebra. >> You've learned a lot about equations, inequalities, fractions and even systems of equations. >> Next we're going to be looking at rational expressions, factoring, radicals and exponents. >> Get ready for some excellent units. >> Good luck.
To start this unit, we'll look at Math Quill so you can input exponents and scientific notation with ease. You might have already discovered how to enter exponents using Math Quill, and even if you haven't here's a quick walk-through on what to do. First you want to type the letter x, this will create an x in your math quill box. Then you want to type carrot symbol, you notice that right before you type the carrot symbol your cursor bar will appear after the x. Immediately after you type the carrot key, the cursor bar will move up and in the position for an exponent, or a superscript. You will then type the number 4 to get the 4 as the exponent. And here's where you can run into trouble. If you keep continue typing, you will be typing exponents or powers. To move the cursor bar back down to the main line, you want to hit the right arrow key. Hitting the right arrow key gets us to the main line and now it continues typing the rest of our expression. So finish entering this expression, and then try these other ones. There's also a challenge problem down here. It contains both an exponent and a fraction.
For the first one, you need to type y, the caret key, the negative symbol and then 5. For the second expression, we'll start by entering z, the caret symbol and then 6. Remember this caret will move our cursor bar from the main line up into the exponent position. We hit the right arrow key to continue back down to the main line, then we type a plus sign, then number 3, and the variable x. To get this exponent of 3, we type our carrot symbol and then the number 3. Our cursor bar would be here at this point, so we need to move it back down to the main line, so we hit the right arrow key. And we finish out by typing y, the arrow key, and then the number 3. For the last expression, we start by typing in the negative sign and then a forward slash. The forward slash will create two empty spots for the numerator and the denominator. To get our numerator we simply type 5y, the carat key, and the number 2. At this point a cursor bar will be in the exponent position. We hit the right arrow key once, which will send our cursor bar back down to the main line of the numerator. To move into the denominator, we hit the right arrow key one more time. Now that our cursor bar is down to the denominator, we type 8 to get our full fraction. Now that we have our fraction, we need to move our cursor bar out of it, we hit the right arrow key one more time and then we can enter in the plus sign x the caret key and the number three. This will gove us our second term. Our cursor bar would be here, so we hit the right arrow key to get back to the main line and then we type the plus sign and 2. This gives us our full expression.
Welcome back. In the last unit, we compared graphs and equations to determine if they intersected, were parallel or if they represented the same line. In this unit, we'll focus on exponent rules and polynomials. By the end of this unit, I hope you'll be able to find patterns while working with exponents and use those patterns to help you add, subtract, multiply and divide polynomials. Only one lesson in this unit will focus on exponents, but we'll use everything we learn here for the rest of the unit with polynomials.
Let's start off by looking at some exponent patterns. Here are some expressions. And here are those same expressions expanded based on the exponents. I have three 3s multiplied together and three 2s multiplied together. Because the exponents were 3 and 2. The same is true down here. I have two 2s and three 5s. If I wanted to simplify these expressions, I would write the final answer with one base and the exponent. For example, for this case I would have three 5s. So, I could write this answer as 3 to the 5th. This second one's a little bit tricky. I have a base of 2 and a base of 5. The bases are different, so I can't combine these two bases and write them with an exponent. I have to keep them separate. I still have 2 twos and 3 fives. When we simplify expressions with exponents, we want to try and combine the exponents that have the same base. So think about what I did here. What did I really do with these exponents to get this answer? And what did we have to do in this case? And I know it seems weird that we have an answer, that's the original question but sometimes that happens. Sometimes we can't combine the powers. We could, of course, multiply each of these out to get a number, but I wasn't focused on the number. I wanted to look at the exponents and what we're doing with them.
Based on what we just covered, try and answer this question. What are the bases and exponents for the expressions, combine any bases if you can. And write your answer in the box.
If you got these answers excellent work. This would be the expanded form for the first problem. We have four 2s and three 2s. So all together I would have 2 to the 7th. I have seven 2s multiplied together. The second problem is very similar to this first problem. Instead of having a 3 as our base, we have an x. We have some unknown number. So we have 3 of those x's multiplied by 2 of those X's. So all together, we have 5x's multiplied. Or X to the 5th. Here's the expanded form for the 3rd expression. Notice that I have x's and y's. So these bases aren't the same. And i can't combine them. I just have to list x to the 3rd and y to the 4th. That's my answer. >> For the last problem, this is the expanded expression. Since I have a product of factors, I can rearrange the factors to group the like terms together. For example, these x's can be moved over here. And these y's can be moved to the right. We can do this because the community property of multiplication. We know 3 times 2 is the same as 2 times 3. If ever I have a bunch of factors, I can rearrange the order of those factors. We'll still wind up with the same number. Now this is a little bit easier to see. I have x to the 7th and y to the 7th. So these are the most simplified expressions with exponents.
Let's try one more question. What do you think is an equivalent expression for 5 to the 8th times 5 to the 4th? Choose the best answer.
The answer is 5 to the 12th. Great job if you got it. For 5 to the 8th, we would write a 5, 8 times. And then we're going to write a 5, 4 times. Here are 8 5's. And here are 4 5's. I have them multiplied all together, so I could use an exponent to represent this repeated multiplication. I have 5 to the 12th.
Notice that in all these cases, we perform the same steps of the same operation with the exponents that have the same base. So what general rule could you write for any base that's non-zero, and its exponents a and b? What would x to the a times x to the b always equal? As a hint, think back to what you did in the last problems. What were you doing with these exponents?
It turns out we were adding the exponents together. Whenever we have a non zero base x. We can simply combine their exponents by adding them. Just like in our first problem we encountered. We had 3 to the third times 3 squared. Well, we knew we had 3 threes and 2 threes. And multiplied all together, 3 to the 5th. So, we added the exponents.
We've looked at multiplying identical bases together, as well as different bases together, now let's look at powers of powers. I have a base raised to a power, and then that quantity, raised to another power. We're going to try and come up with an exponent rule to handle these. This cubed power indicates I should write each of these and write our final answer as two to some power. So what do you think that would be? Can you write the expressions with one base and one exponent here? Do the same for these problems.
Excellent work if you got these correct. When I expand each of these, I would have three 2s, three 2s, and three 2s, That would make nine 2s multiplied all together, so 2 to the 9th. For the second problem, I have 8 to the 4th raised the 3rd power, I multiply 8 to the 4th, 3 times then I can expand each 8 to the one, we have 5 to the 3rd three times, which means we would have nine 5s multiplied all together, 5 to the 9th. For the final problem, we have x squared to the 4th, our powers still apply to the variable. Remember, variables are just numbers, so it would work the same in this case, as it does in these three. First I multiply x squared by itself four times, then I have x multiplied by itself eight times, x to the 8th.
There must be a shorter way, though, than doing all this work. That's a lot of writing, and I don't want to do that every time I work with an exponent. So what else is going on here? Maybe you can find the pattern. Take a look at the exponents on the left and the exponents on the right and see if you can answer this question. What general rule can you find for raising this base and power to another power? Choose the best answer.
It turns out we multiply the exponents together, this choice. Great work if you identified the pattern. If this question gave you some trouble, go back and look at the last slide. You'll notice that the powers on the left will multiply to give you the powers on the right.
Let's look at a more complicated case of powers of powers. Here, I don't just have one base, I have two I'm multiplying 4 cubed and 3 squared. And this quantity will be raised to the 4th power. If I want to simplify this problem and in particular the exponents, I would need to rewrite this. First, I write 4 cubed times 3 squared four times, that would be this. Notice that I have multiplication between every single term, so this is really just a product of factors, and I don't need to list the parenthesis. We can rearrange the product of our factors and I can group the 4 cubes together and the 3 squared's together. If I expanded all this out I would have 12 4s and 8 3s. So I know my final answer is 4 to the 12th times 3 to the 8th. So I know my final answer would be 4 to the 12th and 3 to the 8th. Again, I don't want to have to list all these factors out when I'm doing this work. There must be a shorter way. If I look at how these exponents changed, what did I really do? Well it turns out that I distributed this exponent of four to each of the factors 4 cubed and 3 squared. So I have 4 to the 3rd raised to the 4th power and 3 squared raised to the 4th power. Now we just apply the rule that we learned before. When we raise a power to a power, we multiply the exponents together. So I get 4 to the 12th and 3 to the 8th. Pretty neat.
Let's see if you can put this together. Try and answer this question. What are the bases and exponents of these expressions in simplified form? Write your bases and exponents in each of the boxes. And remember, this 3 has an exponent of 1, when the exponent isn't listed, we assume it to be 1, the same is true for the c. Also for these numbers, don't worry about multiplying them out. For example, if you have the base of 7 in your answer, leave it as a base of 7 with some exponent as the number. As always, try your best and feel free to play around with these exponents, it's okay to make a mistake.
Here are the correct expressions. If you got these, excellent work. I know these last two were pretty tricky. So, it's okay if you didn't get them right on your first try. First, we can distribute the 5th power to the 7 squared and the x squared. Now, we apply our power to our power rule. We multiply 2 times 5 to get power of 5, because they had factors here. Numbers and variables multiplied together. Then next one, we distribute the square to 3 to the 1, n cubed and n squared. And we get this. We multiply the exponents together to get 3 squared. N to the 6th, and n to the 4th. Finally, for the last one, we carry out the same steps. We have a cubed, cubed. B squared cubed and c to the 1 cubed. Raising a power to a power. We get a to the 9th, b to the 6th and c cubed. There's our final answer.
These were the expressions from the last problem, and these were the answers. What was going on here? What's the pattern that we could find? In other words, what do we do with the exponents a and b, when they're raised to another power, c? Choose the best rule that represents what we did here.
We simply distributed or multiplied the c to each of the exponents. So that would be this choice. For the base of 7, we had an exponent of 2. And 2 times 5 makes 10. For x, the exponent was also 2. So, 2 to the 5th was x to the 10th. We were simply multiplying each exponent by 5. The same was true in the second case. 1 times 2 is 2. 3 times 2 is 6. And 2 times 2 is 4. That's how we got 3 squared. M to the 6th times n to the 4th. This was a pretty tricky pattern so, great job if you found the correct answer.
Let's try putting all these properties of powers together to answer this question. What would be this in simplified form? Write your bases and your exponents. This time, for the numbers, I actually want you to multiply them together. For example, if 2 squared times 3 squared was in your answer, I would want you to write that as 4 times 9, or 36. 36 should be the coefficient in front of the box. Don't use the powers. This is a pretty tough problem, so take your time with it. I don't expect you to get it right on the first time, but I do want you to play around with the exponents. Try expanding this, and see what you get.
We have negative 5x cubed times y squared, squared. So I should write this twice. Now that I have a product of factors, I can regroup the basis that are alike. So I have two negative 5, two x cubes, and two y squared's multiplied together. That would have been the same if I would just distributed the square to each of the terms. I would have had two negative 5, two x cubes, and two y squared. Now that I'm here, I can multiply my numbers together, and my bases together. Negative 5 times negative 5 is positive 25 times 2 is 50. For the base of x, I simply add the exponents together, 3 plus 3 makes 6 and 7 makes 13. The y's are a little bit easier, so I just have y to the 4th. All together I have 50 times x to the 13th times x to the 4th. If you reasoned your way somewhat through that or got any of these exponents right, great job. If you got this right on your first try, that's even more impressive, great work.
Here's another one to try out. You can put your answer in this box. And remember, you should have a number as a leading coefficient and then bases are variables raised to exponents.
When starting the problem, we want to recall that the negative 3 has an exponent of 1. Now, we can raise each of these bases to this power of 3. We multiply every exponent on the inside of the parentheses by the exponent on the outside of the parentheses. And we get negative 3 to the 3rd, b to the 12th, c to the and my variables with the same basis together. Finally, I multiply the numbers and I add the exponents. Our final answer is negative 54 times B to the 15th times c to the 11th. Fantastic work if you got it.
In a couple of the problems before, we distributed the parenthesis on the outside to the exponents on the inside of the parenthesis. We can also do this with fractions. If we think all the way back to unit 1, we saw 1 3rd times 1 3rd times 1 3rd. We knew that result was 1 27th. We found this answer by multiplying thought about this. I know the exponent on this base, 1, is a 1. And the exponent on this base of 3 is also a 1. We can distribute this exponent of 3, or multiply it to each of the exponents on the inside of the parentheses. So we've a base of 1 and an exponent of 1 times 3, and a base of 3 with an exponent of 1 times 3. This is really 1 cubed, and this is really 3 cubed, or 1 divided by 27. If we can get to this step more quickly, then we can get to our answer, 1 cubed is 1, and 3 cubed is 27. Here we have a negative sign, and we're going to square it, so we know our answer should be positive. So I'm going to have negative 5 squared and x squared. So all together, we have 25 divided by x squared. This last case of variables works just like this first example. We'll have a to the the parenthesis by the exponent on the outside. We raised a power to a power.
Try simplifying these expressions. Write your answer as a fraction in this box and if your answer is something like 2 squared over 3 squared, I want you to evaluate the number. You would want to write your answer as 4 9ths, because 2 squared is 4 and 3 squared is 9.
Here the correct answer is. Did you get them right? If so nice work. I know this last one is little bit tricky. So don't worry if you didn't get all of these right. For the first one we would have 6 7ths times 6 7ths or 6 squared over 7 squared. So 36 over 49. For the next one I would have negative 3 squared and 4 squared. So 9 divided by 16. And finally, for this last one, I have x divided by simply 1.
We looked at positive powers. So let's check out some negative powers. A negative exponent tells us to write the reciprocal of a base, and write it with a positive power. So here the reciprocal of 3 is 1 3rd and then I write this with a positive power of 2. This negative sign is kind of like an operation. It tells us to write the reciprocal. Of our base. In the second example we need to be careful, we have 5 times x to the negative 3. The exponent on the 5, is a positive 1, so we're just going to list the 5 as a factor. Now, we take the reciprocal of x, which is 1 over x, and then change the negative power to a positive 3. For a fraction, we'll take the reciprocal of the base, which is 3 get 1 9th, 5 divided by x cubed, and 9 16ths. Looking back at our work, we can see that a negative exponent in a numerator moves the factor from the numerator, to the denominator. The base of 3 was the numerator, and it ended in the denominator. The same is true for the x and for the 4.
Let's try some practice with negative powers. What do you think this expression is equal to? Write the correct letter here.
This one would be d. For a negative exponent we take the reciprocal of the base and raise it to a positive power. We know 1 divided by y cubed is the same as 1 divided by y cubed. Next, we multiply our numerators and denominators together to get 7x squared divided by y cubed. This choice. To get to the answer more quickly, think about what a negative exponent does. It moves the base from the numerator to the denominator, just like in d.
Let's try another. What do you think this expression is equal to? Write the letter in the box.
Answer choice e is correct here. Great job if you're figuring out the meaning of negative exponents. We know negative exponents would give us the reciprocal of the base, and then we change each negative exponent to it's positive form. In the end, I'll have 7 times 4 times 1, which is just 7, over x squared times y cubed. And again, our negative exponents move the bases to the denominator with positive exponents.
What about this one? What's this expression equal to?
This one would be C. Excellent work if you figured it out. Here, the negative 7 doesn't mean we do the reciprocal. A negative sign in front of a number just makes it negative. The actual move to the denominator because the power is negative. So we'll have x squared down there. This matches with answer choice C.
Here's the last in our rapid fire series. What do you think this one is?
This one was answer choice a. Excellent work if you figured it out. For the negative exponent, we'll do the reciprocal of our fraction, so we'll have y cubed over 7x squared. Any based race to the 1 power is just itself. So, here's our answer, a.
We've seen what a negative exponent does to something in the numerator, it moves the base to the denominator. So if a negative exponent moves something in the numerator to the denominator, then a negative exponent in the denominator must move that to the numerator. When we first saw negative exponents, we learned that a base raised to a negative exponent, would be the reciprocal of that base raised to its positive exponent. This can also work in the opposite way, like if we have a negative exponent in the denominator. A negative exponent indicates to write the reciprocal of the base to the positive power, so I have 1 divided by 1 dividing by a fraction is the same as multiplying by a reciprocal, so I can rewrite this expression as 1 times 4 squared. And again, notice a negative exponent in the denominator, moves the base to the numerator with its positive exponent. We can use the same reasoning with variables. I see a negative exponent in the denominator, so I know that this base is going to wind up with a positive power in the numerator, y to the 5th. For this last one we want to be careful, 5 has an exponent of 1, and x has an exponent of negative 4, so only the x term is moved to the numerator. When we handle this negative exponent, we rewrite our expression as x to the 4th divided by 5. In general here's the rule we can write for base raised to a negative exponent that appears in the denominator, we simply write it in the numerator, with its positive exponent.
Here's a different type of practice problem. For this one, I want you to write each expression using only positive exponents. Then, I want you to check the boxes for the ones that are equal. Good luck.
The first expression would be 2 to the 4th n cubed. I moved the n from the denominator to the numerator since it has a negative exponent. For this expression, I move the base of m to the denominator and the base of n to the numerator. Here, these 2 bases come to the numerator. So aha, I know these two are equal. This one doesn't match these other expressions, so it's out. For the second one, the m and the n both move to the numerator with positive exponents, but the 2 stays in the denominator, because the exponent is a positive 1. This one would be out, too. For this last one, I have a negative sign in front of the The n to the negative 3, however, moves up to the numerator because the exponent is negative. This doesn't match these, so it's out as well. For the last problem, the two moves up to the numerator because the exponent is negative as well as the m. This definitely matches the other expressions. So here are the three. This was a really tricky quiz, so don't worry if you didn't get it right on the first try. I really wanted you to focus on negative exponents, negative numbers, and negative exponents for numbers. We have to keep in mind these differences when working with problems.
We have one more pattern to cover with bases and exponents and this involves division. I can think about expanding 3 to the 4th and 3 squared, just like before. We can simplify common factors in the numerator and denominator. They make 1. Because we're working with a fraction made up of factors, we know these make 1. So we're left with 3 times 3, or 3 squared. We can think about expanding these variables as well. We simplify common factors in the numerator and in the denominator, and we're left with f to the 3rd. This last one is just like the others before except we needed to do something additional for the two numbers or coefficients. I know 6 divides into 12 and 6 divides into 18. 6 into 12 is 2, and 6 into 18 is 3. So I have 2 3rds times w to the 5th. I've showed you the expanded form, and the simplification, so that way we could look at the pattern. What's happening as we move from the left side to the right side? What do you notice about the exponents? Let's see if you can answer this with a question. What's a rule for the pattern that you see over here? You can write your answer in this box. And be sure that it includes a base of x, and involves your exponents a and b. What'd you do with them?
It turns out that we subtract the exponents. We do a minus b. We can see that in the first example, we have 3 to the 4th divided by 3 squared. So 3 to the 4 minus 2, or 3 squared. The same is true for f. We have 8 minus 5, or f cubed. For the last one, we leave our numbers in the front as a fraction, and then we have w to the 8 minus 3, or w to the 5th. We saw the fraction was reduced to 2
All right, I'm going to throw something challenging at you. I want you to find these expressions simplified and I only want you to write them using positive powers. You can write your answers in these boxes, and they may contain fractions.
When dividing bases, we subtract the exponents. So we have 3 to the 2 minus 4 or denominator with the power of 2. So we'll have 1 over 3 squared or 1 9th. Here I'll subtract the exponents again, f to the 5 minus 8 or f to the negative 3. We know this f will be moved to the denominator with a positive power. So 1 over f cubed. In both this first example and the second example, you want to think, where did you have the higher exponent? It was in the denominator. The 3 to the exponent than f to the 5th. If we look at expanded form, we can see that the 3's will simplify, and we're left with 3 of the 2's in the denominator or 1 9th. The same is true with the f's. 5 of the f's simplify to 1, and then we're left with f cubed in the denominator. This should make sense. I have more f's in the denominator than I do in numerator. We use a similar resoaning to figure out the last one. We should have five more w's in the denominator than the numerator. So I know w to the 5th is that my denominator. Now I need to figure out the other two numbers, 8 divides into 16 and 40 so I'd be left with 2 5th. So I have 2 divided by 5 times w to the 5th. There's my last fraction. I hope you found a lot of different ways to make sense of these exponents. There's no one way or wrong way to figure them out. You can apply the properties of power or look at the expanded forms to get your answers. Just be sure that it's making sense along the way.
Here's your chance for another practice. Try simplifying this expression. Only write the answer with positive exponents. Try using expanded form or the rules you came up with for exponents.
Here's the simplified expression. Great work if you got it correct. We can divide 10 and 15 by 5, so I get 2 3rds in the front. Then I notice that I'll have 3 more g's in my numerator than in my denominator. So g cubed in the numerator. Finally, I'll have 3 more h's in my denominator than I would in my numerator. So I'll have an h cubed, or h to the 3rd, down below. But remember I said there's always more than one way to think about this. We can simplify our fraction in the front, just like from before, so we'll have 2 3rds. Then we'll have g to the 7 minus 4 here, we follow that with h to the 5 minus 8, 7 minus 4 makes 3, and 5 minus 8 is negative 3. G to the 3rd, has a positive exponent, so we make sure we list it in the numerator. H to the negative 3 has a negative exponent, so we know that moves to the denominator, and we change the sign to make it positive. Think about the two ways to do this problem, and choose the best that works for you.
Let's look at another simplification process involving negative exponents. I'm going to show you two ways of doing this problem, and then you can pick which one you prefer. We learned before that a base with a negative exponent can be moved to the denominator. So we'll have 1 divided by 2 cubed times 2 squared, or exponent rules directly. We know when we divide bases, we subtract the exponents. So we have 2 to the negative 2 minus 3, negative 2 minus 3 equals negative 5. And finally we have a base with a negative exponent, so we move the base to the denominator with a positive exponent. Leaving us with 1 divided by 2 to the 5th. And that's exactly the same as we got before.
Let's see if you have a handle on this. What do you think this expression simplifies to? Write your answer here.
Well I know this negative exponent will move to the numerator. So I have y to the 5th times y to the 4th, next we add the exponents together to get y to the would have done y to the 5th minus negative 4. We subtract the second exponent, this also leaves us with y to the 9th, our answer.
Here's one last simplification that I want to show you. Again, there's multiple ways to do that type of problem. So, I'm going to show you two, and you choose what you prefer. In the first way, I'm going to start outside the parenthesis and take care of this negative exponent. I do the reciprocal of the entire inside parenthesis to get 1 over 3a to the 4th, b to the negative 3 squared. Next, I square each factor in the parenthesis. 1 squared is 1. 3 squared is 9. A to the 4th squared is a to the 8th. And b to the negative 3 squared is b to the negative 6. Finally, we move b to the negative 6 to the numerator for our final result. The second way of doing this problem is starting inside the parentheses. We can move any negative exponents to the denominator. So I have 3a to the 4th divided by b cubed, and then all of this is raised to the negative 2. We have a fraction that's a base raised to a negative exponent. So we take the reciprocal of this entire fraction and change the exponent to be positive. Finally, b cubed squared is b to the 6, 3 squared is 9, and a to the 4 squared is a to the 8th. Here's our anwer. Both of these results are the same. So we know our work is correct. In the first case, i started with a negative exponent here. Then I handled the negative exponent on the inside. For the second case, I handled this negative exponent on the inside first, then I did the negative exponent on the outside. You'll have to move terms around either way you do it, just get comfortable with one way.
Here's the last question for the lesson. Try simplifying this expression and write it with positive exponents in the box. Good luck.
Our final answer is a to the 6th, times b to the 8th, divided by 4. You might have written it as 1 4th times a to the 6th, times b to the 8th. This is also correct. You would have just been putting the 1 4th out in front of our two factors. We start by moving this a to the negative 3 to denominator. Next, we take the reciprocal of our base, because we have a negative exponent. So I flip the fraction over, and change my exponent from negative to positive. Finally, we square each of our factors. A cube squared is a to the 6th. B to the 4th squared is b to the 8th. And 2 squared is 4. Here's our final answer.