Mathematics

4.NBT.B.5 Transcript

This is Common Core state standards support video in mathematics, the standard is 4.NBT.B.5.

This standard reads multiply a whole number of up to four digits by a one-digit whole number, and multiply two two-digit numbers, using strategies based on place value and the properties of operations. Illustrate and explain the calculation by using equations, rectangular arrays, and/or area models.

There are standards that are related to this. Back in third grade, we had standard 3.NBT.A.3, which dealt with multiplying one-digit whole numbers by multiples of 10; so that was an initial exposure. Then in the same grade level, we have a couple of related standards 4.NBT.A.2, where students are to read and write multi-digit whole numbers using base-ten numerals, number names, and expanded form, and also compare those numbers. We also have standard 4.NBT.A.3, where students use place value understanding to round multi-digit whole numbers.

This standard that we're dealing with is very important, because if we look ahead, it's really laying the foundation for standard 5.NBT.B.5, where students are expected to fluently multiply multi-digit whole numbers using the standard algorithm. It's also the foundation for standard 5.NBT.B.7, where students will be adding, subtracting, multiplying, dividing decimals up to the hundreds place. We also need to be concerned about addressing the standards for mathematical practice. But if we use the right strategies and activities in this standard, we can cover these: reason abstractly and quantitatively, construct viable arguments and critique the reasoning of others, use appropriate tools strategically, attend to precision, and look for express regularity in repeated reasoning.

Let's look at the first part of the standard where we multiply a whole number of up to four digits by a one-digit whole number. So let's just jump straight to four digits. So let's say we have 3723 times 5. Most of us went straight to the standard algorithm where we were taught 5 times 3 is 15 (and this is terrible terminology), carry the one, then we have 5 times 2 is 10, plus that 1 is 11. So we write down the 1 and carry the 1, and we have 5 times 7 is 35 plus the one that's 36. Write down our 6, and the infamous carry the 3. Then we have 5 times 3 is 15, plus the 3, is 18.

Now in doing the standard algorithm, we lose sight of a lot of things, especially place value. It becomes a procedure where students really don't understand what in the world is going on. It's important, a key piece for this standard is that, we need to use strategies based on place value and the properties of operations. Now it could be done with the standard algorithm, but it's rare. I know that I wasn't shown this, that we had 5 times 3 is 15, and we wrote down our 5 because that was five ones, and then the other ten ones we actually composed them into one ten. We converted the ten one into ten ones, and so that's what this one is. Then we had 5 times 2, but that's really five times to tens, which is ten tens plus the one ten here, that gave us eleven tens. Then we actually took those ten tens and made them one hundred, which is what this is.

So again, it could be done, but it's better to use those properties and use place value prior to ever going to the standard algorithm. So let's take 3723 and break it down by place value. Now we can do the multiplication by 5, but we're doing it separately. So we took 5 times 3 is 15, 5 times 20 is a hundred, and so forth. And then we combine them together to get our product of 18615.

Now it might be a little tough horizontally to actually do the adding at the end, so a good idea would be to, well let's do the same thing except let's put our partial products vertically. Five times 3 is 15, 5 times 20 is a hundred, 5 times seven hundred is 3500, and 5 times 3000 is 15000. Then, we can combine them to be 18615.

Let's take 3723 times 5. What we're doing now is we're going to do exactly what we did before, but we're just going to do it with the understanding of place value without having 3723 expanded out. So we have 5 times 3 is 15, we have 5 times 2, but that's really two tens. So it's 5 times 2, that would be 100, 5 times 7, It's not really 5 times 7; it's 5 times 700, and then we have 5 times 3, but that's not really 5 times 3; it's 5 times 3000. Combine them all together to get our product.

Now let's change it up a little bit, if we take that same exact problem, let's do it horizontally. So what we need to do now is use our place value and change our representation of 3723 to show what we have by place value. We're multiplying everything by five. We use or distributive property now where we multiply by five again by each one of those, so that becomes four different multiplication problems. Then we get our partial products, except now they're horizontal instead of in a vertical format. Do our combinations, and then we get the same solution.

We're focusing on place value and properties of operations, in particular, the distributive property, a very important property that students need more experience with. Now we took sort of a shortcut in that example, but what we were really doing, we also were using equations because the 3723 times 5 is actually what would go on the left-hand side of the work that we did just a little while ago.

Now this last statement that deals with illustrating and explaining the calculation by using equations, rectangular arrays, and/or area models. We have a little bit of a problem here, because 3723 can be thought of as 3723 sets of 5. So we would need 3723 of these, and that would be very tough to actually do physically. So there are some limitations to using physical representations when we're dealing with such large numbers.

So, now let's look at multiplying two two-digit numbers. Let's say we are multiplying 54 times 36. So going back to the standard algorithm, 6 times 4 is 24, we bring down our 4 and infamous carry the 2; 6 times 5 is 30 plus 2 is 32, and then we have a multiplication by three, 3 times 4 is 12, there's our 1, and then 3 times 5 is 15, plus the 1, is 16. Then we combine it to be 1944.

Now of course this raises a lot of issues, a lot of questions with students, they're wondering why did this happen, why do we shift this one place to the left, and so forth. So again it's a procedure that can be very confusing, with little understanding of the foundations as to why these things happen. Let's backtrack, and let's make sure that we use strategies based on place value and our properties of operations.

So just like we did with the four-digit by one-digit multiplication, we're going to break this up by place value as to what it really means; 50 plus 4 times 30 plus 6. So now if we do our multiplications, we have 6 times 4 is 24, 6 times 50 is 300. Now we multiply 30 times 4 is 120, and 30 times 50 would be 1500. Combine them together to get our final product of 1944.

Now if we compare what we just did to the standard algorithm, notice there's our 24 and our 300; there's the 324. And here's where you really can see what really happened, the 120 plus the 1500; that’s 1620. So what happens here when we shift to the left obviously is that this is really a zero.

Okay, let's look at 54 times 36 again. And just like before, let's do this horizontally. We'll change our 54 times 36 to what it means by place value, we break them up to 50 fifty plus 4 times 30 plus 6. Then we're going to do horizontally what we already did vertically; we need to multiply 50 times 30, and we need to multiply 50 fifty times the 6. So we are done with the 50. Now we have to multiply by the 4; 4 times 30 and also 4 times 6. Then it is just a matter of doing our multiplications, getting our partial products, and combining them all together to get our same solution of 1944. Again we didn't do anything that different, we just did it in a horizontal format instead of a vertical format. And it's important that students get this kind of experience also, because later on in algebra, they'll be doing this type of multiplication with the distribute property, the only difference will be that they'll be using valuables, not just numbers.

So let's look at what happens here, let's make some connections. If we go back and compare what we did horizontally and vertically, notice that it's the exact same stuff. Again the only difference is how it's laid out. There's our 24, there's our 120, there's our 300, and there's our 1500; again, no difference in the actual process, just in the looks. So we can slowly, gradually build up to the standard algorithm. We can start this way by having our own numbers actually expressed using place value. Then that could lead to the exact same process except now students are using their understanding of what these digits really represent; and then that connects, overlays a foundation for the standard algorithm. And again making the connections, there's our 324, and there's our 1620.

Now, let's look at the last statement where we're supposed to illustrate and explain using rectangular arrays, and/or models. Fifty-four times 36, okay so we have 36, so what we're really doing here is we're interpreting this as 54 sets of 36. So if we put them together we get 54 sets of 36, so again we need 54 of them, which would be difficult, but depending on what materials you have to work with, it is feasible. So here we have 54 sets of 36.

We can also think more along the lines of the rectangular array, where we have our 36, like so; and we need 54 of them. So we don't lose count here's our rectangular arrays of 36, so there's 10 of them, here's another 10, another 10, another 10, and one more. So we've got 50 of them so far, we need 4 more. So there’s our 54 sets of 36 using rectangular arrays. Now simply by scrunching this together, now we have a rectangle that's 54 by 36. Okay, now this has become an area model, but keep in mind that we're really dealing with the area, which is all the space that's inside of the rectangle.

So again if we look at this strictly from an area model perspective, it would be for example a 54 by 36 rectangle, and mark of our units, fill it in. So here's our physical model of 54 by 36. It would be kind of tough to count all your squares this way. One possibility would be to go ahead and split it up, for example here these are blocks of a hundred, these are different sized blocks, because again this is 6, so you have 6 by 10 right here and so forth, and then over here you have 4 sets of 10. But again students can take these and add them all up to get the final solution; a little tougher, as opposed to just multiplying 54 times 36.

Now, you have to really be aware of the resulting units when you multiply. Here's what I mean—if this is an area model, multiply 23 by 15; the result is 345 square units, these middle one-by-one squares. Now if we did it with a rectangular arrays, we would have 23 sets of 15, so we have 23 of these rectangular arrays of 15 each, but the result is not 345 square units, it's just 345 of these little blocks. So we have 345 of these little individual blocks.

Now when we use little squares or whatever within our rectangular arrays, it does get kind of confusing, because it's really easy to kind of cross over the line and think of it as area instead. So one thing we can do is kind of transition to where what we're using in the rectangular arrays doesn't look like little squares. So here it's a little bit more obvious. I've got 23 sets of 15, and so now it's easier to see that the result is 345 of these objects.

So this standard is very important. It lays a foundation for the standard algorithm for multiplication, but it does it in such a way where we use strategies based on place value and we use the properties of operations so that students understand what really happens in that whole multiplication process.