Question Video: Using the Distributive Property to Rewrite Products as a Sum of Multiplication Expressions | Nagwa Question Video: Using the Distributive Property to Rewrite Products as a Sum of Multiplication Expressions | Nagwa

Question Video: Using the Distributive Property to Rewrite Products as a Sum of Multiplication Expressions Mathematics • 3rd Grade

Look at Ethan’s work. He used 8 = 5 + 3 to write 8 × 3 as the sum of simpler products. 8 × 3 = (5 × 3) + (3 × 3). Find another sum of products that is equal to 8 × 3. [A] (2 × 6) + (6 × 3) [B] (3 × 3) + (6 × 3) [C] (2 × 3) + (6 × 3) [D] (2 × 2) + (6 × 3)

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Video Transcript

Look at Ethan’s work. He used eight equals five plus three to write eight times three as the sum of simpler products. Eight times three equals five times three plus three times three. Find another sum of products that is equal to eight times three. Two times six plus six times three, three times three plus six times three, two times three plus six times three, or two times two plus six times three.

One of the useful things about multiplication is something called the distributive property. This means that if we’re like Ethan and we come across a multiplication that perhaps we don’t know the answer to, we can split it up into the sum of simpler products, in other words, two multiplication facts that are a little bit easier that we can add together to find the fact that we want to know.

Now we can see that the multiplication fact that Ethan wants to find out is eight times three. But perhaps he’s not great with his eight times table because he split up the eight part of eight times three into five and three. And because five and three make eight, instead of working out eight times three, he just needs to work out the answer to five times three and three times three and then add the two together.

And this is what we can see in his number sentence. Eight times three equals five times three plus three times three. Here’s what eight times three might look like as an array. And by splitting the number eight into five plus three, we can see that Ethan split up the larger calculation into smaller ones that are hopefully easier to work out.

Now, this question doesn’t ask us to use Ethan’s method to find the answer to eight times three. In a way, we’re asked to use his method but differently. We need to find another sum of products that’s equal to eight times three. Now, the first thing we can say about what Ethan’s done is he split up the number eight, but he’s kept the number three in his calculation. Can you see he’s working out both five times three and three times three? Both of these simpler products are still multiplying by three, aren’t they? It’s only the eight that’s been split up.

And if we look at our possible answers, only two of them could be correct. Only this one here and this one here show two products that involve multiplying by three both times. We’ve got three times three plus six times three, and then we’ve got two times three plus six times three. The answer has got to be one of these two. So, we need to ask ourselves, “Ethan’s split up the number eight into five plus three. How else could we split this number up?”

For example, eight is the same as four plus four. Then, we get the calculation four times three plus four times three. But that’s not one of our possible answers. How else could we split up the number eight? What about six and two? Then we could answer the question by finding the sum of six times three and two times three. Now, if we look carefully at our possible answers, we can see that one of them shows this addition. The multiplications may be the other way around. But we know we can add two values together in different orders and they’d still make the same answer. We know that eight equals two plus six. And so, we also know that eight times three can be solved by working out two times three plus six times three.

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