Video: Using the Distributive Property of Multiplication to Expand Expressions Containing Brackets

Expand and simplify 7 βˆ’ (3 βˆ’ 𝑦)(𝑦 + 2).


Video Transcript

Expand and simplify seven minus three minus 𝑦, 𝑦 plus two.

In this question, we can see we have two binomials which are multiplied together. We’re going to expand these binomials first and then subtract the answer from seven. We’re going to use the grid method to multiply these binomials. And notice the negative in front of them isn’t included.

Setting up our grid, we have three minus 𝑦 as our first binomial. Our second binomial can be split into the terms 𝑦 and two. And we can write that with or without the plus sign. It doesn’t matter which way round do we put our binomials. Filling in our grid, we start with 𝑦 times three, which is three 𝑦. Next, we have 𝑦 times negative 𝑦, which will give us negative 𝑦 squared. On the next row, we have two times three, which is six. And the final term is calculated by two times negative 𝑦, which is negative two 𝑦.

To take our answer from the grid then, we add together the four products we’ve just calculated. Starting with our largest exponent of 𝑦, we have negative 𝑦 squared. Next, we notice that we have two terms in 𝑦, which we can collect together. So three 𝑦 plus negative two 𝑦 will give us plus 𝑦. And then, we add on our final term from the grid which is plus six. So to answer the question then of seven minus three minus 𝑦, 𝑦 plus two, we replace what we’ve worked out in our expansion, giving us seven minus minus 𝑦 squared plus 𝑦 plus six.

It’s important to include all of these in parentheses since we want to subtract all of these terms. It can be helpful to write the next line where we distribute the negative across all the terms. So we have seven minus minus 𝑦 squared, which is plus 𝑦 squared. A negative times plus 𝑦 will give us negative 𝑦. And finally, a negative times plus six will give us negative six.

To simplify then, we check if there are any like terms. We can see that we have a seven and a negative six, which is equivalent to one, giving us a final answer of 𝑦 squared minus 𝑦 plus one. We could write these terms in any order. But by convention, we usually write them in decreasing exponent values. Here, we have our 𝑦 squared first, then our 𝑦 term, and then our constant.

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