Video: Using the Distributive Property to Expand Algebraic Expressions

Use the distributive property to fully expand (2π‘₯ + 𝑦)(π‘₯𝑦 βˆ’ 2𝑧).

02:51

Video Transcript

Use the distributive property to fully expand two π‘₯ plus 𝑦, π‘₯𝑦 minus two 𝑧.

In this question, the phrase β€œto use the distributive property” means that we can multiply the sum by multiplying each addend separately and then adding the products. We could use a number of methods to expand the brackets. Here, we’re going to look at two of those. The first method is the FOIL method, which is an acronym. We can recall that it stands for the First, the Outer, the Inner, and the Last terms of our binomials.

So with our two binomials set, we start by multiplying the first two terms, which is two π‘₯ times π‘₯𝑦. Next, our outer terms will be two π‘₯ times negative two 𝑧. It’s always worth taking extra care when we have the variable 𝑧, because we don’t want to confuse it with the digit two. Next, our inner terms will be 𝑦 multiplied by π‘₯𝑦. And finally, we add on the product of our last terms, which is 𝑦 times negative two 𝑧.

We now need to simplify our products. The coefficient of our first term is two. π‘₯ multiplied by π‘₯ will be π‘₯ squared. And we have our 𝑦. So the first term simplifies to two π‘₯ squared 𝑦. The coefficient of our second term is found by multiplying plus two by negative two, giving us negative four. And we then have π‘₯ times 𝑧. Our third term simplified will be π‘₯𝑦 squared since we have 𝑦 times 𝑦. And our final term will be negative two 𝑦𝑧. At this stage, we always check if we can simplify our answer by collecting any like terms. And as there are none here, this would be our final answer.

As an alternative method, we could use the area or grid method to expand our binomials. To set up our grid, we split up our binomials into their component terms with one on the row and one on the column. And it doesn’t matter which way round we put them. To fill in each cell in our grid, we multiply the row value by the column value.

So taking our first cell, we have π‘₯𝑦 times two π‘₯, which would be two π‘₯ squared 𝑦. For our next cell, we multiply π‘₯𝑦 by 𝑦 to give us π‘₯𝑦 squared. On our next row then, we have negative two 𝑧 times two π‘₯, which would give us negative four π‘₯𝑧. And our final grid value would be negative two 𝑦𝑧. To find our answer from this grid, we add our four products, giving us two π‘₯ squared 𝑦 minus four π‘₯𝑧 plus π‘₯𝑦 squared minus two 𝑦𝑧, which is the same as we achieved using the FOIL method.

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