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Use the distributive property to fully expand two π₯ plus π¦, π₯π¦ minus two π§.
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In this question, the phrase βto use the distributive propertyβ means that we can multiply the sum by multiplying each addend separately and then adding the products.
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We could use a number of methods to expand the brackets.
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Here, weβre going to look at two of those.
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The first method is the FOIL method, which is an acronym.
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We can recall that it stands for the First, the Outer, the Inner, and the Last terms of our binomials.
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So with our two binomials set, we start by multiplying the first two terms, which is two π₯ times π₯π¦.
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Next, our outer terms will be two π₯ times negative two π§.
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Itβs always worth taking extra care when we have the variable π§, because we donβt want to confuse it with the digit two.
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Next, our inner terms will be π¦ multiplied by π₯π¦.
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And finally, we add on the product of our last terms, which is π¦ times negative two π§.
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We now need to simplify our products.
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The coefficient of our first term is two.
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π₯ multiplied by π₯ will be π₯ squared.
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And we have our π¦.
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So the first term simplifies to two π₯ squared π¦.
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The coefficient of our second term is found by multiplying plus two by negative two, giving us negative four.
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And we then have π₯ times π§.
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Our third term simplified will be π₯π¦ squared since we have π¦ times π¦.
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And our final term will be negative two π¦π§.
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At this stage, we always check if we can simplify our answer by collecting any like terms.
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And as there are none here, this would be our final answer.
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As an alternative method, we could use the area or grid method to expand our binomials.
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To set up our grid, we split up our binomials into their component terms with one on the row and one on the column.
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And it doesnβt matter which way round we put them.
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To fill in each cell in our grid, we multiply the row value by the column value.
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So taking our first cell, we have π₯π¦ times two π₯, which would be two π₯ squared π¦.
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For our next cell, we multiply π₯π¦ by π¦ to give us π₯π¦ squared.
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On our next row then, we have negative two π§ times two π₯, which would give us negative four π₯π§.
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And our final grid value would be negative two π¦π§.
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To find our answer from this grid, we add our four products, giving us two π₯ squared π¦ minus four π₯π§ plus π₯π¦ squared minus two π¦π§, which is the same as we achieved using the FOIL method.