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Expand and simplify seven minus three minus π¦, π¦ plus two.
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In this question, we can see we have two binomials which are multiplied together.
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Weβre going to expand these binomials first and then subtract the answer from seven.
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Weβre going to use the grid method to multiply these binomials.
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And notice the negative in front of them isnβt included.
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Setting up our grid, we have three minus π¦ as our first binomial.
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Our second binomial can be split into the terms π¦ and two.
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And we can write that with or without the plus sign.
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It doesnβt matter which way round do we put our binomials.
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Filling in our grid, we start with π¦ times three, which is three π¦.
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Next, we have π¦ times negative π¦, which will give us negative π¦ squared.
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On the next row, we have two times three, which is six.
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And the final term is calculated by two times negative π¦, which is negative two π¦.
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To take our answer from the grid then, we add together the four products weβve just calculated.
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Starting with our largest exponent of π¦, we have negative π¦ squared.
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Next, we notice that we have two terms in π¦, which we can collect together.
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So three π¦ plus negative two π¦ will give us plus π¦.
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And then, we add on our final term from the grid which is plus six.
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So to answer the question then of seven minus three minus π¦, π¦ plus two, we replace what weβve worked out in our expansion, giving us seven minus minus π¦ squared plus π¦ plus six.
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Itβs important to include all of these in parentheses since we want to subtract all of these terms.
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It can be helpful to write the next line where we distribute the negative across all the terms.
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So we have seven minus minus π¦ squared, which is plus π¦ squared.
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A negative times plus π¦ will give us negative π¦.
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And finally, a negative times plus six will give us negative six.
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To simplify then, we check if there are any like terms.
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We can see that we have a seven and a negative six, which is equivalent to one, giving us a final answer of π¦ squared minus π¦ plus one.
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We could write these terms in any order.
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But by convention, we usually write them in decreasing exponent values.
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Here, we have our π¦ squared first, then our π¦ term, and then our constant.