Video: Using an Area Model to Multiply Polynomials

Which of the following equations is represented by the given area model? [A] (4𝑦 βˆ’ 2π‘₯)(2𝑦 + (π‘₯/2) βˆ’ 3) = 8𝑦² βˆ’ 2π‘₯𝑦 βˆ’ 12𝑦 βˆ’ π‘₯Β² βˆ’ 6π‘₯ [B] (4𝑦 βˆ’ 2π‘₯)(2𝑦 + (2/π‘₯) βˆ’ 3) = 8𝑦² βˆ’ 4π‘₯𝑦 βˆ’ 12𝑦 βˆ’ 4 + 6π‘₯ + (8𝑦/π‘₯) [C] (4𝑦 βˆ’ 2π‘₯)(2𝑦 + (π‘₯/2) βˆ’ 3) = 8𝑦 βˆ’ 2π‘₯𝑦 βˆ’ 12𝑦 βˆ’ π‘₯ + 6π‘₯ [D] (4𝑦 βˆ’ 2π‘₯)(2𝑦 + (π‘₯/2) βˆ’ 3) = 8𝑦² βˆ’ 2π‘₯𝑦 βˆ’ 12𝑦 βˆ’ π‘₯Β² + 6π‘₯ [E] (4𝑦 βˆ’ 2π‘₯)(2𝑦 + (π‘₯/2) βˆ’ 3) = 8𝑦 βˆ’ 2π‘₯𝑦 βˆ’ 12𝑦 βˆ’ π‘₯ βˆ’ 6π‘₯

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Video Transcript

Which of the following equations is represented by the given area model? Option A) four 𝑦 minus two π‘₯ times two 𝑦 plus π‘₯ over two minus three equals eight 𝑦 squared minus two π‘₯𝑦 minus 12𝑦 minus π‘₯ squared minus six π‘₯. Option B) four 𝑦 minus two π‘₯ times two 𝑦 plus two over π‘₯ minus three equals eight 𝑦 squared minus four π‘₯𝑦 minus 12𝑦 minus four plus six π‘₯ plus eight 𝑦 over π‘₯. Option C) four 𝑦 minus two π‘₯ times two 𝑦 plus π‘₯ over two minus three equals eight 𝑦 minus two π‘₯𝑦 minus 12𝑦 minus π‘₯ plus six π‘₯. Option D) four 𝑦 minus two π‘₯ times two 𝑦 plus π‘₯ over two minus three equals eight 𝑦 squared minus two π‘₯𝑦 minus 12𝑦 minus π‘₯ squared plus six π‘₯. Option E) four 𝑦 minus two π‘₯ times two 𝑦 plus π‘₯ over two minus three equals eight 𝑦 minus two π‘₯𝑦 minus 12𝑦 minus π‘₯ minus six π‘₯.

So, let’s have a look at the question and the area model. An area model is a diagram that allows us to represent the multiplication of terms as a rectangle. We take the values to be multiplied as the length and the width. And we would find the area of these two values by multiplication of the so-called length and the width. So, for example, if we wanted to multiply the terms three and two π‘₯, then the term inside the rectangle would be six π‘₯ since that’s equal to three times two π‘₯.

Let’s have a look at our diagram and see if we can find any of the missing values. The missing value in our highlighted cell would be found by multiplying negative two π‘₯ and negative three. We know that if we multiply negative two by negative three, this would give us a positive value of six. So, negative two π‘₯ times negative three will give us six π‘₯, which we can write into our area model. Let’s see what other values we can work out.

If we consider that in our area model we’re multiplying two values to give us an answer, this means that in our model we’re looking for two known values and an unknown, which could be two lengths and an unknown area or unknown area value and one missing length. So, let’s have a look at the missing length highlighted along the top. Here, we need to calculate negative two π‘₯ times what equals negative four π‘₯𝑦. If we look at our coefficients, then negative two times two must give us negative four. If we look at our π‘₯-values, we have an π‘₯ on the left-hand side and an π‘₯ on the right-hand side, so we don’t multiply in anymore.

On the right-hand side, we have a 𝑦 term, so we must multiply by two 𝑦. So, we can add this term to our area model. The next missing length can be found since we know that that value times negative three would give us negative 12𝑦. And the answer must be four 𝑦, since four times negative three would give us negative 12. And we must have a 𝑦 term on the left-hand side and the right-hand side. Next, we can calculate the missing area value formed from two 𝑦 times four 𝑦, which is eight 𝑦 squared.

Our next missing length, can be found by calculating negative two π‘₯ times what will give us negative π‘₯ squared. Considering our coefficients, this means negative two times what will give us negative one, which means that our missing coefficient must be a half. Considering our π‘₯ terms then, we have π‘₯ times what will give us π‘₯ squared, leaving us with π‘₯. And so, our missing term must be a half π‘₯. We can leave it as a half π‘₯ or we can write it in the slightly neater format of π‘₯ over two. And so, our final missing value will be found by calculating four 𝑦 times π‘₯ over two, which will be four π‘₯𝑦 over two.

We can notice that two will go into the numerator and the denominator, giving us a simplified answer of two π‘₯𝑦, which we can fill into the table. So, now, we have all the values in our area model, we can determine the two original expressions that were being multiplied. The first expression would be composed of four 𝑦 and negative two π‘₯, which we write in parentheses as four 𝑦 minus two π‘₯. The second expression is composed of two 𝑦, π‘₯ over two, and negative three, giving us two 𝑦 plus π‘₯ over two minus three in parentheses.

We notice that the question has asked us for an equation and not for an expression. So, we must also evaluate the result of our expressions being multiplied. And we can do this using our area model. We need to take each individual area and add them together to get a total area. This will be eight 𝑦 squared plus two π‘₯𝑦 minus 12𝑦 minus four π‘₯𝑦 minus π‘₯ squared plus six π‘₯. We notice that we have two terms in π‘₯𝑦, so we can simplify this expression further. And so, now, we have an answer that matches option D, four 𝑦 minus two π‘₯ times two 𝑦 plus π‘₯ over two minus three equals eight 𝑦 squared minus two π‘₯𝑦 minus 12𝑦 minus π‘₯ squared plus six π‘₯.

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