# Video: Multiplying Two Algebraic Expressions

Find 𝐴𝐵 given that 𝐴 = 7𝑧² − 4𝑧 + 5 and 𝐵 = 3𝑧² − 4𝑧 + 1.

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### Video Transcript

Find 𝐴𝐵 given that 𝐴 equals seven 𝑧 squared minus four 𝑧 plus five and 𝐵 equals three 𝑧 squared minus four 𝑧 plus one.

The first thing in this question is to recognize that 𝐴𝐵 means 𝐴 times 𝐵. That means we’re going to need to take everything in our 𝐴 term and multiply it by everything in our 𝐵 term. So this means that we need to multiply all of seven 𝑧 squared minus four 𝑧 plus five times all of three 𝑧 squared minus four 𝑧 plus one. And we can write the brackets in to remind us that it’s all of the terms in each expression. If we didn’t have the brackets, we could get confused and think that we were just multiplying the five by the three 𝑧 squared. But that wouldn’t be correct. So we could approach this kind of multiplication in two different ways.

In the first method, we’re going to take each term in 𝐴 and multiply it by everything in 𝐵. So we can start by writing 𝐴𝐵 equals seven 𝑧 squared times all of three 𝑧 squared minus four 𝑧 plus one. And then since the next term in 𝐴 is a negative four 𝑧, we could write this as negative or minus four 𝑧 times all of three 𝑧 squared minus four 𝑧 plus one. And finally, our last term, five, becomes plus five times all of three 𝑧 squared minus four 𝑧 plus one. And in the next stage, we start to simplify our terms. When we have a number outside a bracket and a number inside a bracket, that means we need to multiply those two terms together.

So let’s begin with our seven 𝑧 squared times three 𝑧 squared. Well, we know that seven times three will give us 21. But what happens with the 𝑧 squared times the 𝑧 squared? Well, we need to use the rules from multiplying with exponents. That say we had a value 𝑥 to the power of 𝑎 times 𝑥 to the power of 𝑏. When we multiply those we would get 𝑥 to the power of 𝑎 plus 𝑏. So that means when we have 𝑧 square times 𝑧 squared, that will give us 𝑧 to the power of four. So now, we’ll find 21𝑧 to the power of four.

Let’s go on to our next terms. We’re multiplying seven 𝑧 squared by negative four 𝑧. Since when we have a value 𝑧 on its own with no power, this is the same as 𝑧 to the power of one. So when we multiply seven 𝑧 squared by negative four 𝑧, we get negative 28𝑧 to the power of three. And multiplying the final terms in the brackets, we have seven 𝑧 squared times one which will give us plus seven 𝑧 squared. Moving on to the next bracket, we’re going to multiply negative four 𝑧 by three 𝑧 squared. And we must remember in this bracket that we’re multiplying everything by negative four and not just four which will give us negative 12𝑧 to the power of three because that 𝑧 on its own is still 𝑧 to the power of one.

Next, we have negative four 𝑧 times negative four 𝑧 which will give us a plus, a positive value, 16𝑧 squared, Since when we multiply two negatives, we always get a positive value. And then finally in that bracket, we have negative four 𝑧 times plus one which will give us negative four 𝑧. And then for our last bracket, we have five times three 𝑧 squared, which will give us 15𝑧 squared. Then five times negative four 𝑧 will give us negative 20𝑧. And then we have five times plus one, which gives us plus five. In the next stage, we’re going to simplify by collecting like terms. That means we’re going to take all the terms that have 𝑧 to the fourth power. And then we’re going to collect separately all the terms that have 𝑧 to the third power. And then all the terms that 𝑧 squared, and so on.

So let’s start by saying how many terms that have 𝑧 to the fourth power. In this case, there’s just one, so we can rewrite this. Next, we’re going look for the terms that have 𝑧 to the third power. We can see that we have negative 28𝑧 to the third power and negative 12𝑧 to the third power. And we must remember to include the signs when we’re considering these. So we can simplify negative 28𝑧 to the third power take away 12𝑧 to the third power as negative 40𝑧 to the third power. Next, considering our 𝑧 square terms, we have plus seven 𝑧 squared plus 16𝑧 squared plus 15𝑧 squared, giving us plus 38𝑧 squared. And next, for our terms with just 𝑧, we have a negative four 𝑧 and we have a negative 20 𝑧.

So simplifying this will give us negative 24𝑧. And finally then, our constant that remains is plus five. So we can rewrite that, giving us 21𝑧 to the fourth power minus 40𝑧 to the third power plus 38𝑧 squared minus 24𝑧 plus five. Since we have all parts that are different and a constant, then we can’t simplify this anymore. As a second alternative method, we’re going to use a multiplication grade to calculate 𝐴 times 𝐵. Here, we have all the terms of 𝐴 in separate columns across the horizontal rows and all the terms of 𝐵 in separate rows. But it wouldn’t matter if we had 𝐴 and 𝐵 the other way round.

So in our first empty cell, we’re calculating three 𝑧 squared times seven 𝑧 squared. We know that three times seven is 21. And we use our rules of exponents again to calculate 𝑧 squared times 𝑧 squared is 𝑧 to the fourth power. In the next empty cell, we calculate three 𝑧 squared times negative four 𝑧, which gives us negative 12𝑧 to the third power, remembering, of course, the negative signs are very important. And in the last empty cell, we have three 𝑧 squared times five, which gives us 15𝑧 squared. In the next row, we have negative four 𝑧 times seven 𝑧 squared, which gives us negative 28𝑧 to the third power. In the next cell we have negative four 𝑧 times negative four 𝑧 which will give us 16𝑧 squared, remembering our two negatives multiplied will give us a positive value. And our final entry in that row will be negative four 𝑧 times five, which is negative 20𝑧.

And in the final row, our entries will be seven 𝑧 squared, negative four 𝑧, and five. At this point in our method, we can see that all the entries in the grade are the same as in the second line of our first method. But the multiplication grade does allow a much needed a way to write these. In order to get our answer from the grades, we add together all the values and collect the like terms together. So we have one 𝑧 to the fourth power term. That’s 21 𝑧 to the fourth power. Then we can simplify our 𝑧 to the third power terms, negative 28𝑧 to the third power and negative 12𝑧 to the third power, giving us negative 40𝑧 to the third power.

Then we can simplify our next three terms, 15𝑧 squared plus 16𝑧 squared plus seven 𝑧 squared, which gives us plus 38𝑧 squared. Our two terms in 𝑧 negative, four 𝑧 and negative 20𝑧, will give us negative 24𝑧. And finally, we add on our remaining term plus five. So either method would lead us to the final answer for 𝐴𝐵, 21𝑧 to the fourth power minus 40𝑧 to the third power plus 38𝑧 squared minus 24𝑧 plus five.