### Video Transcript

Find π΄π΅ given that π΄ equals seven π§ squared minus four π§ plus five and π΅ equals three π§ squared minus four π§ plus one.

The first thing in this question is to recognize that π΄π΅ means π΄ times π΅. That means weβre going to need to take everything in our π΄ term and multiply it by everything in our π΅ term. So this means that we need to multiply all of seven π§ squared minus four π§ plus five times all of three π§ squared minus four π§ plus one. And we can write the brackets in to remind us that itβs all of the terms in each expression. If we didnβt have the brackets, we could get confused and think that we were just multiplying the five by the three π§ squared. But that wouldnβt be correct. So we could approach this kind of multiplication in two different ways.

In the first method, weβre going to take each term in π΄ and multiply it by everything in π΅. So we can start by writing π΄π΅ equals seven π§ squared times all of three π§ squared minus four π§ plus one. And then since the next term in π΄ is a negative four π§, we could write this as negative or minus four π§ times all of three π§ squared minus four π§ plus one. And finally, our last term, five, becomes plus five times all of three π§ squared minus four π§ 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 letβs begin with our seven π§ squared times three π§ squared. Well, we know that seven times three will give us 21. But what happens with the π§ squared times the π§ squared? Well, we need to use the rules from multiplying with exponents. That say we had a value π₯ to the power of π times π₯ to the power of π. When we multiply those we would get π₯ to the power of π plus π. So that means when we have π§ square times π§ squared, that will give us π§ to the power of four. So now, weβll find 21π§ to the power of four.

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

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

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

So simplifying this will give us negative 24π§. And finally then, our constant that remains is plus five. So we can rewrite that, giving us 21π§ to the fourth power minus 40π§ to the third power plus 38π§ squared minus 24π§ plus five. Since we have all parts that are different and a constant, then we canβt simplify this anymore. As a second alternative method, weβre going to use a multiplication grade to calculate π΄ times π΅. Here, we have all the terms of π΄ in separate columns across the horizontal rows and all the terms of π΅ in separate rows. But it wouldnβt matter if we had π΄ and π΅ the other way round.

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

And in the final row, our entries will be seven π§ squared, negative four π§, 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 π§ to the fourth power term. Thatβs 21 π§ to the fourth power. Then we can simplify our π§ to the third power terms, negative 28π§ to the third power and negative 12π§ to the third power, giving us negative 40π§ to the third power.

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