### Video Transcript

Part a) Simplify negative two π multiplied by five π. Part b) Simplify π multiplied by π multiplied by π multiplied by π. And part c) Simplify 12π minus 15π over three.

So for part a, if weβre actually going to simplify negative two π multiplied by five π, the best way to do this is actually to separate it into two parts. First of all, weβre gonna multiply the coefficients or the numerical values.

So therefore, we get negative two multiplied by five. And then weβre gonna multiply our π and π terms. So weβve got negative two multiplied by five multiplied by π multiplied by π. And this is going to give us negative 10, and thatβs because negative two multiplied by five gives us negative 10, because a negative multiplied by a positive is a negative, and then multiplied by ππ, because if weβre multiplying two terms, we just write it as ππ.

And then to use the standard notation, we actually rewrite this as negative 10ππ, because we donβt need the multiply sign between the 10 and the ππ. So therefore, we can say that if you simplify negative two π multiplied by five π, you get negative 10ππ. Okay, letβs move on to part b.

So in part b, what we need to do is simplify π multiplied by π multiplied by π multiplied by π. Well, the first thing to notice is that π is actually the same as π to the power of one. We just donβt write the power of one.

Itβs then at this point we remind ourselves of one of the index laws. The index law tells us that if we have two numbers with the same base but different powers, then if we multiply them together, then all we do is add the powers. So, for instance, if we had π to the power of π multiplied by π to the power of π, this is equal to π to the power of π plus π.

Well, we if we look at part b, weβve actually got four terms, each with the same base, because itβs π. So therefore, what weβre going to do is actually add all of their powers. So as weβve already stated, π is the same as π to the power of one. So therefore, weβre gonna have π to the power of one plus one plus one plus one, which will give us the result π to the power of four. So therefore, we can say that if we simplify π multiplied by π multiplied by π multiplied by π, we get π to the power of four. Okay, so now letβs move on to part c.

In part c, we need to simplify 12π minus 15π over three. So the first thing we want to do in this question is actually divide our fraction into two parts. We have 12π minus 15π over three. We can divide this into two other fractions, so 12π over three minus 15π over three. And the reason we can do this is because actually if we subtract one fraction from another, when they have the same denominator, itβs just the numerators we actually deal with. So 12π over three minus 15π over three would just become 12π minus 15π over three.

So therefore, our first term is going to be four π, and thatβs because 12π divided by three, well, we divide 12 by three, which is four, so weβre left with four π. And then weβre going to have minus five π, and thatβs because we have 15π divided by three. Well, 15 divided by three is just five. So therefore, we can say that if you simplify 12π minus 15π over three, the result is four π minus five π.