Video: GCSE Mathematics Foundation Tier Pack 4 β€’ Paper 2 β€’ Question 10

GCSE Mathematics Foundation Tier Pack 4 β€’ Paper 2 β€’ Question 10

03:48

Video Transcript

Part a) Simplify negative two π‘˜ multiplied by five 𝑙. Part b) Simplify π‘š multiplied by π‘š multiplied by π‘š multiplied by π‘š. And part c) Simplify 12𝑝 minus 15π‘ž over three.

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

So therefore, we get negative two multiplied by five. And then we’re gonna multiply our π‘˜ and 𝑙 terms. So we’ve got negative two multiplied by five multiplied by π‘˜ multiplied by 𝑙. And this is going to give us negative 10, and that’s because negative two multiplied by five gives us negative 10, because a negative multiplied by a positive is a negative, and then multiplied by π‘˜π‘™, because if we’re multiplying two terms, we just write it as π‘˜π‘™.

And then to use the standard notation, we actually rewrite this as negative 10π‘˜π‘™, because we don’t need the multiply sign between the 10 and the π‘˜π‘™. So therefore, we can say that if you simplify negative two π‘˜ multiplied by five 𝑙, you get negative 10π‘˜π‘™. Okay, let’s move on to part b.

So in part b, what we need to do is simplify π‘š multiplied by π‘š multiplied by π‘š multiplied by π‘š. Well, the first thing to notice is that π‘š is actually the same as π‘š to the power of one. We just don’t write the power of one.

It’s 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‘Ž to the power of 𝑏 multiplied by π‘Ž to the power of 𝑐, this is equal to π‘Ž to the power of 𝑏 plus 𝑐.

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

In part c, we need to simplify 12𝑝 minus 15π‘ž over three. So the first thing we want to do in this question is actually divide our fraction into two parts. We have 12𝑝 minus 15π‘ž over three. We can divide this into two other fractions, so 12𝑝 over three minus 15π‘ž 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, it’s just the numerators we actually deal with. So 12𝑝 over three minus 15π‘ž over three would just become 12𝑝 minus 15π‘ž over three.

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

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.