### Video Transcript

Part a) Factorise two 𝑥 squared minus three 𝑥 plus one. Part b) Solve the inequality four 𝑥 minus three is greater than or equal to seven 𝑥
plus one. And part c) Given that 𝑛 is an integer, find all the values of 𝑛, with three 𝑛
plus one being greater than or equal to negative four but less than four.

So we’re gonna start with part a, and we have this word “factorise.” So it’s asking us to actually factorise this expression. Factorise means putting into brackets. And also, as we actually know that we’re looking at a quadratic in the form 𝑎𝑥
squared plus 𝑏𝑥 plus 𝑐, we know that when we factorise, we’re gonna have two
brackets. And also, if we look at the expression we’ve got, we’ve got two 𝑥 squared, which
means then we know what we’re gonna have at the beginning of each bracket. And that’s two 𝑥 and 𝑥, because these are the only pair of factors that will
actually make two 𝑥 squared, cause two 𝑥 multiplied by 𝑥 gives us two 𝑥
squared.

Okay, so what about the other terms in our brackets? Well, the second terms of each bracket have to have a product of positive one. Okay, so let’s think about what possible combinations there are. Well, because we’re looking to actually have a product of positive one, there’re
actually only two pairs of factors that will give us this is, either negative one
multiplied by negative one or positive one multiplied by positive one. But which one is it?

Well, to help us decide now what we do is we look at the coefficient of our 𝑥
term. We can see here that it’s negative three. So this tells us that, actually, our factors have to be the negative one and negative
one. And actually we can double-check this, because if we think about what would happen if
we had two 𝑥 multiplied by negative one, that will give us negative two 𝑥. So the coefficient will be negative two. And then if we had negative one multiplied by 𝑥, this will give us negative 𝑥 or
negative one 𝑥.

So this time, the coefficient would be negative one for our 𝑥 term. So if we have negative two plus negative one, that gives us negative three. And that’s what we were looking for as a coefficient of our 𝑥 term. So there we go. We’ve actually found the factors of two 𝑥 squared minus three 𝑥 plus one.

So now we actually found these through inspection, actually working out what fits
this particular question. However, there is also a method we can use. It’s actually very structured, and some students might prefer this. So what I’m gonna do is also show you this method.

Well, this other method is something that we’d use if we have a coefficient of 𝑥
squared greater than one, like we do in this case. We have two 𝑥 squared. So what we consider is our expression, and we consider what 𝑎, 𝑏, and 𝑐 would
be.

So in this case, our 𝑎 would be two, our 𝑏 would be negative three, and our 𝑐
would be positive one. And then the first thing we do is we actually multiply our 𝑎 and our 𝑐. So when we do that, we’re gonna have two multiplied by one, which gives us an answer
of two. So now what we need to do is actually look for a pair of factors whose product is
gonna be equal to 𝑎𝑐 — so in this case it’s gonna be equal to two — but whose sum
is equal to 𝑏 — so in this case it’s gonna be negative three.

Well, the two factors that are actually gonna fit this in this example are negative
two and negative one. That’s because the product of negative two and negative one is two, because if you
multiply negative by negative you get a positive. And also if you have negative two plus negative one, it’s gonna give us negative
three.

Okay, so now what do we do next? Well, what we now do is use these factors to actually rewrite our coefficient of
𝑥. And we actually split it into both parts. So what we now have is two 𝑥 squared minus two 𝑥 minus, well I’ve put here one in
brackets 𝑥 just so we can see what’s happening, then plus one. And now what we’re gonna do is actually factorise pairs of terms.

So first of all, we’re gonna factorise two 𝑥 squared minus two 𝑥. So first of all, we know that, actually, we’re gonna take two 𝑥 outside the bracket,
because this is actually a factor of two 𝑥 squared and two 𝑥. And then inside the brackets, it’s going to be 𝑥 minus one. And that’s because two 𝑥 multiplied by 𝑥 is two 𝑥 squared, and two 𝑥 multiplied
by negative one is actually gonna give us negative two 𝑥.

Okay, so now let’s move on and factorise the next two terms. So now, for the next two terms, we’re gonna have negative one outside the bracket,
cause negative one is actually a factor of negative one 𝑥 and positive one. And therefore, inside the bracket, we’re gonna have 𝑥 minus one. That’s because negative one multiplied by 𝑥 gives us negative 𝑥 or negative one 𝑥,
and negative one multiplied by negative one gives us positive one.

Also, we know to choose these factors because what should happen at this point is we
should have the same factor in each bracket. If we didn’t have this, then we know that, actually, something’s gone wrong. We’d have a look at how we’ve been factorising, because we should have the same
factor in each bracket, so we can actually do the final stage.

So therefore, fully factorised, we get two 𝑥 minus one. And we get that because that’s the two 𝑥 that was actually in front of the first
bracket and the negative one which is in front of the second bracket. And then this is multiplied by the 𝑥 minus one, which is what we had inside the
bracket.

Now if we check this, yes it’s the same as what we found with the first method. So therefore, we can definitely say that if you factorise two 𝑥 squared minus three
𝑥 plus one, you get two 𝑥 minus one multiplied by 𝑥 minus one. Okay, great! We finished part a. Let’s move on to part b.

So for part b, what we’re gonna do is actually solve an inequality. And that inequality is four 𝑥 minus three is greater than or equal to seven 𝑥 plus
one. And when we’re solving the inequality, we actually carry out steps much in the same
way we would as if we were actually solving an equation.

So the first thing we do is we inspect our inequality. And we want to see where the most 𝑥s are, so the highest coefficient of 𝑥. And we can see that, actually, that’s on the right-hand side of our inequality. So therefore, the first step we’re gonna do is actually subtract four 𝑥 from each
side of the inequality. So therefore, if we actually do this, what we’re left with is negative three is
greater than or equal to three 𝑥 plus one.

So now what we want to do is actually subtract one from each side of the
inequality. So therefore, we’ll have numbers on the left-hand side and our 𝑥 term on the
right-hand side. And if we subtract one from each side, we’re gonna have negative four is greater than
or equal to three 𝑥. And that’s because we had negative three minus one, which gives us negative four.

Okay, and the final step is to actually divide three by three, which is gonna give us
negative four over three or negative four-thirds is greater than or equal to 𝑥. So therefore, we can say that the solution to the inequality four 𝑥 minus three is
greater than or equal to seven 𝑥 plus one is 𝑥 is less than or equal to negative
four-thirds.

Okay, so now we’ve solved part b, let’s move on to part c. So now the first stage when actually solving part c is to solve the inequality three
𝑛 plus one is greater than or equal to negative four or less than four. Well, in order to solve it, what we’re going to actually do is find out what our
bounds are for 𝑛.

So the first thing we’re gonna do is actually subtract one from each term. And when we do that, we get three 𝑛 is greater than or equal to negative five, but
less than three. So then the next stage is actually divide each of our terms by three, cause we want
to find out, like I said, what single 𝑛 the bounds of that would be. And when we actually do that, what we get is that 𝑛 is greater than or equal to
negative five-thirds, or negative five over three, but less than one.

So great! We’ve actually solved the inequality. So is that the problem finished? Well, no, cause if we look at the question, what it wants us to do is actually find
all the values of 𝑛 which actually satisfy our inequality.

Now to help us actually find all the values of 𝑛 that are gonna satisfy our
inequality, I’ve actually drawn this number line. But actually there’s gonna be so multiple values of 𝑛, so how are we gonna work out
which ones to include? Well, this is when we need to make sure that we look at the question again because
we’re given another piece of useful information and that is that 𝑛 is an
integer.

So therefore, we know that the 𝑛 values have to be a whole number. So this cuts it down drastically. So now let’s see what our possible values of 𝑛 can be. Well, if we look at our number line, we’ve got a coloured-in dot above negative five
over three. And that’s because if we look at the inequality, it says that 𝑛 can be greater than
or equal to negative five over three. But we’ve got an open circle above one, and that’s because it cannot be one. It needs to be less than one.

So therefore, we can say that the only integers that are possible values of 𝑛 are
negative that actually satisfy the inequality three 𝑛 plus one is greater than or
equal to negative four but less than four are negative one or zero.