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Hello, my name is Mr. Clasper.

And today we're going to learn how to factorise a quadratic with a leading coefficient, which is greater than one.

Let's begin with a recap on expanding brackets by using a multiplication grid.

When we multiply two x by x, we get two x squared.

When we multiply five by x, we get five x, multiplying two x by three would give us six x, and when we multiply five by three, we get a constant of 15.

This would give us an expression of two x squared, plus five x, plus six x, plus 15.

However, when we simplify this, we can collect our like terms by adding six x and five x, which should give us 11x.

So our expression is two x squared plus 11x, plus 15.

Let's factorise this expression.

We know we need two numbers which have a product of five as this is the only way that we can generate our constant of five in the expression.

Looking at factor pairs of five.

We could only use the numbers one and five as five is a prime number.

So let's try this.

If we choose two x plus five and x plus one, this would give us two x squared, plus five x, plus two x, plus five.

However, when we add our like terms together, we get seven x and not 11x.

So this combination will not work.

Let's try it the other way round.

So if we multiply these two, we would get two x squared, plus x, plus 10x, plus five.

That would give us a total of 11x, which means that when we factorise this, we should get two x plus one all multiplied by x plus five.

Let's try another example.

This time we're looking for two numbers which have a product of 15.

So factor pairs of 15 will give us one and 15, or we could also use three and five.

So if we choose one and 15, this combination would give us two x squared plus x plus 30x plus 15.

However, when we simplify this, we will get 31x and not 13x.

So this combination is not going to work.

Let's try it the other way around.

So if we use this combination, we would get two x squared, plus 15x, plus two x, plus 15.

However, again, when we simplify, we do not get 13x, this time we would get 17x.

So one and 15 will not be suitable for this example.

Let's try three and five.

So if we try it in this order, we would get two x squared plus five x plus six x plus 15.

However, once again, we don't get 13x when we simplify.

Let's move the constant.

If we expand in this order, we would get two x squared, plus three x, plus 10x, plus 15.

And as we can see, we now have a sum of 13x.

So this means that our expression is equivalent to two x plus three all multiplied by x plus five.

Here's a question for you to try.

Pause the video to complete your task and resume once you're finished.

And here is the solution.

So when we look, we needed two numbers which have a product of 15, and we need to make a sum of 11x.

So that meant that our final answer was two x plus five multiplied by x plus three.

Here's another question for you to try.

Pause the video to complete your task, and resume once you're finished.

And here is your next solution.

So again, we needed two numbers with a product of 10, but we needed to generate 17x with our sum when collecting our like terms. So that meant our final answer was three x plus two and x plus five.

Our next example involves negative numbers.

In this case, we're going to need two numbers which have a product of negative three.

So let's have a look at the factor pairs for negative three.

We could choose negative one and three, or we could choose positive one and negative three.

Let's try negative one and three.

So if we use them in this order, we would get two x squared minus x, plus six x, minus three.

However, when we simplify this, we would get five x and not minus x, which is what we want.

So this combination will not work.

Let's switch the numbers.

If we expand this way, we would get two x squared, plus three x, minus two x, minus three.

But again, this will give us a sum of one x, and we need negative x or negative one x.

This means that negative one and three are not the pair of constants that we need to use.

Let's try one and negative three.

So if we try one and negative three, when we expand we would get two x squared, plus x, minus six x, minus three.

However, when we simplify this, we would get two x squared, minus five x, minus three, which is not what we want as we need negative one x.

Lets switch the numbers one more time.

If we expand this where we would get two x squared, minus three x, plus two x, minus three, and we can see there that we also get a sum of negative one x.

So this is our correct answer.

Here's some questions for you to try.

Pause the video to complete your task, and resume once you're finished.

And here are your answers.

So if we look at part e when we factorise this we should get three x, minus one, multiplied by x plus two.

As we can see in the grid below.

And for part f we should get five x plus two, and x minus seven as our final response.

Here's another question for you to try.

Pause the video to complete your task, and resume once you're finished And here is the solution.

So looking into this question, we could have looked for some clues.

So for example, with the last expression on the left, six x plus one, multiplied by two x plus three.

This would have to have a constant of three when we expand it, which leaves us with only one option on the right hand side.

Here's your last question.

Pause the video to complete your task, and resume once you're finished.

And this is our solution.

So when we look at this, Ashraf is actually wrong because six x could also be generated by multiplying three x by two x.

And then if we use a multiplication grid, you would discover that when we factorise this, we actually get three x plus one, and two x plus five.

And that brings us nicely to the end of our lesson.

I hope you've enjoyed it.

Hopefully see you soon.