Video: Pack 5 β€’ Paper 1 β€’ Question 16

Pack 5 β€’ Paper 1 β€’ Question 16

05:39

Video Transcript

Write three π‘₯ squared minus five π‘₯ minus two over two π‘₯ squared minus seven π‘₯ plus six in the form π‘Žπ‘₯ plus 𝑏 over 𝑐π‘₯ plus 𝑑, where π‘Ž, 𝑏, 𝑐, and 𝑑 are integers. You must show your working.

Looking at the original form of this fraction, we see that both the numerator and the denominator are quadratic expressions in π‘₯, whereas in the form we’re requested to change it to, both the numerator and denominator are linear expressions in π‘₯. What this suggests is that we need to factorize both of the quadratic expressions. And then, hopefully, one of the linear factors will cancel from the numerator and the denominator to leave just a single linear expression in the numerator and a linear expression in the denominator.

Let’s begin with the numerator: the quadratic three π‘₯ squared minus five π‘₯ minus two. There aren’t huge number of possibilities here, as the coefficient of π‘₯ squared three is a prime number and so is the number part of the constant term, two. The three π‘₯ squared must be broken down into three π‘₯ and π‘₯. So these are the terms at the front of the two brackets.

Notice that the sign of the constant term is negative, which means that the two numbers we’ll multiply together to achieve this must have different signs. The possibilities are negative two and one or two and negative one. And to work out which pair it is, we can use a bit of trial and error. Our first put positive one in the first bracket with three π‘₯ and negative two in the second bracket with π‘₯.

We can check whether this is the right option by expanding the brackets out again using the FOIL method. I know that I’m gonna get three π‘₯ squared when I multiply the first terms together: three π‘₯ multiplied by π‘₯ is three π‘₯ squared. Multiplying the outer terms together, I have three π‘₯ multiplied by negative two which is negative six π‘₯. Multiplying the inner terms together, positive one multiplied by π‘₯ is positive π‘₯. Finally, multiplying the last terms together positive one multiplied by negative two gives negative two, which we knew we would achieve.

If we now look at the number of π‘₯s in our expansion, we have negative six π‘₯ plus π‘₯ which is negative five π‘₯ and this is the correct coefficient of π‘₯, which tells us we’ve got the correct factorization. If you’ve been unlucky and tried a different combination of two and negative one first, then you wouldn’t have got the correct number of π‘₯s when you expanded and this would tell you that you need to try a different combination.

Now, let’s consider factorizing the quadratic expression in the denominator: two π‘₯ squared minus seven π‘₯ plus six. Now, there aren’t a lot of possibilities for the π‘₯ terms in the brackets. Two is a prime number. So we split two π‘₯ squared out into two π‘₯ and π‘₯. The factors of six β€” the constant term β€” are six and one or two and three. We notice that the constant term is positive, but the coefficient of π‘₯ is negative, which means that the two numbers we’re looking for must both be negative numbers. Our options are negative six and negative one or negative two and negative three.

Again, you can use a bit of trial and error in order to work out which way around the two numbers go and in which bracket. But there’s a bit of a trick here. Remember we’re expecting that one of the brackets in the numerator will cancel with one of the brackets in the denominator. And therefore, we can look at our factorized version of the numerator for a bit of a hint. We’re expecting one of the brackets to be the same. We don’t have the option of recreating the bracket of three π‘₯ plus one, but we can recreate the bracket of π‘₯ minus two. This would make the other bracket two π‘₯ minus three.

Let’s expand using the FOIL method and check whether this works. Two π‘₯ multiplied by π‘₯ gives two π‘₯ squared. Two π‘₯ multiplied by negative two gives negative four π‘₯. Negative three multiplied by π‘₯ is negative three π‘₯. And finally, negative three multiplied by negative two is positive six. We knew that we’d get the two π‘₯ squared and the positive six correctly. And if we look at the π‘₯ terms, we have negative four π‘₯ minus three π‘₯ which is negative seven π‘₯, the correct number of π‘₯s, which means we’ve got the right factorization.

Now that we factorized both the numerator and denominator, we can substitute the factorized forms back into this fraction. We have three π‘₯ plus one multiplied by π‘₯ minus two in the numerator and two π‘₯ minus three multiplied by π‘₯ minus two in the denominator. The common factor of π‘₯ minus two in both the numerator and denominator can be cancelled out, giving a simplified fraction three π‘₯ plus one over two π‘₯ minus three, which is now in the required form as both the numerator and denominator are linear expressions.

We weren’t explicitly asked to list the values of π‘Ž, 𝑏, 𝑐, and 𝑑. But comparing our fraction with the requested form, we see that π‘Ž is equal to three, 𝑏 is equal to one, 𝑐 is equal to two, and 𝑑 is equal to negative three.

Remember that tip: because we’re looking to find a common factor that we can cancel from both the numerator and the denominator, we can use the factorized form of the first quadratic to make factorizing the second a little bit easier.

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