Video: AQA GCSE Mathematics Higher Tier Pack 4 β€’ Paper 2 β€’ Question 24

Write the following expression in the form (π‘Žπ‘₯ + 𝑏)/(𝑐π‘₯ + 𝑑), where π‘Ž, 𝑏, 𝑐, and 𝑑 are integers. (4π‘₯Β² βˆ’ 1)/(π‘₯Β² + π‘₯ βˆ’ 2) Γ— (π‘₯ + 2)/(2π‘₯ βˆ’ 1) Γ— (π‘₯Β² + 4π‘₯ βˆ’ 5)/(π‘₯Β² + 7π‘₯ + 10)

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Video Transcript

Write the following expression in the form π‘Žπ‘₯ plus 𝑏 over 𝑐π‘₯ plus 𝑑, where π‘Ž, 𝑏, 𝑐, and 𝑑 are integers. Four π‘₯ squared minus one over π‘₯ squared plus π‘₯ minus two times π‘₯ plus two over two π‘₯ minus one times π‘₯ squared plus four π‘₯ minus five over π‘₯ squared plus seven π‘₯ plus 10.

The process for multiplying fractions is multiplying all the numerators by each other and then multiplying all the denominators by each other. Before we get to π‘Žπ‘₯ plus 𝑏 over 𝑐π‘₯ plus 𝑑, we want to try and factorise some of the terms.

First, let’s factorise four π‘₯ squared minus one. This time, we notice that four π‘₯ squared is a square value. Two π‘₯ times itself is four π‘₯ squared. One is also a square. One multiplied by itself is a square. And that means we’re taking the difference of squares.

To factorise the difference of squares, we’ll have two π‘₯ and one. The key to this factorisation is that one of the terms will be plus one and one of them will be minus one. Four π‘₯ squared minus one can be rewritten as two π‘₯ plus one times two π‘₯ minus one.

Next, we can factor the denominator π‘₯ squared plus π‘₯ minus two. This time, we’re not dealing with the difference of squares. Since our first term does not have a coefficient, we can use the π‘₯ as our first value. π‘₯ times π‘₯ will equal π‘₯ squared. From there, we need to find two values that multiply together to equal negative two and their sum is positive one.

What factors multiply together to equal negative two? One and negative two or negative one and positive two. And now we need to check their sum. We’re looking for the two values whose sum is positive one. One plus negative two equals negative one. Negative one plus two is positive one. So we plug in negative one and two. π‘₯ squared plus π‘₯ minus two is factorised as π‘₯ minus one times π‘₯ plus two.

Moving on to the next fraction, π‘₯ plus two is already factorised, as is two π‘₯ minus one. Next, we’ll factorise π‘₯ squared plus four π‘₯ minus five. Our first term is π‘₯ squared. We can break that up into π‘₯ times π‘₯. From there, we need to find two factors that multiply together to equal negative five and when added together sum to positive four. The factors of negative five, one times negative five and negative one times five. We need the one whose sum is positive four. Negative one plus five equals positive four. So we’ll plug in negative one and positive five. π‘₯ squared plus four π‘₯ minus five is factorised as π‘₯ minus one times π‘₯ plus five.

Finally, we need to factorise π‘₯ squared plus seven π‘₯ plus 10. Our leading value is π‘₯ squared, which is π‘₯ times π‘₯. We need to look for values that multiply together to equal positive 10 and sum to positive seven. What values multiply together to equal 10? One and 10 or two and five.

However, negative one and negative 10 also multiply together to equal positive 10, as do negative two and negative five. We should consider which of these sets add to equal positive seven. Two plus five equals seven. And so we’ll use positive two and positive five. π‘₯ squared plus seven π‘₯ plus 10 is factorised as π‘₯ plus two times π‘₯ plus five.

What we need to do now is substitute all the factorisations in. π‘₯ plus two over two π‘₯ minus one doesn’t change. Here’s substitution for the first fraction, and here’s substitution for the third fraction. At this point, we can do some simplifying.

We have two π‘₯ minus one in our numerator and a two π‘₯ minus one in the denominator. These two values cancel. π‘₯ plus two in the numerator, π‘₯ plus two in the denominator cancels. π‘₯ minus one in the denominator, π‘₯ minus one in the numerator cancels. π‘₯ plus five in the numerator, π‘₯ plus five in the denominator cancels.

We have two π‘₯ plus one remaining in the numerator and π‘₯ plus two remaining in the denominator. The product of these three fractions is two π‘₯ plus one over π‘₯ plus two. At this point, we need to consider, is two π‘₯ plus one over π‘₯ plus two in the form π‘Žπ‘₯ plus 𝑏 over 𝑐π‘₯ plus 𝑑? The value for π‘Ž would be the integer two. 𝑏 would be equal to one. The value for 𝑐 might not be immediately obvious. But π‘₯ is the same thing as one π‘₯, which makes 𝑐 an integer, the value one. And 𝑑 is equal to two. Two π‘₯ plus one over π‘₯ plus two is in the form π‘Žπ‘₯ plus 𝑏 over 𝑐π‘₯ plus 𝑑. Therefore, the expression is simplified to two π‘₯ plus one over π‘₯ plus two.

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