Video: AQA GCSE Mathematics Higher Tier Pack 4 β€’ Paper 1 β€’ Question 17

Consider the identity (2π‘₯ βˆ’ 5) (π‘₯ + 3) + π‘Žπ‘₯ + 𝑏 = 2π‘₯Β² + 4π‘₯ βˆ’ 3. Work out the values of π‘Ž and 𝑏.

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Video Transcript

Consider the identity two π‘₯ minus five multiplied by π‘₯ plus three plus π‘Žπ‘₯ plus 𝑏 is identically equal to two π‘₯ squared plus four π‘₯ minus three. Work out the values of π‘Ž and 𝑏.

An identity is a statement which is always true no matter what the value of the variable, in this case π‘₯, takes. It’s denoted by this sign here, an equal sign with an extra horizontal line to indicate that the statement is always true.

We’re asked to work out the values of π‘Ž and 𝑏. And to do this, we’ll begin by expanding and simplifying the left-hand side of this identity. We’ll use the FOIL method to expand the brackets. First, we multiply two π‘₯ by π‘₯, giving two π‘₯ squared. Then we multiply the outer terms together. Two π‘₯ multiplied by three gives positive six π‘₯. Next, the inner terms, negative five multiplied by π‘₯ gives negative five π‘₯. And finally, the last terms, negative five multiplied by three gives negative 15. We can then bring down the plus π‘Žπ‘₯ plus 𝑏 from the previous line and also the right-hand side of this identity.

Next, we can simplify the like terms in our expansion. Positive six π‘₯ minus five π‘₯ just leaves positive one π‘₯ or π‘₯. So we have two π‘₯ squared plus π‘₯ minus 15 plus π‘Žπ‘₯ plus 𝑏 is identically equal to two π‘₯ squared plus four π‘₯ minus three.

Now we’re actually going to go a little bit further with our simplification because we still have like terms. We have plus π‘₯ and then plus π‘Žπ‘₯. We can combine these two terms together with a coefficient of one plus π‘Ž, because when we expand this bracket, we’ll get π‘₯ plus π‘Žπ‘₯. We’ll also combine our constant term. We have negative 15 plus 𝑏, which we can write as 𝑏 minus 15. We then bring down the right-hand side of the identity.

Now here’s a key fact that we need to know in order to answer this question. If this statement is an identity, then it means that the coefficient of π‘₯ squared, π‘₯, and the constant term must be the same on both sides of this identity in order for it to be true for all values of π‘₯.

We can see, for example, that the coefficient of π‘₯ squared is two on each side. In order to find the values of π‘Ž and 𝑏, we need to compare the coefficients of π‘₯ and the constant term on the two sides of the equation. If we compare the coefficients of π‘₯ first of all, we have one plus π‘Ž on the left side of this identity and four on the right. So we have the equation one plus π‘Ž equals four, which we can solve to find the value of π‘Ž. We just need to subtract one from each side, giving π‘Ž equals three.

Finally, we compare the constant terms. We have 𝑏 minus 15 on the left and negative three on the right. So we have an equation that we can solve for 𝑏. We add 15 to each side of the equation, giving 𝑏 equals 12. So by expanding, simplifying, and then comparing coefficients, we found the values of π‘Ž and 𝑏. π‘Ž is equal to three, and 𝑏 is equal to 12.

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