A household tracks its combined gas and electricity bill using algebra, where \(n\) is the number of months since a new tariff began. Over that time, the ga...
Question 1 Report
A household tracks its combined gas and electricity bill using algebra, where \(n\) is the number of months since a new tariff began. Over that time, the gas bill, in pounds, is modelled by \(3n^2 + 5n - 2\) and the electricity bill, in pounds, is modelled by \(n^2 - 4\).
- Show that the total combined bill, in pounds, is \(4n^2 + 5n - 6\). (2)
- Factorise \(4n^2 + 5n - 6\) fully. (2)
- Hence, or otherwise, simplify fully \(\dfrac{4n^2+5n-6}{n^2-4}\). (3)
Answer Details
Combining two separate bill formulas means adding them term by term, and once the total is a single quadratic, it can be factorised and used to simplify a fraction built from it.
(a) Adding the gas and electricity formulas: \( (3n^2+5n-2)+(n^2-4) = 3n^2+n^2+5n-2-4 = 4n^2+5n-6 \), as required [2 marks].
(b) For \( 4n^2+5n-6 \), the product \( 4 \times (-6)=-24 \) and the required sum is \(5\); the numbers \(8\) and \(-3\) satisfy \( 8 \times (-3)=-24 \) and \( 8+(-3)=5 \). Splitting the middle term and grouping gives \( 4n^2+8n-3n-6 = 4n(n+2)-3(n+2) = (4n-3)(n+2) \) [2 marks].
(c) The denominator \( n^2-4 \) is a difference of two squares, factorising as \( (n-2)(n+2) \). The fraction becomes \( \dfrac{(4n-3)(n+2)}{(n-2)(n+2)} \), and the common factor \( (n+2) \) cancels, leaving \( \dfrac{4n-3}{n-2} \) [3 marks].
Factorising both parts of a fraction before attempting to cancel anything is essential here: \( 4n^2+5n-6 \) and \( n^2-4 \) share the factor \( (n+2) \), which is only visible once both are written as products rather than sums.
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