The answer
(a) \(a = -4\), \(b = 3\)
(b) \(x = 2\), \(x = -2 + \sqrt7\), \(x = -2 - \sqrt7\)
O-Level A-Math 2024 Paper 2 Question 6 · Verified worked solution by the Genius Plus Academy teaching team
The question
(a) \(x^3 + x^2 + ax + 2b\) and \(x^3 + 4x^2 - ax + b\) both have factor \(x + 3\). Find \(a\) and \(b\). [4]
(b) Solve \(x^3 + 2x^2 - 11x + 6 = 0\) in exact form. [5]
(a) Both expressions are \(0\) at \(x = -3\). First: \(-27 + 9 - 3a + 2b = 0 \Rightarrow -3a + 2b = 18\). Second: \(-27 + 36 + 3a + b = 0 \Rightarrow 3a + b = -9\). Adding: \(3b = 9 \Rightarrow b = 3\), then \(3a = -9 - 3 \Rightarrow a = -4\).
(b) Testing \(x = 2\): \(8 + 8 - 22 + 6 = 0\), so \((x - 2)\) is a factor. Dividing, \(x^3 + 2x^2 - 11x + 6 = (x - 2)(x^2 + 4x - 3)\). Then \(x^2 + 4x - 3 = 0 \Rightarrow x = \dfrac{-4 \pm \sqrt{16 + 12}}{2} = -2 \pm \sqrt7\). So \(x = 2\), \(-2 + \sqrt7\), \(-2 - \sqrt7\).
Answer: (a) \(a = -4\), \(b = 3\)
(b) \(x = 2\), \(x = -2 + \sqrt7\), \(x = -2 - \sqrt7\)
Same structure, different numbers
Swap the constants, dress a quadratic as a length, hide a derivative inside an integral, and a student sees a brand new problem. The structure underneath is the same, and so is the method. Once a student can name the structure, a whole row of questions that look different start to open the same way.
That is where marks really leak: in choosing the method, not in the algebra that follows. We call it Lock and Key, name the lock, then the key follows.
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Genius Plus Academy · O-Level & IP Mathematics
Our O-Level A-Math tuition trains the same recognise-the-structure method these worked solutions show, taught by a team that has marked these papers for years. It runs within our weekly Secondary Math programme, Sec 1 to 4 and IP.
It is a factor theorem (two simultaneous conditions) question from Polynomials & partial fractions, worth 9 marks: 4 + 5.
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