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12 (a) Prove that (2x + 1) is a factor of p(x) - AQA - A-Level Maths Mechanics - Question 12 - 2018 - Paper 1
Question 12
12 (a) Prove that (2x + 1) is a factor of p(x). 12 (b) Factorise p(x) completely. 12 (c) Prove that there are no real solutions to the equation $$30 \sec^2 x + 2 ... show full transcript
Worked Solution & Example Answer:12 (a) Prove that (2x + 1) is a factor of p(x) - AQA - A-Level Maths Mechanics - Question 12 - 2018 - Paper 1
Step 1
Prove that (2x + 1) is a factor of p(x)
Answer
To prove that (2x + 1) is a factor of p(x), we will use the Factor Theorem. This theorem states that for a polynomial p(x), if p(a) = 0 for some a, then (x - a) is a factor of p(x).
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We start by substituting x = -\frac{1}{2} into p(x):
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Calculate each term:
- First term: (30(-\frac{1}{2})^3 = 30 \cdot -\frac{1}{8} = -\frac{30}{8} = -\frac{15}{4})
- Second term: (-7(-\frac{1}{2})^2 = -7 \cdot \frac{1}{4} = -\frac{7}{4})
- Third term: (-7(-\frac{1}{2}) = \frac{7}{2} = \frac{14}{4})
- Fourth term: 2 (= 2 = \frac{8}{4})
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Adding these results together:
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Since p(-1/2) = 0, by the Factor Theorem, (2x + 1) is a factor of p(x).
Step 2
Factorise p(x) completely
Answer
To factorise p(x) completely, knowing (2x + 1) is a factor, we can perform polynomial long division or synthetic division.
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Divide p(x) by (2x + 1).
- The quotient should yield a quadratic polynomial: (p(x) = (2x + 1)(ax^2 + bx + c)).
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The result from the division gives:
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Now factor the quadratic if possible:
- The factors of 15x^2 - 5x + 2 can be found using the quadratic formula or by factoring directly if simple. After factoring, we find:
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Therefore, the complete factorisation of p(x) is: .
Step 3
Prove that there are no real solutions to the equation
Answer
To prove there are no real solutions to the equation:
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Rearrange the equation: Start by multiplying both sides by 7 to eliminate the fraction:
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Substitute (\sec x = \frac{1}{\cos x}) into the equation:
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Multiply through by (\cos^2 x) to eliminate the denominators:
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Rearranging gives a cubic equation:
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Analyze the function's range:
- As (\cos x) varies from -1 to 1, use the intermediate value theorem or graphing to show the cubic does not cross zero, indicating no real solutions exist.