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12 (a) Prove that (2x + 1) is a factor of p(x) 12 (b) Factorise p(x) completely - AQA - A-Level Maths Pure - 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) 12 (b) Factorise p(x) completely - AQA - A-Level Maths Pure - Question 12 - 2018 - Paper 1

Step 1

Prove that (2x + 1) is a factor of p(x)

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Answer

To prove that (2x + 1) is a factor of p(x), we need to use the Factor Theorem.

First, let us substitute ( x = -\frac{1}{2} ) into ( p(x) ):

$p(-\frac{1}{2}) = 30(-\frac{1}{2})^3 - 7(-\frac{1}{2})^2 - 7(-\frac{1}{2}) + 2$

Calculating each term:

The first term: ( 30(-\frac{1}{2})^3 = 30 \times -\frac{1}{8} = -\frac{15}{4} )
The second term: ( -7(-\frac{1}{2})^2 = -7 \times \frac{1}{4} = -\frac{7}{4} )
The third term: ( -7(-\frac{1}{2}) = \frac{7}{2} = \frac{14}{4} )
The constant term: ( 2 = \frac{8}{4} )

Now, combining these results:

$p(-\frac{1}{2}) = -\frac{15}{4} - \frac{7}{4} + \frac{14}{4} + \frac{8}{4}$

This simplifies to:

$p(-\frac{1}{2}) = 0$

Since ( p(-\frac{1}{2}) = 0 ), we can conclude that ( 2x + 1 ) is indeed a factor of ( p(x) ).

Step 2

Factorise p(x) completely.

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Answer

To factorise ( p(x) ) completely, we will first use the result from part (a) where we established that ( (2x + 1) ) is a factor.

The polynomial can then be expressed as:

$p(x) = (2x + 1)(\text{quadratic factor})$

We can perform polynomial long division of ( p(x) ) by ( (2x + 1) ). After performing the division, we find:

$p(x) = (2x + 1)(15x^2 - 11x + 2)$

Next, we will factor the quadratic ( 15x^2 - 11x + 2 ).

The quadratic can be factorised as:

$15x^2 - 11x + 2 = (3x - 1)(5x - 2)$

Thus, the complete factorisation of ( p(x) ) is:

$p(x) = (2x + 1)(3x - 1)(5x - 2)$

Step 3

Prove that there are no real solutions to the equation

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Answer

Given the equation:

$30 \sec^2 x + 2 \cos x = 7 \sec x + 1$

We can rearrange this equation to isolate terms:

$30 \sec^2 x - 7 \sec x + 2 \cos x - 1 = 0$

Next, we substitute ( \sec x = \frac{1}{\cos x} ) and rewrite the equation:

$30 \frac{1}{\cos^2 x} - 7 \frac{1}{\cos x} + 2 \cos x - 1 = 0$

Multiplying through by ( \cos^2 x ) to eliminate the fractions yields:

$30 - 7 \cos x + 2 \cos^3 x - \cos^2 x = 0$

This can be rearranged to:

$2 \cos^3 x - \cos^2 x - 7 \cos x + 30 = 0$

Analyzing this cubic equation, we see that the derivative must be examined for critical points:

$\frac{d}{dx} (2 \cos^3 x - \cos^2 x - 7 \cos x + 30)$

Exploring the range of ( \cos x ) (which is between (-1) and (1)), we see the function does not intersect the x-axis, indicating there are no real solutions. Therefore, we conclude that there are no real solutions to the equation.

12 (a) Prove that (2x + 1) is a factor of p(x) 12 (b) Factorise p(x) completely - AQA - A-Level Maths Pure - Question 12 - 2018 - Paper 1

Worked Solution & Example Answer:12 (a) Prove that (2x + 1) is a factor of p(x) 12 (b) Factorise p(x) completely - AQA - A-Level Maths Pure - Question 12 - 2018 - Paper 1

Prove that (2x + 1) is a factor of p(x)

Factorise p(x) completely.

Prove that there are no real solutions to the equation

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