Homework 5

Homework 5#

Released: 4pm Monday Week 8. Due: 4pm Monday Week 9.

Thank you to everyone who completed this work!

Your work is now marked with personalised feedback. Please take some time to view it on Crowdmark and compare with the more general comments and model solution below.

General feedback#

There were up to 3 marks available for part (i), and I was pleased to see most people got these. In some cases, marks were lost because of confusing presentation — remember to get in touch if you are having issues with this and would like more personalised feedback on your written mathematics.

Part (ii) was out of 7, and generally submissions to this part were much harder to make sense of. If you got a low score for this part, it is especially important that you review comments on Crowdmark and get in touch if you would like support improving your written maths. Remember, we can only award marks if we can see mathematical reasoning in your work — this applies to the written exam as well as to written homework.

High marks were awarded for using a correct method, correct mathematical notation, with appropriate explanation, written in a clear, easy to follow fashion. Method marks were lost if we couldn’t follow your argument, and/or for notational issues. A full model solution with mark scheme can be viewed below.

If you would like to look at worked examples similar to part (ii), I recommend reviewing Lectures 21 (11/11/25) and Lecture 26 (16/12/25). Handwritten notes and encore lecture recordings are available on Blackboard in the usual places.

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Question#

(i) Use the factor theorem to fully factorise \(2x^3-4x^2-2x+4\). (3 marks)

(ii) Using algebra, find all the values of \(x \in \mathbb{R}\) for which \(-2x^3 + 4x^2 + 2x -4 \geq 0\). (7 marks)

Model solution with mark scheme#

Hidden in case you want to re-attempt the question as revision.

Click to view solution

(i) Let \(p(x) = 2x^3 - 4x^2 - 2x +4\). Observe that \(p(1)=2-4-2+4=0\). Therefore, by the factor theorem, \((x-1)\) is a factor of \(p(x)\). (1 mark)

Also, \(p(-1)=-2-4+2+4=0\), and so \((x+1)\) is also a factor of \(p(x)\). (1 mark)

Clearly 2 is factor of \(p(x)\), and so we get

\[ p(x) = 2(x-1)(x+1)(x+c), \]

for some \(c\in\mathbb{R}\). The constant term of \(p(x)\) is \(4\), and therefore we must have \(c=2\). Hence

\[ p(x) = 2(x+1)(x-1)(x-2). \]

(1 mark)

(ii) We have been asked to solve

\[ -p(x) = -2x^3 + 4x^2 + 2x -4 \geq 0, \]

which is equivalent to solving \(p(x)\leq 0\). Using the factorisation from the first part, we will solve

\[ 2(x+1)(x-2)(x-1) \leq 0, \]

or in other words,

\[ (x+1)(x-2)(x-1) \leq 0. \]

(1 mark)

For a product of three terms to be negative, either exactly one factor is negative and the other two are positive, OR all three factors are negative. (1 mark)

  • Case 1: If all three factors are negative: then we must have \(x+1 \leq 0\) and \(x-2 \leq 0\) and \(x -1 \leq 0\).
    That is, \(x \leq -1\) and \(x \leq 2\) and \(x \leq 1\), which means that \(x \leq -1\). (1 mark)

  • Case 2: If one factor is negative and other two are positive, then there are three possibilities:

    • Case 2a: \(x+1 \leq 0\) and \(x -1 \geq 0\) and \(x-2 \geq 0\).
      That is, \(x \leq -1\) and \(x \geq 1\) and \(x \geq 2\). This is clearly impossible.

    • Case 2b: \(x-1 \leq 0\) and \(x-2 \geq 0\) and \(x+1 \geq 0\).
      That is, \(x \leq 1\) and \(x \geq 2\) and \(x \geq -1\). This is also impossible.

    • Case 2c: \(x-2 \leq 0\) and \(x+1 \geq 0\) and \(x-1 \geq 0\).
      That is, \(x \leq 2\) and \(x \geq -1\) and \(x \geq 1\), or, in other words, \(1 \leq x \leq 2\).

(3 marks)

Hence the solution to the inequality

\[ -2x^3 + 4x^2 + 2x -4 \geq 0 \]

is \(x \leq -1\) or \(1 \leq x \leq 2\). (1 mark)

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