Notes

M1: Sets the sum of the roots = 21 and the product of the roots = 91

M1: Forms an equation in one variable using their sum of roots and product of roots equations

A1: Correct simplified quadratic equation

dM1: Dependent on the previous method mark, solves their equation to find real roots, may be a cubic if slips earlier

ddM1: Dependent on previous method mark. Finds the value of the other variable

dddM1: Dependent on previous method mark. Uses their values of \(\alpha\) and \(\beta\) to find the value of the roots

A1: Correct roots

Alternative

M1: Sets the sum of the roots = 21

M1: Simplifies and deduces one root

A1: One root = 7

dM1: Factorises into linear and quadratic, achieves correct first and last term

ddM1: Factorises their quadratic, which must be factorisable

dddM1: Uses their factorised quadratic to find the remining 2 roots

A1: Correct roots 1 and 13

Notes

B1: Identifies the correct values for the product and triple pair sum. If seen these must be correct, but they may be implied if not explicitly extracted, e.g. \(\displaystyle\sum \dfrac{3}{\alpha_i} = 3\dfrac{\sum \alpha_i\alpha_j\alpha_k}{\alpha\beta\gamma\delta} = 3 \times \dfrac{5}{6} = \dfrac{5}{2}\) does not technically contain an incorrect step if one realises the 2’s will cancel. A good indication will be what happens in part (b) – if there they realise the 2’s are involved then allow b.o.d. for a calculation such as here given in (a).
Note \(\sum \alpha\beta\gamma\delta\) is a correct notation for the product of roots.

M1: Uses the correct identity and their values of the product and triple sum. (The values need not be correct for this mark.)

A1: Correct value following correct identity with no indication of incorrect values for triple sum and product (see comment on the B mark).

Alt (a)

Alt (a): Some may use a transformation. These can be scored as

B1: Makes a correct substitution into the equation to solve the problem. This will probably be \(x = \dfrac{3}{w}\) but note that \(x = \dfrac{1}{w}\) can also be used.

M1: Multiplies through by \(w^4\) and extracts the correct sum of roots from the new equation. If using \(x = \dfrac{1}{w}\) they must also multiply through by the 3 to gain this mark.

A1: Correct answer from correct work on the relevant coefficients (the others need not be seen).

Notes

B1: Identifies the correct values for the sum and pair sum. Allow when first seen – some will list all these in part (a), which is fine for this mark.

M1: Attempts to find the identity for \((\alpha + \beta + \gamma + \delta)^2\) in terms of the sum of squares and pair sum (seen or implied). Allow attempts where the “2” is incorrect with no other method shown.

A1: Correct identity (seen or implied).

dM1: Dependent on previous method mark. Substitutes the sum and pair sum of the roots into their identity and forms and solves a 3TQ for \(A\) (usual rules).

A1: Correct values of \(A\)

Notes

B1: Correct values stated.

Notes

(b) Note question requires use of (a) so other methods will score no marks.

M1: Use a correct identity with an attempt to substitute their values into the correct places (allowing a slip) to find the value required. If identity is not shown it may be implied by the working.

A1ft: Correct value (follow through their part (a)) Note that this means the error. \(\Sigma\alpha_i = 3,\ \Sigma\alpha_i\beta_j = 12,\ \Pi\alpha_i = -7\) will score M1A1ft for the correct answer albeit from incorrect values.

Notes

M1: Attempts to expand the product fully (allow sign slips and at most one incorrect or missing term).

A1: Correct expansion in terms of product, pair sum and sum - must be seen grouped or implied by substitution of values seen.

A1: Correct value.

Notes

M1: Use a correct identity with an attempt to substitute their values into the correct places (allowing a slip) to find the value required. If identity is not shown it may be implied by the working.

A1ft: Correct value from correct working (follow through their part (a)). Note this means the error \(\Sigma\alpha_i = -\dfrac{3}{2}\) will still give the A1ft here if correct identity is used.

Notes

B1: Correct sum and pair sum, they may be seen anywhere in the candidates working.

M1: Uses a correct identity and substitutes in their sum and pair sum to find a value.

A1: Correct value following B1, if uses \(\alpha + \beta + \gamma = -\dfrac{3}{2}\) this can score B0 M1 A0 cso

Notes

B1: Correct value for the product (may be seen anywhere in the candidates working) or for using \(x = \dfrac{3}{w}\) in the given equation.

M1: Uses a correct identity and substitutes in their pair sum and product to obtain a value or multiplies through by \(w^3\) to identify at least the required terms and finds their new sum.

A1: Correct value from correct pair sum and product cso

Notes

M1: Correct strategy for obtaining the required value by expanding, must reach an expression for the form \(A \pm B(\alpha + \beta + \gamma) \pm C(\alpha\beta + \alpha\gamma + \beta\gamma) \pm (\alpha\beta\gamma)\) may not be factorised for example.
\(A \pm B\alpha \pm B\beta \pm B\gamma \pm C\alpha\beta \pm C\alpha\gamma \pm C\beta\gamma \pm (\alpha\beta\gamma)\)
or
Attempts the correct linear transformation of the given equation and expands.
or
Uses \(\mathrm{f}(x) = A(x - \alpha)(x - \beta)(x - \gamma)\) to find a value for \(A\)

M1: Uses their sum, pair sum and product to obtain a value. Allow recovery from a sign slip as long as substituting into an expression of the form \(A \pm B(\alpha + \beta + \gamma) \pm C(\alpha\beta + \alpha\gamma + \beta\gamma) \pm (\alpha\beta\gamma)\).
This would be A0 even if the correct answer is achieved.
or
Simplifies to obtain at least the required terms to find a value for the new product. Ignore the other terms whether correct or not.
Or
Uses \(\dfrac{\mathrm{f}(5)}{2}\)

A1: Correct value with no errors seen cso

Notes

B1: Selects the method of making a connection between \(z\) and \(w\) by writing \(z = \dfrac{w + 1}{3}\). Other variables may be used

M1: Applies the process of substituting their \(z = \dfrac{w + 1}{3}\) into \(z^4 + 5z^2 - 30 = 0\)

M1: Manipulates their equation into the form \(w^4 + aw^3 + bw^2 + cw + d = 0\) having substituted their \(z\) in terms of \(w\). Note that the “= 0” can be missing for this mark.
A1: At least two of \(a\), \(b\), \(c\), \(d\) correct. Note that the “= 0” can be missing for this mark.
A1: Fully correct equation including “= 0” Must be in terms of \(w\)

Alternative

(Corrected from the printed mark scheme: the new triple sum is printed with \(+6(p + q + r + s)\); the correct coefficient is \(+9(p + q + r + s)\). The value −94 is unchanged because \(p + q + r + s = 0\).)

(i) Alternative
B1: Selects the method of giving four correct equations containing \(p\), \(q\), \(r\) and \(s\)
M1: Applies the process of finding at least three of the new sum, new pair sum, new triple sum and new product. Condone slips but the intention is clear and uses their values.
M1: Applies \(w^4 - (\text{new sum})w^3 + (\text{new pair sum})w^2 - (\text{new triple sum})w + (\text{new product}) = 0\).
Condone use of any variable for this mark.
Note that the “= 0” can be missing for this mark.
A1: At least two of \(a\), \(b\), \(c\), \(d\) correct. Note that the “= 0” can be missing for this mark.
A1: Fully correct equation including “= 0” Must be in terms of \(w\)

Notes

M1: Uses the sum and product to form two equations in \(\alpha\) and \(\beta\). Condone product = \(\dfrac{81}{4}\) for this mark

Note: \(4\alpha - \beta = -\dfrac{n}{4}\) or \(4\alpha - \beta = 81\) is M0

A1: Correct equations need not be simplied

M1: Solves simultaneous equations to find a value for \(\alpha\) or \(\beta\)

M1: Uses their values for \(\alpha\) and \(\beta\) to find the roots using \(\alpha,\ 2\alpha,\ \alpha - \beta\). Condone third root as \(\beta\)

A1: Correct roots

Candidates may use their own notation for the roots

Condone confusion with \(\beta\) for M1 but A0

Notes

M1: Finds the pair sum for their numerical roots and multiplies by 4

Alternative multiplies out three brackets \((x - \text{their } \alpha)(x - \text{their } 2\alpha)(x - (\alpha - \beta))\) to achieve the form \(4x^3 + \ldots\)

A1: Correct value from correct roots only

Notes

Alternative using sum, pair sum and product of roots:

Notes

Allow a variable other than \(w\) to be used for the first 4 marks.
The “= 0” is not required until the final mark.

B1: Selects the method of making a connection between \(x\) and \(w\) by writing \(x = w - 2\)

M1: Applies the process of substituting their \(x = \text{"}w - 2\text{"}\) into the equation for all occurrences of \(x\).

M1: Depends on having attempted substituting either \(x = w - 2\) or \(x = w + 2\) into the equation. This mark is for manipulating their resulting equation into the required form so must have gathered terms. Condone poor squaring/cubing of brackets as long as a cubic expression is obtained.

A1: At least two of \(p\), \(q\) and \(r\) correct.

A1: Correct final equation (including “= 0”). Must be an equation in \(w\).

Note if they say e.g. \(x = w - 2\) and then substitute \(w + 2\), it is possible to score B1 M0 M1
Note if they say e.g. \(x = w + 2\) and then substitute \(w - 2\), allow recovery

Alternative:
B1: Selects the method of giving three correct equations for the sum, pair sum and product in terms of \(\alpha\), \(\beta\) and \(\gamma\). Note that the correct values may be seen embedded when they attempt the new sum, pair sum and product e.g. \((\alpha + 2)(\beta + 2)(\gamma + 2) = \alpha\beta\gamma + 2(\alpha\beta + \alpha\gamma + \beta\gamma) + 4(\alpha + \beta + \gamma) + 8 = \underline{-6} + 2(\underline{-12}) + 4(\underline{7}) + 8\)

M1: Applies the process of finding the new sum, pair sum and product. Mark positively here and allow slips provided they are attempting \(\alpha + 2 + \beta + 2 + \gamma + 2\), \((\alpha + 2)(\beta + 2) + (\alpha + 2)(\gamma + 2) + (\beta + 2)(\gamma + 2)\) and \((\alpha + 2)(\beta + 2)(\gamma + 2)\)

M1: In this method, this mark is for choosing \(p = -\)(their new sum), \(q =\) their new pair sum, \(r = -\)(their new product) or forming \(w^3 - (\text{new sum})w^2 + (\text{new pair sum})w - (\text{new product})\)
(corrected from the printed mark scheme: the first term is printed as \(w^2\))

A1: At least two of \(p\), \(q\) and \(r\) correct. As values or seen in their equation.

A1: Correct final equation (including “= 0”). Must be an equation in \(w\).

In all methods, the final A mark depends on all the previous marks.

Notes

B1: Identifies the correct values for the sum and pair sum. This may be implied by substituting into an equation, it must be clear

M1: Uses the correct identity and values of their sum and their pair sum to find a value of \(p\)

A1*: \(p = 12\) cso there is no need to see a reason

Notes

M1: Establishes a correct identity

M1: Uses their identity and their pair sum and their product of roots to find a value of \(q\). Condone a slip but the intention must be clear.

A1: \(q = 3\) Allow this mark from incorrect sign of both pair sum and product

Alternative

M1: Uses \(x = \dfrac{1}{w}\) the substitution

M1: Simplifies to an quartic equation of the form \(aw^3 + bw^2 + cw + d = 0\) and uses \(\dfrac{14}{3} = -\dfrac{b}{a}\) to find a value for \(q\)

A1: \(q = 3\)

Notes

M1: Attempts to multiply out the three brackets.

A1: Correct expansion.

dM1: Dependent on previous method. Substitutes in the value of their sum, pair sum and the value of their product as appropriate. Condone a slip but the intention must be clear

A1: Correct value

Alternative

M1: Substitutes \((x + 1)\) or \(x = 1\) into the cubic with their value of \(q\). Allow the use of different letters e.g. \((w + 1)\)

A1ft: Correct constant terms, follow through on their value of \(q\)

dM1: Applies \(-\dfrac{\text{`their constant'}}{4}\)

A1: Correct value

Notes

B1: Correct sum and pair sum of roots seen or implied. Must realise the pair sum is zero.
Note: These values can be seen anywhere in the candidate’s solution

M1: Uses correct expression for the sum of squares.

A1: \(\dfrac{25}{9}\). Allow this mark from incorrect sign on sum of squares (but they will score B0 if the sign is incorrect).

Notes

B1: Correct triple sum and product of roots seen or implied. May be stated in (i). Alternatively, this may be scored for sight of \(x = \dfrac{2}{w}\) used as a transformation in the equation.
Note: These values can be seen anywhere in the candidate’s solution

M1: Substitutes their values into \(2 \times \dfrac{\sum\alpha_i\alpha_j\alpha_k}{\alpha\beta\gamma\delta} = \ldots\) In the alternative it is for rearranging the equation to a quartic in \(w\) and uses to find the sum of the roots.

A1: \(\dfrac{7}{3}\) Allow this mark from incorrect sign of both triple sum and product (but they will score B0 if the sign is incorrect).

Notes

M1: A correct method to find the value used – may recognise structure as scheme, may expand the expression in stages, or may attempt to use a linear transformation \((3 - x)\) or e.g. \((3 - w)\) in original equation. Condone slips as long as the intention is clear.

dM1: Dependent on previous method mark. Uses at least 2 values of their sum of roots etc. in their expression. If using a linear shift this is for expanding to find the coefficient of \(x^4\) and constant term and attempts product of roots by dividing the constant term by the coefficient of \(x^4\).

A1: 121.

Notes

(Corrected from the printed mark scheme: the printed scheme has \(\dfrac{87 + c}{19}\) in both places; dividing through by 16 gives \(\dfrac{87 + c}{16}\).)

B1: Selects the method of making a connection between \(x\) and \(w\) by writing \(w = 4x - 1\) or \(x = \dfrac{w + 1}{4}\)

M1: Applies the process of substituting their \(x = \dfrac{w + 1}{4}\) into \(ax^3 + bx^2 - 19x - b = 0\) or \(w = 4x - 1\) into \(w^3 - 9w^2 - 97w + c = 0\). Must be substitution of the correct variable into the opposing equation but may be scored if the initial linear equation is incorrect (e.g. \(x = 4w - 1\) into the first equation). Note that the “ = 0 ” can be missing for this mark.

M1: Expands the brackets and collects terms in their equation (in \(x\) or \(w\)). Note that the “ = 0 ” can be missing for this mark.

M1: A complete method for finding a value for \(a\) or \(b\). See scheme, it involves dividing through by an appropriate factor for their equation to balance the \(w^3\) or \(-19x\) terms, then equating other coefficients and solving equations if necessary.

M1: A complete method for finding a value for \(c\). They must have divided through by an appropriate factor as per the previous M before attempting to compare the constant coefficient (and use their \(a\) and \(b\) if appropriate).

A1: \(a = 4 \quad b = -12 \quad c = 105\)

Alternative

B1: Selects the method of giving at least two correct equations containing \(\alpha\), \(\beta\) and \(\gamma\)

M1: Applies the process of finding the new sum to generate an equation in \(a\) and \(b\). Must be substituting in the correct places.

M1: Attempts the new pair sum to generate another equation connecting \(a\) and \(b\). Must be substituting in the correct places.

M1: Solves their equations to find a value for \(a\) or \(b\).

M1: Uses the new product with their values to find values for \(a\), \(b\) and \(c\)

A1: \(a = 4 \quad b = -12 \quad c = 105\)

Notes

B1: Correct values for the product and pair sum of the roots

M1: A complete method to find the sum of \(\dfrac{1}{\alpha} + \dfrac{1}{\beta} + \dfrac{1}{\gamma}\). Must substitute in their values of the product and pair sum

A1: correct value 4

Note: If candidate does not divide by 3 so that \(\alpha\beta\gamma = -1\) and \(\alpha\beta + \alpha\gamma + \beta\gamma = -4\) the maximum they can score is B0 M1 A0

Notes

M1: A correct method to find the value of the new pair sum and the value of the new product

M1: Applies \(x^3 - (\text{part (a)})x^2 + (\text{their new pair sum})x - (\text{their new product})\,(= 0)\)

A1: Fully correct equation, in any variable, including = 0

Alternative

M1: Realises the connection between the roots and substitutes into the cubic equation

M1: Manipulates their equation into the form \(x^3 + ax^2 + bx + c = 0\)

A1: Fully correct equation in any variable, including = 0

Notes

M1: Equates sum of roots to 8 and obtains an equation in just \(\alpha\).

A1: Obtains a correct equation in \(\alpha\).

M1: Forms a three term quadratic equation in \(\alpha\) and attempts to solve this equation by either completing the square or using the quadratic formula to give \(\alpha = \ldots\)

A1: \(\alpha = 2 \pm \mathrm{i}\sqrt{2}\)

M1: Any correct method for finding the remaining root. There are various routes possible. See scheme for common ones.
Allow this mark if \(-24\) is used as the product.
See note below for a less common approach.

A1: Third root found with all three roots correct. Note \(\alpha\) and \(\beta\) need not be identified.

Note for (a) final M – it is possible to find the second and third roots using only one initial root (e.g. if second root forgotten or error leads to only one initial root being found).
Product of roots \(= \alpha\beta\left(\alpha + \dfrac{12}{\alpha} - \beta\right) = 24 \Rightarrow \alpha\beta^2 - (\alpha^2 + 12)\beta + 24 = 0\), substitutes in \(\alpha\) and attempts to solve the quadratic in \(\beta\) to achieve remaining roots. The final M can be gained once three roots in total have been obtained. (This is unlikely to be seen as part of a correct answer.) Allow if \(-24\) has been used for the product.

Notes

M1: Any correct method of finding \(p\). For example, applies the factor theorem, process of finding the pair sum of roots, or uses the roots to form \(\mathrm{f}(z)\).

A1: \(p = 22\) by correct solution only. Note: this can be found using only their complex roots from (a) (e.g. by factor theorem)

Notes

B1: Selects the method of making a connection between \(x\) and \(w\) by writing \(x = w - 3\)

M1: Applies the process of substituting their \(x = aw \pm b\) into \(2x^3 + 6x^2 - 3x + 12\ (= 0)\). So accept e.g. if \(x = \dfrac{w}{3}\) is used.

M1: Depends on having attempted substituting either \(x = w - 3\) or \(x = w + 3\) into the equation. This mark is for manipulating their resulting equation into the form \(pw^3 + qw^2 + rw + s\ (= 0)\) \((p \ne 0)\). The “\(= 0\)” may be implied for this.

A1: At least three of \(p\), \(q\), \(r\) and \(s\) are correct in an equation with integer coefficients. (need not have “\(= 0\)”)

A1: Correct final equation, including “\(= 0\)”. Accept integer multiples.

Alternative (ALT 1)

B1: Selects the method of giving three correct equations each containing \(\alpha\), \(\beta\) and \(\gamma\).

M1: Applies the process of finding sum roots, pair sum and product.

M1: Applies \(w^3 - (\text{their sum roots})w^2 + (\text{their pair sum})w - (\text{their product})\ (= 0)\). Must be correct identities, but if quoted allow slips in substitution, but the “\(= 0\)” may be implied.

A1: At least three of \(p\), \(q\), \(r\) and \(s\) are correct in an equation with integer coefficients. (need not have “\(= 0\)”)

A1: Correct final equation, including “\(= 0\)”. Accept multiples with integer coefficients.

Note: may use another variable than \(w\) for the first four marks, but the final equation must be in terms of \(w\)

Notes: Do not isw the final two A marks – if subsequent division by 2 occurs then mark the final answer.

Notes

B1: Identifies the correct values for all 3 expressions (can score anywhere). Allow notation such as \(\sum p,\ \sum pq\) for the sum and pair sum.

M1: Uses a correct identity for the sum

A1ft: Correct value (follow through their 2, 4 and 5)

Alternative for part (i)

B1: Obtains the correct cubic in “\(y\)”

M1: Uses a correct method

A1ft: Correct value (follow through their 2, 4 and 5)

Notes

M1: Attempt to expand – must have an expression that involves the sum, pair sum and product

A1: Correct expansion

A1: Correct value

Alternative for part (ii)

M1: Substitutes \(x + 4\) for \(x\) in the given cubic

A1: Calculates the correct constant term

A1: Correct value

Notes

M1: Establishes a correct identity that is in terms of the sum, pair sum and product and substitutes to reach a numerical expression for \(p^3 + q^3 + r^3\)

A1: Correct value

Alternative for part (iii)

Notes

B1: Selects the method of making a connection between \(z\) and \(w\) by writing \(z = \dfrac{w - 1}{2}\)

M1: Applies the process of substituting their \(z = \dfrac{w - 1}{2}\) into \(z^3 - 3z^2 + z + 5 = 0\)

(Allow \(z = 2w + 1\))

M1: Manipulates their equation into the form \(w^3 + pw^2 + qw + r\,(= 0)\) having substituted their \(z\) in terms of \(w\). Note that the “= 0” can be missing for this mark.

A1: At least two of \(p\), \(q\), \(r\) correct. Note that the “= 0” can be missing for this mark.

A1: Fully correct equation including “= 0”

The first 4 marks are available if another letter is used instead of \(w\) but the final answer must be in terms of \(w\).

Alternative (ALT 1)

B1: Selects the method of giving three correct equations containing \(\alpha\), \(\beta\) and \(\gamma\)

M1: Applies the process of finding the new sum, new pair sum, new product

M1: Applies \(w^3 - (\text{new sum})w^2 + (\text{new pair sum})w - (\text{new product})\,(= 0)\)
or identifies \(p\) as –(new sum) \(q\) as (new pair sum) and \(r\) as –(new product)

A1: At least two of \(p\), \(q\), \(r\) correct.

A1: Fully correct equation including “= 0”

The first 4 marks are available if another letter is used instead of \(w\) but the final answer must be in terms of \(w\).

Typical solution

Typical solution

Typical solution

Typical solution

Translation \(\begin{bmatrix} -1 \\ 0 \end{bmatrix}\)

Typical solution

Let \(y = x - 1\)

So the new equation is

So \(\mathrm{p}(x) = x^3 + 3x^2 + 2x - 7\)

Typical solution

Let \(y = 3x\)

Then \(x = \dfrac{y}{3}\)

Typical solution

\(-\dfrac{1}{5}\)

Typical solution

\(\therefore\) the critical values include \(-3\) and 8

Typical solution

Typical solution

Typical solution

Typical solution

Typical solution

The roots of the quadratic are \(\sinh\theta_1\) and \(\sinh\theta_2\)

So \(\sinh\theta_1 + \sinh\theta_2 = 1\)

Typical solution

Typical solution

Typical solution

Typical solution

Substituting in the original equation

Typical solution

(i)

(ii)

Typical solution

Typical solution

positive root only as \(\alpha \gt \beta\)

Typical solution

Let the roots be \(\alpha - D\), \(\alpha\) and \(\alpha + D\)

\(\Sigma\alpha\): \(\quad \alpha - D + \alpha + \alpha + D = -\left(\dfrac{-12}{4}\right)\)

\(\Sigma\alpha\beta\):

Typical solution

\(b \lt 0 \text{ and } c \lt 0 \Rightarrow \alpha \gt 0 \text{ and } \beta \gt 0\)

Typical solution

The coefficient of \(x^2\) is zero \(\therefore \alpha + \beta + \gamma = 0\)

Typical solution

max point \(= (-1, 2)\) and min point \(= (1, -2)\)

Typical solution

(i)

Let \(w = x + 1\)

constant term \(= p + 2\)

(ii)

0 and 2

Typical solution

(i)

But \(\alpha + \beta + \gamma = -k\) and

So

(ii)

But \(\alpha + \beta + \gamma = -k\)

And \(\alpha\beta + \beta\gamma + \gamma\alpha = 0\)

And \(\alpha\beta\gamma = -9\)

So \(\alpha^2\beta^2 + \beta^2\gamma^2 + \gamma^2\alpha^2 = -18k\)

Typical solution

(i)

But \(\alpha + \beta + \gamma = -k\) And \(\alpha\beta + \beta\gamma + \gamma\alpha = 0\)

so \(k = -\dfrac{40}{9}\)

(ii)

So \(s = 232\)

Typical solution

But as \(\alpha\), \(\beta\) and \(\gamma\) are real then each of \((\alpha - \beta)^2\), \((\gamma - \alpha)^2\) and \((\beta - \gamma)^2\) must be non-negative.

AG

Notes

B1: for any two correct

Notes

M1: For expanding \((\alpha\beta + \beta\gamma + \gamma\alpha)^2\) - must be equivalent to nine terms but allow an error in at most two terms only. Or for using the correct identity \(\sum\alpha^2\beta^2 = \left(\sum\alpha\beta\right)^2 - 2\alpha\beta\gamma\sum\alpha\)

A1FT: Correct equation/expression for \(\alpha^2\beta^2 + \beta^2\gamma^2 + \gamma^2\alpha^2\) using their results from part (a)

Notes

B1: correct expansion of \((\alpha + \beta + \gamma)^2\)

B1: www but condone from \(\alpha + \beta + \gamma = -\frac{4}{5}\) stated in part (a) – this mark is independent of the first B1 mark

Notes

M1: For a four term cubic equation/expression with two correct coefficients (not including the cubic term) FT their answers to part (b) and (c). Condone missing = 0. Can use any single unknown, including \(x\)

A1: Must include = 0. Can use any single unknown, including \(x\) but must be integer coefficients
SC B1 for \(x = (\pm)\sqrt{u}\) substitution with correct equation with integer coefficients

Notes

M1: Expressing \(\alpha^2 + \beta^2\) in terms of the symmetric functions
“Using answers to part a”, so some evidence of how these are used is needed

A1: \(-2.75\)

M1: Expressing \(1/\alpha + 1/\beta\) in terms of the symmetric functions

A1: Exact equivalent only \((-0.6)\)

Notes

M1: Attempt to find the three Vieta's formulae for the original equation.
This mark can be awarded if 2 of the 3 are correct and the 3^rd has been attempted or if all are correct but for signs.

B1: Showing that the new product of roots is the same as the original.

M1: Both considered and at least one correctly written in desired form using Vieta's expressions
May be embedded in expressions for \(\sum\alpha'\beta'\) or \(\sum\alpha'\)

B1: For correct \(\sum\alpha'\beta'\) in terms of Vieta's expressions i.e. \(\sum\alpha'\beta' = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha)\)
Can be awarded before numerical value found
Correct numerical value implies correct \(\sum\alpha'\beta'\) (but must be identified as \(\sum\alpha'\beta'\))

B1: For correct \(\sum\alpha'\) in terms of Vieta's expressions i.e. \(\sum\alpha' = \dfrac{(\alpha\beta + \beta\gamma + \gamma\alpha)^2 - 2\alpha\beta\gamma(\alpha + \beta + \gamma)}{\alpha\beta\gamma}\)
Can be awarded before numerical value found
Correct numerical value implies correct \(\sum\alpha'\) (but must be identified as \(\sum\alpha'\))

A1: Any non-zero integer multiple. Must be a cubic equation (i.e. it must have “=0”) with integer coefficients but can be in any unknown.
Correct answer from finding roots on calculator only scores 0/6.

Notes

M1: For \(\alpha\beta\gamma = -\left(-\tfrac{3}{2}\right)\).

M1: Appropriate substitution of the form \(w = (\alpha\beta\gamma)x\) with their value of \(\alpha\beta\gamma\) (need not be substituted into given cubic for this mark). Condone reciprocal e.g. \(x = \tfrac{3}{2}w\).

A1: Allow any integer multiple but must have integer coefficients. Must be an equation (so must = 0 or terms on both sides of an equation) with all terms simplified. Condone if in terms of \(x\).

Alternative method

M1: For at least one of \(\alpha\beta\gamma, \alpha + \beta + \gamma, \alpha\beta + \beta\gamma + \gamma\alpha\) correct. Can be implied by one correct coefficient in new cubic equation.

M1: For at least two of \((\alpha\beta\gamma)^4\), \((\alpha\beta\gamma)^2(\alpha\beta + \beta\gamma + \gamma\alpha), \alpha\beta\gamma(\alpha + \beta + \gamma)\) correct – not from incorrect values of \(\alpha\beta\gamma, \alpha + \beta + \gamma, \alpha\beta + \beta\gamma + \gamma\alpha\).

A1: Allow any integer multiple but must have integer coefficients. Must be an equation (so must = 0 or terms on both sides of an equation) with all terms simplified. Condone if in terms of \(x\).

Notes

M1: Rewrite the expression in terms of the standard symmetric functions
Need to see \(\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta\) oe
No need to see \(\alpha^2\beta^2 = (\alpha\beta)^2\)

A1: cao
Accept eg \(-0.7708\dot{3}\) but not rounded, incorrect or incomplete decimal form.
SC B1 for correct answer if B0B0M0
Accept any equivalent fraction

Notes

M1: Substitute \(z = w - 1\)

M1: Expanding with at least \((w - 1)^4\) seen

A1: Convincingly shown AG. Must include \(= 0\) on last line. Brackets must be fully expanded in working or evidence of collection of like terms
A0 if variable used is not \(w\).

Alternative method

M1: Substitute \(w = z + 1\) into end result

Alternative method using symmetry of roots

B1: For sums from original equation and finding the sum of the new roots

B1: For showing convincingly the sum of new roots in pairs

B1: For the last two

Notes

M1: Solving quadratic equation in \(w^2\) (or using their variable)

M1: Square rooting their \(w^2\), including \(\pm\), as long as their \(w^2\) not both non-negative and real.

A1: cao
Answers with no working is 0

Notes

B1: Correct substitution stated or used

M1: Attempt to expand their \((u - 2)^3\) following a linear substitution. 4 terms using \(\binom{n}{r}2^ru^{n-r}\)
\(u^3 - 6u^2 + 12u - 8\)
Binomial coefficients might not be correct or evaluated. Might be by expanding brackets.

M1: Attempt to expand their \((u - 2)^4\) following a linear substitution. 5 terms using \(\binom{n}{r}2^ru^{n-r}\)
\(u^4 - 8u^3 + 24u^2 - 32u + 16\)
Binomial coefficients might not be correct or evaluated. Might be by expanding brackets.

M1: Forming (LHS of) equation in \(u\) (could be using their expansions)

A1: Final answer can be in terms of \(x\)
Final answer needs to be an equation with coefficients simplified.
ISW attempts to cancel down following correct equation

Notes

B1: Or \((z - 1)\left(z - \mathrm{e}^{\frac{2}{3}\pi\mathrm{i}}\right)\left(z - \mathrm{e}^{\frac{4}{3}\pi\mathrm{i}}\right)(= 0)\) Or all three values stated

B1: Or one of \(M = \frac{1}{2}\mathrm{e}^{\pi\mathrm{i}}, L = \frac{1}{2}\mathrm{e}^{\frac{1}{3}\pi\mathrm{i}}, N = \frac{1}{2}\mathrm{e}^{-\frac{1}{3}\pi\mathrm{i}}\)

M1: Attempt at polynomial relating to \(LMN\).
\(\left(z - \frac{1}{2}\mathrm{e}^{\pi\mathrm{i}}\right)\left(z - \frac{1}{2}\mathrm{e}^{\frac{1}{3}\pi\mathrm{i}}\right)\left(z - \frac{1}{2}\mathrm{e}^{-\frac{1}{3}\pi\mathrm{i}}\right)\) suffices for this mark.

M1: Attempt product of their two cubic factors.

A1: A0 without justification of \(8z^3 + 1 = 0\). Must see \(= 0\).

Alternative method

B1: Finding A, B, C

B1: Finding L, M, N

M1: Combining to give a 6^th degree polynomial

M1: Multiplying out some terms to give quadratics or cubics

A1: Must see = 0

Alternative methods

By rotational symmetry Because \(\left(\mathrm{e}^{\frac{2}{3}\pi\mathrm{i}}\right)^3 = 1\)

Calculation of vertices of \(L\), \(M\) and \(N\). Use of \(1 + \omega + \omega^2 = 0\) (and \(\omega^3 = 1\)) to simplify eg \((z - \frac{1}{2}(\omega + 1))(z - \frac{1}{2}(\omega^2 + 1))\ ))(z - \frac{1}{2}(\omega + \omega^2))\) \(= (z + \frac{1}{2}\omega^2))(z + \frac{1}{2}\omega)\ ))(z + \frac{1}{2})\) etc or to find sum/product of roots.

Notes

2nd B1: (ie sum of new roots \(= -3\)) or \(\dfrac{b}{a} = 3\)

M1: Correctly writing either \(\alpha'\beta'\gamma'\) or \(\Sigma\alpha'\beta'\) in a form involving the basic symmetrical forms

A1: Must be an equation with integer coefficients. Allow use of any unknown.

Alternative method

B1: Can be implied by 2^nd B1.

M1: Rearrange and substitute into original equation

A3: for fully correct equation in the correct form.
Award A1 only for coefficients of \(u^3\) and one other correct in any form
Award A2 only for coefficients of \(u^3\) and two others correct in any form

Notes

B1: Quantity must be either identified as, or used as, sum of new roots
NB for reference: \(\sum\alpha = \dfrac{3}{5},\ \sum\alpha\beta = -\dfrac{2}{5},\ \alpha\beta\gamma = -\dfrac{9}{5}\)

M1: (1st) For expressing product of new roots in terms of old roots

A1: (1st) For expressing product of new roots in terms of old roots
Condone \((9/5)^2\) if seen. Do not condone \(-9/5^2\) or \(-(9/5)^2\) unless recovered

M1: (2nd) Finding sum of products and rewriting into symmetric form

A1: (3rd) Or any non-zero integer multiple
Needs to be an equation.

Alternative method

B1: (2nd) SOI

Notes

M1: Attempt to obtain identity and substitute
Condone missing 2 and sign errors

Notes

M1: For the substitution
“=0” not necessary here but needed for A1

A1: Allow in terms of \(z\). Allow ft from their \(k\) in (b)

Alternate method

M1: For calculating the sum, product and sum of product of pairs of reciprocals of \(\alpha, \beta, \gamma\)

Notes

M1: (1st) Attempt to expand using binomial. 4 terms.
Follow through on their \(u = x + 1\)

M1: (2nd) Substituting into equation. Allow if no “= 0” here. Must have an attempt at expanding \((u - 1)^3\) and \((u - 1)^2\)
Follow through on their \(u = x + 1\)

A1: Must be an equation
For correct equation found using sums and products of roots allow SC2 (Method required was dictated in question)
Only allocate marks using main scheme, or SC method

Notes

M1: Use of \(\alpha\beta\gamma = \dfrac{3}{2}\) to find the 3rd root.
Alternatively, find \(x^2 - 2x + 3\) and divide

Notes

M1: Multiply out (can be seen in (b))

Alternative method

M1: Use of symmetry forms for roots

(corrected from the printed mark scheme: the first line is printed as \(\alpha + \beta + \gamma = -\dfrac{a}{2}\); the coefficient is \(p\))

Notes

M1: Either \(\alpha\) or \(\beta\) seen in mod/arg form

A1: For both of them – accept exponentials

M1: Derivation of \(\alpha^n\) or \(\beta^n\)

Notes

B1: For at least 2 correct

B1ft: \(2 \times\) their \(\Sigma\alpha\)
\(A\), \(B\) and \(\Gamma\) are the roots of the new equation

M1: (1st) Attempting to find the new \(\Sigma\alpha\beta\) and expanding to a symmetrical form

M1: (2nd) Opening brackets convincingly and writing in symmetrical form
Other correct forms are possible eg \(\alpha^2\beta + \alpha^2\gamma + \beta^2\alpha + \beta^2\gamma + \gamma^2\alpha + \gamma^2\beta + 2\alpha\beta\gamma\)

A1: (3rd) Or any non-zero integer multiple
Must be integer coefficients

Alternative method

B1*: Used or stated

B1dep*: Might occur before the first B1 in candidates working
Note only the sum of roots needed here

B1**: Show that when \(x\) is one of the roots of the original equation, \(u\) is one of the roots of the new equation

B1 dep**: Show that \(x = -\dfrac{3}{5} - u\) gives the other two required roots

M1: Use substitution \(x = -\dfrac{3}{5} - u\) and expand \(\left(-\dfrac{3}{5} - u\right)^3\) and \(\left(-\dfrac{3}{5} - u\right)^2\)
\(\left(-\dfrac{3}{5} - u\right)^3 = -\left(u^3 + \dfrac{9}{5}u^2 + \dfrac{27}{25}u + \dfrac{27}{125}\right)\)

A1: (2nd) Or any non-zero integer multiple
Must be integer coefficients. Must have “=0”

Notes

B1: Correct substitution chosen

M1: Oe. Attempting to make substitution
or preparation for substitution by removing odd powers.
eg \(x^2(3x^2 - 5)^2 = (2x^2 + 4)^2\)…

M1: Rearranging and squaring bs to remove the square root(s)
…and then substituting \(u(3u - 5)^2 = (2u + 4)^2\)

A1: Rearranging to answer. Equation can be in \(x\)

Alternative method

B1: Must include one intermediate step

M1: Writing the expression in terms of standard symmetrical forms
NB \(\sum\alpha = \frac{2}{3},\ \sum\alpha\beta = -\frac{5}{3},\ \alpha\beta\gamma = \frac{4}{3}\)

M1: Writing the expression in terms of standard symmetrical forms
Condone without factorisation of “2”

A1: Substituting in and rearranging to answer
NB \(\sum\alpha^2 = \frac{34}{9},\ \sum\alpha^2\beta^2 = 1\)

Notes

M1: Their \(\alpha^2\beta^2 + \beta^2\gamma^2 + \gamma^2\alpha^2\) from part (a) over \(\pm\dfrac{4}{3}\). Strict ft

Notes

DR: Detailed reasoning required for this question
See appendix for Special Cases for this question (below).

M1: (1st) Any one of \(\alpha\beta\gamma\), \(\alpha\beta + \beta\gamma + \gamma\alpha\), \(\alpha + \beta + \gamma\)

A1: (1st) All three correct

M1: (2nd) Expansion showing or implying 6 terms including squares and cross-terms, symmetrical in \(\alpha\), \(\beta\) and \(\gamma\). Might have missing 2

A1: (2nd) Correct, useful form
Could see early substitution (see alternative scheme)

Alternate

B1: Soi by correct substitution – must be used

M1*: (1st) Substitution and dealing with \(\left(\sqrt{u}\right)^3\) term

M1*: (2nd) Rearrangement with \(\sqrt{u}\) terms collected on one side and squaring. LHS does not need to be factorised before squaring

A1: Correct expansion and rearranging. Could be in terms of “\(u\)”. Or any non-zero integer multiple. Must be an equation.
Must see the correct entire cubic for this mark if using this method.

A1ft dep(*): Correct coefficient ratio identified. Must gain method marks and have a cubic.
Cubic might not be completely correct but coefficients of \(x^3\) and \(x\) must be.

Appendix: Special cases for question 5 using the substitution method

Working is shown in part (b) and no working shown in part (a).

1. Working in part (b) is fully correct. Then award:
5/5 if a reference to the working in part (b) is made, the correct cubic equation is written down, there is a comment stating that the roots of the new cubic are \(\alpha^2, \beta^2, \gamma^2\) and the correct answer is given.
4/5 as above but no comment about the roots of the new cubic equation being \(\alpha^2, \beta^2, \gamma^2\) or there is no reference to the working in part (b) (i.e. one element of explanation is missing).
3/5 if the correct cubic equation is written down and correct answer of \(\dfrac{13}{25}\)
3/5 if a reference is made to working in part (b) but the cubic equation is not re-written and no comment about the roots being \(\alpha^2, \beta^2, \gamma^2\) (and correct answer of \(\dfrac{13}{25}\) is given)

2. Working in part (b) is not fully correct.
3/5 if a reference to the working in part (b) is made, the same cubic equation from part (b) is written down, there is a comment stating that the roots of the new cubic are \(\alpha^2, \beta^2, \gamma^2\) and the correct follow through coefficient ratio is given.
2/5 as above but no comment about the roots of the new cubic equation being \(\alpha^2, \beta^2, \gamma^2\) or there is no reference to the working in part (b) (i.e. one element of explanation is missing).
1/5 if same cubic equation as in part (b) is written down and the correct follow through coefficient ratio is given.

3. If \(\dfrac{13}{25}\) appears as an answer in part (a) with no supporting working or comments then 0/5

Notes

See appendix for Special Cases for this question (below).

M1: Expansion showing or implying 6 terms including squares and cross-terms, symmetrical in \(\alpha\), \(\beta\) and \(\gamma\). Might have 2 missing.
\(\sum\alpha^2 = \left(\tfrac{2}{5}\right)^2 - 2 \times \tfrac{3}{5}\)

A1: (1st) Seen or implied

A1: (2nd) Or any non-zero integer multiple. Must be = 0. CAO

Alternate

B1: Soi by correct substitution - must be used

M1: (1st) Substitution and dealing with \(\left(\sqrt{u}\right)^3\) term

M1: (2nd) Rearrangement with \(\sqrt{u}\) terms collected on one side and squaring. LHS does not need to be factorised before squaring

A1: Correct expansion and rearranging. Could be in terms of “\(u\)”. Or any non-zero integer multiple. Must be = 0. CAO

Appendix: Special cases for question 5 using the substitution method

Working is shown in part (a). If working shown in part (b) mark as mark scheme above. If little or no working then mark part (b) as:

1. Working in (a) is fully correct. Then award:
4/4 If correct cubic equation is re-written and there is a reference to the working in part (a) such as “See working above” or arrows etc.
3/4 If correct cubic equation re-written and there is no reference to the working in part (a)
0/4 If an incorrect cubic is written down, or “=0” missing

2. Working in (a) is not fully correct. Then award:
0/4 if B1 M1 M1 is not awarded in part (a) (i.e. at least three marks given for (a))
3/4 if at least three marks awarded in part (a) and reference is made to the working in part (a) and the same cubic as in part (a) is re-written in part (b) with and “=0”
2/4 if at least three marks awarded in part (a) and the same cubic as in part (a) is re-written in part (b) with and “=0” but no reference is made to the working in part (a)

Notes

M1: Attempt both sum and product of new roots in terms of original roots.
DR so finding roots M0

A1: For one of 12/5 or 36/5

A1: For both, correctly interpreted as quadratic.

Notes

B1: SOI
Other letters can be used as the variable. If “\(x\)” used allow B1 for sight of \(2\left(\frac{1}{x}\right)^3 + 3\left(\frac{1}{x}\right)^2 - 5\left(\frac{1}{x}\right) + 4\ (= 0)\)

M1: For substituting \(\dfrac{1}{u}\) into the given equation and attempting to multiply by \(u^3\)
If no \(u^3\) outside brackets then need to see at least two terms multiplied by \(u^3\).

A1: Or multiple of this (with integer coefficients). Condone \(x\) as variable. Must have “\(= 0\)”

Alternative

M1: (1st) Allow one sign slip

M1: (2nd) At least 2 correct

A1: oe Must be integer coefficients
Must have “= 0”

Notes

M1: sum of roots, allow a sign error. No missing terms except for those which have cancelled.

A1: any correct quadratic equation e.g. \(\alpha^2 + 2 = \alpha\); soi by correct \(\alpha\)

M1: product of roots, allow a sign error. No missing terms except for those which have cancelled.

M1: using product of roots two at a time, soi by correct numerical work. Allow only a sign error on RHS, not on \(c\). No missing terms except for those which have cancelled. Condone missing brackets. Or for substituting a root correctly into equation if \(d\) already found.

M1: using product of roots three at a time, soi by correct numerical work. Allow only a sign error on RHS, not on \(d\). No missing terms except for those which have cancelled. Condone missing brackets. Or for substituting a root correctly into equation if \(c\) already found.

Alternative method 1

M1 expanding; M1 comparing coefft of \(x^3\) and constants

(corrected from the printed mark scheme: the roots of \(z^2 - z + 2 = 0\) are printed as \(\frac{1}{2} \pm \frac{7}{2}\mathrm{i}\); they are \(\frac{1}{2} \pm \frac{\sqrt{7}}{2}\mathrm{i}\), as in the main scheme)

Alternative method 2

M1: sum of roots, allow a sign error

A1: any correct quadratic equation e.g. \(\alpha^2 + 2 = \alpha\)

M1: product of roots, allow a sign error

M1: at least one quadratic factor identified

M1: multiplying their quadratic factors

Alternative method 3

M1: sum of roots, allow a sign error

A1: Any correct quadratic equation e.g. \(\alpha^2 + 2 = \alpha\)

M1: product of roots, allow a sign error

M1: substituting any root into equation correctly

M1: equating real and imaginary parts leading to two equations

Notes

B1: Allow: to form a triangle, only one of the roots is real

A1: Allow a correct diagram to illustrate symmetry about real axis

Notes

M1: Allow M1 for sum of roots \(= -4\)

M1: Forming equation in \(a\) and \(b\) using given area of the triangle

Alternative solution

M1: Long division (oe) to get \(z^2 - 6z \ldots\)

M1: Solving quadratic

A1: roots \(3 \pm \mathrm{i}k\) or Re(roots) \(= 3\)

M1: or \(\frac{1}{2}(3 + 2).2k = 10 \Rightarrow k = 2\)

Notes

M1: correct substitution chosen and rearranged, condone a rearrangement slip

M1: substitution made, condone one slip only

B1: correctly expanding cubic, soi. Do not ft.

A1: must be an equation with integer coefficients

Alternative method

B1: Soi

M1: using sum and products to correctly find at least one of sum of roots, products of pairs or triple product for transformed equation

A1: two of \(\frac{3}{2}, \frac{3}{8}, -\frac{5}{16}\)

A1: must be an equation with integer coefficients

Notes

M1: condone rearrangement slips

M1: substituting for \(x\) in the cubic

A1: expanding \((2y + 1)^3\) correctly

A1: must be an equation

Alternative solution

B1: sum of new roots = 0

M1: attempt to find sum of pairs or product of new roots

A1: either correct (from correct working)

A1: must be an equation

Notes

B1ft: ft their cubic in \(y\) (but not solved BC)

(Corrected from the printed mark scheme: the list of roots is printed as “1. 1 + 4i, 1 − 4i”; the full stop is a comma.)

Alternative solution

M1: factorising by inspection or long division

Notes

B1: May be embedded

B1: May be embedded

M1: oe Accept numerical values in terms of their 1 and 4, i.e. \(1^2 - 2 \times 4\). No slips allowed.

Alternative method

M1: Correct substitution chosen and attempted

M1: Rearranging and squaring both sides to remove square root

A1: Correct equation

Notes

B1: (2nd) or \(\alpha\beta + 2\alpha + 2\beta = 7\) or \(\alpha\beta + 2(\alpha + \beta) = 7\)

M1: (1st) substitution which could lead to a quadratic in \(\alpha\) or \(\beta\) or \(x\)

A1: (1st) could be in \(\alpha\) or \(\beta\)

M1: (2nd) a method to solve their quadratic

Notes

M1: (1st) for combining fractions correctly

M1: (2nd) \(\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta\) (soi)

A1: must be simplified (accept \(\frac{k - 4}{2}\))

Alternative solution

B1: by formula or completing the square

M1: combining fractions

A1: (1st) expanding correctly

A1: (2nd) must be simplified (accept \(\frac{k - 4}{2}\))

Notes

M1: Solving their quadratic to find \(\alpha\)

Notes

M1: Allow in terms of \(\alpha\) and \(\beta\), no missing terms

M1: Allow in terms of \(\alpha\) and \(\beta\), no missing terms

Alternative solution

Alternative solution

Notes

M1: (1st) product of roots = −3 (condone 3 for M1)

M1: (2nd) sum of roots = −2 (condone 2 for M1)

M1: (3rd) getting quadratic in \(\alpha\) or \(k = -2\) found \(\Rightarrow z^2 - z + 1\) by factorising

A1: (2nd) \(\Rightarrow z = \dfrac{1}{2} + \dfrac{\sqrt{3}}{2}\mathrm{i}\) or \(\dfrac{1}{2} - \dfrac{\sqrt{3}}{2}\mathrm{i}\)

[ … ] not required for final A1

(corrected from the printed mark scheme: the last line printed only “so roots are \(\frac{1}{2} + \frac{\sqrt{3}}{2}\mathrm{i}\) and \(\frac{1}{2} - \frac{\sqrt{3}}{2}\mathrm{i}\)”, leaving out the third root \(\beta = -3\) found above)

Notes

Alternative solution

(Corrected from the printed mark scheme: the expansion in the alternative solution is printed as \(x^4 + \left(\beta + \frac{1}{\beta}\right)x^3 + \left(1 - \alpha^2\right)x^2 - \alpha^2\left(\beta + \frac{1}{\beta}\right)x - \alpha^2\); the signs of the \(x^3\) and \(x\) terms are corrected above.)

Notes

M1: Unsimplified

M1: simplified

Alternative solution

A1: Award for either \(x^2\) or \(x\) coefficient correct

Notes

A1: (2nd) or \(5\beta^2 - 10\beta + 25 = 0\)

M1: (3rd) Or equivalent use of formula

A1: (3rd) SCB1 if ow

Alternative solution

M1: (3rd) Factorising

M1: (4th) Solving the resulting quadratic

Notes

B1: soi

M1: substituting their \((3 - y)/2\) for \(x\)

Alternative solution 1

M1: Attempt to find sum and product of new roots

Alternative solution 2

B1: Any exact form

Notes

M1: substituting for \(x\) (not \(2y - 1\))

Alternative method

B1: or by expanding \((x - 2\alpha + 1)(x - 2\beta + 1)(x - 2\gamma + 1)\); must expand fully for first B1

B1ft: must be an equation

Notes

M1: or completing the square
no working M0, allow 1 slip

B1: (2nd) or \(2\left(\cos\dfrac{5\pi}{3} + \mathrm{i}\sin\dfrac{5\pi}{3}\right)\)

Notes

Alternative solution

M1: attempt to substitute into \(x^3 + \lambda = 0\)
no working M1A0A0

A1: (1st) oe (eg \(\beta\) a root as complex conjugate)

Alternative solution

M1: condone \(\alpha\beta\gamma = \lambda\)

A1: cao
condone no check of coefft of \(x\)

Notes

M1: product of roots used

A1: \(\beta = 2\)

M1: sum of roots used; or \((x - 2)(x^2 + x + 1) = 0\) M1A1

A1: or equivalent quadratic (with \(\beta = 2\))

M1: solving their quadratic; \(x = \dfrac{-1 \pm \sqrt{-3}}{2} = -\dfrac{1}{2} \pm \dfrac{\sqrt{3}}{2}\mathrm{i}\)

Notes

M1: \(k\) = product of root pairs; or \((x - 2)(x^2 + x + 1) \Rightarrow k = -1\)

A1: or by direct substitution; or by factor theorem

Notes

M1: subst \(x = \dfrac{y + 3}{2}\)
or \(x = (1 \pm \sqrt{23}\mathrm{i})/6\)

A1: (1st) expanding bracket correctly
new roots \((-8 \pm \sqrt{23}\mathrm{i})/3\)

A1: (2nd) must have ‘\(=0\)’
B3 if fully correct

Alternative solution

M1: attempting sum and product of \(2\alpha - 3\) and \(2\beta - 3\)

A1: (1st) sum = −16/3, product = 29/3

A1: (2nd) must have ‘\(= 0\)’

Notes

B1: soi

B1ft: (1st) ft their \(2 + \mathrm{i}\)

B1cao: must include ‘\(= 0\)’
any variable, e.g. \(x\)

Alternative

M1: (1st) \((z - 2 + \mathrm{i})(z - 2 - \mathrm{i})\)

A1: (1st) \(= z^2 - 4z + 5\)

M1: (2nd) their \((z^2 - 4z + 5) \times (z - 3)\)