Roots of Polynomials

One of the most elegant ideas in Further Mathematics is that you can know a great deal about the roots of a polynomial without ever solving it. The coefficients already encode the sum of the roots, the product of the roots, and every other symmetric function of them. The bridge between coefficients and roots is a set of identities usually called Vieta's formulae, and once you can move fluently in both directions — coefficients $\to$ symmetric functions, and symmetric functions $\to$ a new polynomial — a whole family of exam questions opens up.

This lesson builds that fluency from first principles: where the formulae come from, how to combine them to reach $\sum \alpha^2$ , $\sum \alpha^3$ , $\sum \tfrac{1}{\alpha}$ and the like, and how to construct the polynomial whose roots are a transformation of the original ones.

Where this sits in AQA 7367

This is the roots of polynomials / symmetric functions strand of the compulsory pure content examined on Paper 1 and Paper 2 (each 2 hours, 100 marks, 33⅓% of the A-Level). It builds directly on GCSE/A-Level factor and remainder work and on sum/product of quadratic roots, and it feeds forward into the summation of series and complex numbers strands (where conjugate roots and roots of unity are exactly symmetric-function questions in disguise). The routine evaluation of $\sum \alpha^2$ or the construction of a transformed equation is largely AO1; the "show that" and "hence deduce" framings that examiners hang on this topic carry the AO2 reasoning and communication marks.

Core theory: why the coefficients know the roots

Start with the fact you already trust for quadratics. If $\alpha$ and $\beta$ are the roots of $ax^2 + bx + c = 0$ , then the polynomial factorises as $a(x - \alpha)(x - \beta)$ . Expanding,

$a(x - \alpha)(x - \beta) = a\big(x^2 - (\alpha + \beta)x + \alpha\beta\big) = ax^2 - a(\alpha + \beta)x + a\alpha\beta.$

Comparing this identically with $ax^2 + bx + c$ forces

$-a(\alpha + \beta) = b \;\Rightarrow\; \alpha + \beta = -\frac{b}{a}, \qquad a\alpha\beta = c \;\Rightarrow\; \alpha\beta = \frac{c}{a}.$

The same factor-and-compare argument generalises to any degree. For the cubic $ax^3 + bx^2 + cx + d = 0 = a(x-\alpha)(x-\beta)(x-\gamma)$ , it is worth expanding the product carefully once so the pattern is unmistakable. Multiplying the first two brackets,

$(x-\alpha)(x-\beta) = x^2 - (\alpha+\beta)x + \alpha\beta,$

then multiplying by $(x-\gamma)$ :

$\big[x^2 - (\alpha+\beta)x + \alpha\beta\big](x-\gamma) = x^3 - (\alpha+\beta+\gamma)x^2 + (\alpha\beta+\beta\gamma+\gamma\alpha)x - \alpha\beta\gamma.$

Multiplying through by $a$ and comparing with $ax^3 + bx^2 + cx + d$ gives $\alpha+\beta+\gamma = -\tfrac{b}{a}$ , $\alpha\beta+\beta\gamma+\gamma\alpha = \tfrac{c}{a}$ , $\alpha\beta\gamma = -\tfrac{d}{a}$ . The same expansion, carried one bracket further, yields the quartic relations. The pattern is worth seeing all at once.

Vieta's formulae by degree

Polynomial	$\sum \alpha$	$\sum \alpha\beta$	$\sum \alpha\beta\gamma$	product of all roots
$ax^2 + bx + c$	$-\dfrac{b}{a}$	—	—	$\alpha\beta = \dfrac{c}{a}$
$ax^3 + bx^2 + cx + d$	$-\dfrac{b}{a}$	$\dfrac{c}{a}$	—	$\alpha\beta\gamma = -\dfrac{d}{a}$
$ax^4 + bx^3 + cx^2 + dx + e$	$-\dfrac{b}{a}$	$\dfrac{c}{a}$	$-\dfrac{d}{a}$	$\alpha\beta\gamma\delta = \dfrac{e}{a}$

Here $\sum \alpha\beta$ means the sum of all distinct pairwise products, and $\sum \alpha\beta\gamma$ the sum of all distinct triple products. These are the elementary symmetric polynomials $e_1, e_2, e_3, \ldots$ .

The sign pattern. Reading along a row, the signs alternate $-, +, -, +, \ldots$ . A clean way to remember it: the elementary symmetric function $e_k$ equals $(-1)^k$ times the coefficient of $x^{n-k}$ , divided by the leading coefficient $a$ . So the even symmetric functions (product of pairs, product of all four) come out positive, the odd ones negative.

Building higher symmetric functions

Examiners rarely ask for $\sum\alpha$ directly — they ask for something like $\sum\alpha^2$ , $\sum\tfrac{1}{\alpha}$ or $\sum\alpha^3$ , which you must build from the elementary ones. The three identities below cover almost every case.

Sum of squares. Square the sum and subtract off the cross-terms:

$\sum \alpha^2 = \Big(\sum \alpha\Big)^2 - 2\sum \alpha\beta.$

For two roots this is $\alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta$ ; for three, $\alpha^2+\beta^2+\gamma^2 = (\alpha+\beta+\gamma)^2 - 2(\alpha\beta+\beta\gamma+\gamma\alpha)$ . The reason is simply that $(\sum\alpha)^2$ expands to $\sum\alpha^2 + 2\sum\alpha\beta$ .

Sum of reciprocals. Put the reciprocals over a common denominator:

$\sum \frac{1}{\alpha} = \frac{\sum \alpha\beta}{\alpha\beta\gamma} \quad(\text{cubic}), \qquad \frac{1}{\alpha}+\frac{1}{\beta} = \frac{\alpha+\beta}{\alpha\beta} \quad(\text{quadratic}).$

The numerator is always "one symmetric function up" from the denominator — for a cubic, $e_2$ over $e_3$ .

Sum of cubes. For three roots,

$\sum \alpha^3 = \Big(\sum\alpha\Big)^3 - 3\Big(\sum\alpha\Big)\Big(\sum\alpha\beta\Big) + 3\alpha\beta\gamma.$

For two roots the analogue is $\alpha^3 + \beta^3 = (\alpha+\beta)^3 - 3\alpha\beta(\alpha+\beta)$ . Both follow from expanding $(\sum\alpha)^3$ and re-grouping; you do not need Newton's identities in their full generality at A-Level, but it is worth knowing these are the first two Newton–Girard results.

A fourth identity that exam questions love is the sum of squares of pairwise products, needed whenever you square the roots. For a cubic,

$\sum \alpha^2\beta^2 = (\alpha\beta)^2 + (\beta\gamma)^2 + (\gamma\alpha)^2 = \Big(\sum\alpha\beta\Big)^2 - 2\,\alpha\beta\gamma\sum\alpha.$

This is just the sum-of-squares identity applied to the three quantities $\alpha\beta, \beta\gamma, \gamma\alpha$ : their sum is $\sum\alpha\beta$ , and the cross-terms $\alpha\beta\cdot\beta\gamma + \cdots$ factor as $\alpha\beta\gamma(\alpha+\beta+\gamma)$ . You will use it in the specimen question below to build the equation with roots $\alpha^2, \beta^2, \gamma^2$ . The general principle behind all four identities is the same: any symmetric function of the roots can be rebuilt from the elementary ones $\sum\alpha, \sum\alpha\beta, \alpha\beta\gamma$ — the only skill is knowing which combination to use.

Worked examples (with mark scheme)

Example 1 — squares and reciprocals of a cubic's roots

The equation $x^3 - 6x^2 + 11x - 6 = 0$ has roots $\alpha, \beta, \gamma$ . Find $\alpha^2 + \beta^2 + \gamma^2$ and $\dfrac{1}{\alpha} + \dfrac{1}{\beta} + \dfrac{1}{\gamma}$ .

Write down Vieta's formulae first (here $a = 1$ ):

$\sum\alpha = 6,\qquad \sum\alpha\beta = 11,\qquad \alpha\beta\gamma = 6. \quad (\text{M1: state all three})$

$\sum\alpha^2 = \Big(\sum\alpha\Big)^2 - 2\sum\alpha\beta = 6^2 - 2(11) = 36 - 22 = 14. \quad (\text{M1 apply identity; A1})$

$\sum \frac{1}{\alpha} = \frac{\sum\alpha\beta}{\alpha\beta\gamma} = \frac{11}{6}. \quad (\text{A1})$

(M1 for stating the symmetric functions; M1 for using $\sum\alpha^2 = (\sum\alpha)^2 - 2\sum\alpha\beta$ ; A1 for $14$ ; A1 for $\tfrac{11}{6}$ . As a check, this cubic factorises as $(x-1)(x-2)(x-3)$ , and indeed $1+4+9 = 14$ and $1+\tfrac12+\tfrac13 = \tfrac{11}{6}$ ✓.)

Two habits from this example generalise to every "roots of polynomials" question. First, write the Vieta data down before doing anything else — the line " $\sum\alpha = 6, \sum\alpha\beta = 11, \alpha\beta\gamma = 6$ " earns a method mark in its own right and orients the rest of the work. Second, never try to find the roots: the whole power of the method is that $\sum\alpha^2$ and $\sum\tfrac1\alpha$ come straight from the coefficients, even for cubics that do not factorise nicely. Here the cubic happens to have roots $1, 2, 3$ , which lets us verify the answer, but the method would be identical (and the verification unavailable) for a cubic with irrational or complex roots.

Example 2 — forming a new equation by symmetric functions

The equation $2x^2 - 5x + 1 = 0$ has roots $\alpha, \beta$ . Find the quadratic whose roots are $\alpha^2$ and $\beta^2$ .

$\alpha + \beta = \frac{5}{2}, \qquad \alpha\beta = \frac{1}{2}. \quad (\text{M1})$

New sum of roots:

$\alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta = \frac{25}{4} - 1 = \frac{21}{4}. \quad (\text{M1; A1})$

New product of roots:

$\alpha^2\beta^2 = (\alpha\beta)^2 = \frac{1}{4}. \quad (\text{A1})$

A monic quadratic with these as sum and product is $x^2 - (\text{sum})x + (\text{product}) = 0$ :

$x^2 - \frac{21}{4}x + \frac{1}{4} = 0 \;\Longrightarrow\; 4x^2 - 21x + 1 = 0. \quad (\text{A1 clear fractions})$

(M1 for the original symmetric functions; M1/A1 for the new sum; A1 for the new product; A1 for a correctly cleared final equation. Multiplying through by 4 to remove fractions is good exam practice.)

Example 3 — a transformation via substitution

The cubic $x^3 - 3x + 1 = 0$ has roots $\alpha, \beta, \gamma$ . Find the cubic with roots $2\alpha, 2\beta, 2\gamma$ .

The fastest route is substitution. If the new roots are $y = 2x$ , then $x = \dfrac{y}{2}$ ; substitute into the original equation:

$\left(\frac{y}{2}\right)^3 - 3\left(\frac{y}{2}\right) + 1 = 0 \;\Rightarrow\; \frac{y^3}{8} - \frac{3y}{2} + 1 = 0. \quad (\text{M1 substitute } x = y/2)$

Multiply through by $8$ :

$y^3 - 12y + 8 = 0. \quad (\text{A1})$

(M1 for the correct substitution; A1 for the cleared cubic. Sanity check via symmetric functions: original $\sum\alpha = 0$ , so $\sum 2\alpha = 0$ — matches the absent $y^2$ term; original $\alpha\beta\gamma = -1$ , so $(2\alpha)(2\beta)(2\gamma) = 8(-1) = -8 = -(\text{constant})$ ✓.)

The two routes to a transformed equation — substitution and recomputing symmetric functions — are worth comparing, because the right choice saves time. Substitution ( $x = f^{-1}(y)$ ) is fastest for linear transformations of the roots: scalings $k\alpha$ (substitute $x = y/k$ ) and shifts $\alpha + c$ (substitute $x = y - c$ ), as here and in practice problem P3. The symmetric-functions route is unavoidable for non-linear transformations such as $\alpha^2$ or $\tfrac1\alpha$ , where there is no single substitution that maps each old root to its new one cleanly. A quick test: if the new roots are $g(\alpha)$ for an invertible $g$ , try substitution; otherwise, build the new $\sum, \sum\sum, \prod$ directly. Recognising which tool fits is part of what the examiner is assessing.

Specimen-style exam question

(specimen-style — not from any past paper)

The cubic equation $x^3 + px^2 + qx + r = 0$ has roots $\alpha, \beta, \gamma$ . It is given that $\sum\alpha = 4$ , $\sum\alpha\beta = 1$ and $\alpha\beta\gamma = -6$ . (a) Write down the values of $p$ , $q$ and $r$ . (b) Find $\alpha^2 + \beta^2 + \gamma^2$ . (c) Hence find the cubic equation (with integer coefficients) whose roots are $\alpha^2, \beta^2, \gamma^2$ .

(a) Matching the monic cubic, $p = -\sum\alpha = -4$ , $q = \sum\alpha\beta = 1$ , $r = -\alpha\beta\gamma = 6$ . So the equation is $x^3 - 4x^2 + x + 6 = 0$ .

(b) $\displaystyle \sum\alpha^2 = \Big(\sum\alpha\Big)^2 - 2\sum\alpha\beta = 4^2 - 2(1) = 14.$

(c) For the new roots $\alpha^2, \beta^2, \gamma^2$ we need the three elementary symmetric functions:

$\sum \alpha^2 = 14 \quad(\text{from (b)}),$ $\sum \alpha^2\beta^2 = \Big(\sum\alpha\beta\Big)^2 - 2\,\alpha\beta\gamma\sum\alpha = 1^2 - 2(-6)(4) = 1 + 48 = 49,$ $\alpha^2\beta^2\gamma^2 = (\alpha\beta\gamma)^2 = (-6)^2 = 36.$

So the required cubic is $y^3 - (\sum\alpha^2)y^2 + (\sum\alpha^2\beta^2)y - \alpha^2\beta^2\gamma^2 = 0$ , namely

$y^3 - 14y^2 + 49y - 36 = 0.$

The middle identity $\sum\alpha^2\beta^2 = (\sum\alpha\beta)^2 - 2\alpha\beta\gamma\sum\alpha$ is the "sum of squares" identity applied to the pairwise products $\alpha\beta, \beta\gamma, \gamma\alpha$ ; it is the step that separates a top answer from a stalled one.

The structure of part (c) is the canonical "new equation from squared roots" task, and it is worth internalising the recipe: to build the cubic with roots $\alpha^2, \beta^2, \gamma^2$ you need its three elementary symmetric functions, which are $\sum\alpha^2$ (sum), $\sum\alpha^2\beta^2$ (pair-sum) and $\alpha^2\beta^2\gamma^2 = (\alpha\beta\gamma)^2$ (product). Each is obtained from the original symmetric functions by a standard identity — $\sum\alpha^2 = (\sum\alpha)^2 - 2\sum\alpha\beta$ , $\sum\alpha^2\beta^2 = (\sum\alpha\beta)^2 - 2\alpha\beta\gamma\sum\alpha$ , and squaring the product. Assemble them into $y^3 - (\text{sum})y^2 + (\text{pair-sum})y - (\text{product}) = 0$ , minding the alternating signs. Candidates who memorise "squared roots need these three identities" rarely get stuck; those who try to find the roots first almost always do.

Synoptic links

A-Level Maths (sum/product of quadratic roots, factor theorem). Everything here is that idea promoted to higher degree; $\alpha+\beta = -b/a$ and $\alpha\beta = c/a$ are the degree-2 special case.
Complex numbers. For a real-coefficient polynomial, non-real roots occur in conjugate pairs — itself a statement about symmetric functions (the coefficients are real, so symmetric functions of the roots are real). Roots of unity satisfy $z^n - 1 = 0$ , whose symmetric functions you can read straight off: $\sum z_k = 0$ , $\prod z_k = (-1)^{n-1}$ .
Summation of series. Newton's identities link power sums $\sum\alpha^k$ to the elementary symmetric functions, mirroring the standard-results algebra of the next lesson.
Matrices (eigenvalues). The characteristic polynomial's coefficients are symmetric functions of the eigenvalues: $\operatorname{tr}(A) = \sum\lambda$ and $\det(A) = \prod\lambda$ . The same Vieta's machinery reappears there.

Mark-scheme literacy

"Write down" (e.g. the values of the symmetric functions) signals one mark for a stated result — quote $-b/a$ directly without re-deriving it.
"Hence" demands you use the previous part. In the specimen question, part (c) must reuse $\sum\alpha^2 = 14$ from (b); recomputing from scratch can forfeit the "hence" mark.
"Show that" requires every line of working visible and an explicit final statement; a bare correct answer scores the A-mark but loses the M-marks for unseen method.
Follow-through (ft) is common here: if you mis-state one symmetric function but then apply the identities correctly, the method marks can follow through to a "correct-by-your-values" answer.
Give integer coefficients when asked — always clear fractions (multiply through), as in Examples 2 and 3.

Grade-band model answers

Question: "The cubic $x^3 - 6x^2 + 11x - 6 = 0$ has roots $\alpha, \beta, \gamma$ . Show that $\alpha^3 + \beta^3 + \gamma^3 = 36$ ."

Mid-band response. $\sum\alpha = 6,\ \sum\alpha\beta = 11,\ \alpha\beta\gamma = 6.$ $\sum\alpha^3 = 6^3 - 3(6)(11) + 3(6) = 216 - 198 + 18 = 36.$ Examiner-style commentary: Correct symmetric functions and the right identity, reaching $36$ for full marks on the computation. It quotes the cubic-sum identity without justification, which is acceptable here but leaves the candidate exposed if they misremember a coefficient.

Stronger response. As above, but the identity is stated and named first: "Using $\sum\alpha^3 = (\sum\alpha)^3 - 3(\sum\alpha)(\sum\alpha\beta) + 3\alpha\beta\gamma$ …", then the substitution. Examiner-style commentary: Stating the general identity before substituting is the AO2 communication examiners look for in a "show that" — the reader can follow the logic, not just the arithmetic.

Top-band response. As Stronger, plus a verification: "This cubic factorises as $(x-1)(x-2)(x-3)$ , so the roots are $1, 2, 3$ and $1^3 + 2^3 + 3^3 = 1 + 8 + 27 = 36$ , confirming the result." Examiner-style commentary: The independent check via the explicit roots demonstrates genuine command and would catch any slip in the identity. This self-checking habit is what protects accuracy marks across a whole paper.

Common misconceptions

" $\sum\alpha\beta$ is the product of all the roots." No — for a cubic it is the sum of the three pairwise products $\alpha\beta + \beta\gamma + \gamma\alpha$ . The product of all roots is $\alpha\beta\gamma$ .
"The sign of $\sum\alpha$ is $+b/a$ ." It is $-b/a$ . The alternating signs trip many candidates; only the even symmetric functions are positive.
" $\alpha^2 + \beta^2 = (\alpha+\beta)^2$ ." You must subtract the cross-term: $(\alpha+\beta)^2 - 2\alpha\beta$ . Forgetting the $-2\alpha\beta$ is the single most common error in the topic.
"For $2x^2 - 5x + 1$ , $\alpha+\beta = 5$ ." You must divide by the leading coefficient: $\alpha+\beta = -(-5)/2 = \tfrac52$ , not $5$ . Vieta's formulae need the polynomial in the form with $a$ explicit.
"To get the equation with roots $2\alpha,\dots$ , substitute $x = 2y$ ." It is the inverse: new root $y = 2x$ means $x = y/2$ . Substituting the wrong way scales in the wrong direction.
"A repeated root counts once in the symmetric functions." It counts with multiplicity. A double root $\alpha = \beta$ contributes to $\sum\alpha$ twice and to $\alpha\beta$ as $\alpha^2$ .
"You must find the roots to find $\sum\alpha^2$ ." The whole point is that you never solve the polynomial — symmetric functions come straight from the coefficients. This matters most when the roots are irrational or complex and cannot be found neatly; the method still works unchanged.
"The new equation's constant term is $+(\text{product of new roots})$ ." For a monic cubic $y^3 + py^2 + qy + s$ , the constant is $s = -(\text{product})$ — the alternating sign applies to the new equation just as to the old. Reassembling with the wrong sign is a frequent final-step error.

Common errors

Sign slips on the alternating pattern — writing $\alpha\beta\gamma = +d/a$ for a cubic instead of $-d/a$ .
Not dividing by $a$ when the leading coefficient is not 1 (Example 2's $2x^2 - \cdots$ ).
Dropping the $-2\alpha\beta$ (or $-2\sum\alpha\beta$ ) in the sum-of-squares identity.
Substituting the transformation the wrong way round ( $x = 2y$ instead of $x = y/2$ ).
Forgetting to clear fractions, leaving $x^2 - \tfrac{21}{4}x + \tfrac14 = 0$ when integer coefficients were asked for.
Mismatching the constant-term sign when reassembling the new polynomial — the product of roots enters as $-(\text{product})$ for a monic cubic's constant.

Going further (stretch)

The identities you used are the first cases of Newton's identities (the Newton–Girard formulae), which give every power sum $p_k = \sum \alpha^k$ recursively in terms of the elementary symmetric functions $e_1, e_2, \ldots$ . For three roots they read

$p_1 = e_1,\qquad p_2 = e_1 p_1 - 2e_2,\qquad p_3 = e_1 p_2 - e_2 p_1 + 3e_3,$

and in general, for $k \le n$ ,

$p_k = e_1 p_{k-1} - e_2 p_{k-2} + \cdots + (-1)^{k-1} k\, e_k.$

These let you climb to $\sum\alpha^4$ , $\sum\alpha^5$ , $\ldots$ without ever expanding a fourth power by hand — exactly the kind of efficient recursion a STEP or MAT question rewards. A typical Oxbridge-flavoured task: given a cubic with $\sum\alpha = 0$ , show that $p_3 = 3e_3$ and hence that the three roots of $x^3 + px + q = 0$ satisfy $\alpha^3 + \beta^3 + \gamma^3 = -3q$ . (With $e_1 = 0$ , Newton's recursion collapses to $p_3 = 3e_3 = 3(-q) = -3q$ .)

A second deep idea is the Fundamental Theorem of Symmetric Polynomials: every symmetric polynomial in the roots can be written as a polynomial in the elementary symmetric functions $e_1, \ldots, e_n$ — and therefore in the coefficients. This is why questions like "find $\sum\alpha^2\beta^2$ " always have an answer expressible in the coefficients: it is guaranteed in advance, not a happy accident. The theorem even tells you the result will be a polynomial (no division needed) when the symmetric function is itself a polynomial in the roots — which is why $\sum\alpha^2$ comes out polynomial in the coefficients, whereas $\sum\tfrac1\alpha$ (not polynomial in the roots) involves dividing by $\alpha\beta\gamma$ .

A third connection reaches into the rest of the course. The discriminant of a polynomial — the quantity that detects repeated roots — is itself a symmetric function: for a quadratic, $(\alpha - \beta)^2 = (\alpha+\beta)^2 - 4\alpha\beta = \tfrac{b^2 - 4ac}{a^2}$ , recovering the familiar $b^2 - 4ac$ . For a cubic the analogous $(\alpha-\beta)^2(\beta-\gamma)^2(\gamma-\alpha)^2$ is a (more elaborate) symmetric function of the roots, hence expressible in $b, c, d$ ; computing it is a satisfying STEP-style challenge. And the same elementary-symmetric machinery reappears verbatim for matrix eigenvalues, where the characteristic polynomial's coefficients are $\operatorname{tr}(A) = \sum\lambda$ and $\det(A) = \prod\lambda$ — so the algebra you drill here pays off directly in the matrices strand. Recognising one idea (coefficients are symmetric functions of roots) behind quadratics, cubics, discriminants and eigenvalues is exactly the kind of unifying insight that marks out a strong Further-Maths candidate.

Additional A* practice (with worked answers)

P1. The cubic $x^3 + 2x^2 - 5x - 6 = 0$ has roots $\alpha, \beta, \gamma$ . Find $\sum\alpha$ , $\alpha\beta\gamma$ and $\sum\alpha^2$ . Answer: $\sum\alpha = -2$ , $\sum\alpha\beta = -5$ , $\alpha\beta\gamma = 6$ . Then $\sum\alpha^2 = (-2)^2 - 2(-5) = 4 + 10 = 14$ . (It factorises as $(x-2)(x+1)(x+3)$ : $4+1+9 = 14$ ✓.)

P2. The roots of $x^2 - 3x + 1 = 0$ are $\alpha, \beta$ . Find $\alpha^3 + \beta^3$ . Answer: $\alpha+\beta = 3$ , $\alpha\beta = 1$ . $\alpha^3 + \beta^3 = (\alpha+\beta)^3 - 3\alpha\beta(\alpha+\beta) = 27 - 3(1)(3) = 18.$

P3. Given that $\alpha, \beta, \gamma$ are the roots of $x^3 - 4x^2 + x + 6 = 0$ , find the cubic whose roots are $\alpha+1, \beta+1, \gamma+1$ . Answer: New roots $y = x + 1 \Rightarrow x = y - 1$ . Substitute: $(y-1)^3 - 4(y-1)^2 + (y-1) + 6 = 0$ . Expanding, $(y^3 - 3y^2 + 3y - 1) - 4(y^2 - 2y + 1) + (y - 1) + 6 = y^3 - 7y^2 + 12y + 0 = 0$ , i.e. $y^3 - 7y^2 + 12y = 0$ . (Check: original roots are $-1, 2, 3$ ; shifted roots $0, 3, 4$ give $y(y-3)(y-4) = y^3 - 7y^2 + 12y$ ✓.)

P4. The quartic $x^4 - 2x^3 + 3x^2 - 4x + 5 = 0$ has roots $\alpha, \beta, \gamma, \delta$ . Find $\sum\alpha$ and $\sum\alpha^2$ . Answer: $\sum\alpha = -(-2)/1 = 2$ ; $\sum\alpha\beta = 3$ . So $\sum\alpha^2 = (\sum\alpha)^2 - 2\sum\alpha\beta = 4 - 6 = -2$ . (A negative sum of squares is fine — the roots are not all real.)

P5. The cubic $x^3 + px + q = 0$ (no $x^2$ term) has roots $\alpha, \beta, \gamma$ . Show that $\alpha^2 + \beta^2 + \gamma^2 = -2p$ . Answer: Here $\sum\alpha = 0$ (no $x^2$ term) and $\sum\alpha\beta = p$ . Hence $\sum\alpha^2 = (\sum\alpha)^2 - 2\sum\alpha\beta = 0 - 2p = -2p.$ Note this forces $p \le 0$ for the roots to be all real, since a sum of real squares cannot be negative — a neat consequence worth observing.

Board-alignment footer

Aligned to AQA A-Level Further Mathematics 7367, compulsory pure (Papers 1 & 2): relations between roots and coefficients for quadratics, cubics and quartics, symmetric functions of roots, and forming equations with transformed roots. Directly equivalent to Edexcel (9FM0) Core Pure "roots of polynomials" and OCR (H245) "polynomials and rational functions" — the Vieta machinery is identical across boards.

Visual summary

$\boxed{\;\sum\alpha = -\frac{b}{a},\quad \sum\alpha\beta = \frac{c}{a},\quad \alpha\beta\gamma = -\frac{d}{a};\qquad \sum\alpha^2 = \Big(\sum\alpha\Big)^2 - 2\sum\alpha\beta\;}$

Task	Method
Read symmetric functions	Vieta's formulae from coefficients (mind the $-,+,-$ signs)
$\sum\alpha^2$	$(\sum\alpha)^2 - 2\sum\alpha\beta$
$\sum 1/\alpha$ (cubic)	$\dfrac{\sum\alpha\beta}{\alpha\beta\gamma}$
$\sum\alpha^3$ (cubic)	$(\sum\alpha)^3 - 3(\sum\alpha)(\sum\alpha\beta) + 3\alpha\beta\gamma$
New equation, roots $f(\alpha)$	substitute $x = f^{-1}(y)$ , then clear fractions

Recap. The coefficients of a polynomial are its symmetric functions of roots, up to sign. Master the three Vieta relations and the building identities ( $\sum\alpha^2$ , $\sum 1/\alpha$ , $\sum\alpha^3$ , $\sum\alpha^2\beta^2$ ), and you can answer any "roots of polynomials" question without ever solving the equation. Choose substitution for linear transformations of the roots and the symmetric-functions route for non-linear ones, mind the alternating signs when reassembling a new equation, and verify with a quick numerical check whenever the roots happen to be findable.

Roots of Polynomials (Symmetric Functions of Roots)