4.05a - Roots and coefficients of polynomials | OCR A-Level Further Mathematics A

Read the sums and products of polynomial roots directly from the coefficients, even when the roots themselves are awkward or non-real. You will derive the relationships for quadratics, cubics and quartics, then combine them to find less obvious symmetric expressions without solving an equation.

Why coefficients contain root information

Can the squares of the four roots of $x^4 + 2x^2 + 5 = 0$ add up to a negative number? You will answer this without finding a single root. The key is that coefficients record combinations of roots, rather like a summary that keeps the useful information and leaves out the difficult details.

You already know

A root $\alpha$ of a polynomial gives a factor $x - \alpha$ . A polynomial of degree $n$ has $n$ complex roots when repeated roots are counted with their multiplicity. With real coefficients, non-real roots occur in conjugate pairs.

Start with roots $2$ and $-3$ . A polynomial is monic if its leading coefficient is $1$ . Their monic polynomial is

(x - 2)(x + 3) = x^2 + x - 6

The coefficient of $x$ is the negative of their sum, and the constant is their product. With general roots the same expansion gives

\begin{aligned} a(x - \alpha)(x - \beta) &= a\{x^2 - (\alpha + \beta)x + \alpha\beta\} \\ &= ax^2 + bx + c && \text{compare coefficients} \end{aligned}

This is an identity in $x$ , so matching powers have equal coefficients.

Symmetric expression in the roots

An expression is symmetric if interchanging the names of any two roots leaves it unchanged. For example, $\alpha + \beta$ and $\alpha\beta$ are symmetric in two roots. The expression $\alpha - \beta$ is not. Repeated roots are included once for each occurrence: the roots of $(x - 2)^2$ are $2$ , $2$ , so their sum is $4$ .

Dividing the coefficient comparisons by the leading coefficient removes any overall scaling of the polynomial.

Quadratic root relationships

For $ax^2 + bx + c = 0$ , where $a \ne 0$ , with roots $\alpha$ , $\beta$ , counted with multiplicity:

\alpha + \beta = -\frac{b}{a} \qquad \alpha\beta = \frac{c}{a}

These hold for real and non-real roots.
Not in the formula booklet: learn it.

You can now read the required information without using the quadratic formula.

Worked example 1: a monic quadratic

Question. The roots of $x^2 - 5x + 7 = 0$ are $\alpha$ and $\beta$ . Find their sum and product.

Plan. Compare the coefficients with $x^2 - (\alpha + \beta)x + \alpha\beta$ .

\begin{aligned} -(\alpha + \beta) &= -5 && \text{coefficient of } x \\ \alpha + \beta &= 5 && \text{change both signs} \\ \alpha\beta &= 7 && \text{constant coefficient} \end{aligned}

Why this step? The sum appears with a minus sign in the factor expansion; the product does not.

Answer. $\alpha + \beta = 5$ and $\alpha\beta = 7$ .

Check. The roots $(5 \pm \sqrt{3}\,i)/2$ have sum $5$ and product $(25 + 3)/4 = 7$ , even though neither is real.

Now change only the leading coefficient to $2$ . The coefficients must be compared after allowing for that factor.

Worked example 2: a non-monic quadratic

Question. The roots of $2x^2 - 5x + 7 = 0$ are $\alpha$ and $\beta$ . Find their sum and product.

Plan. Divide the equation by $2$ , then compare its monic coefficients.

\begin{aligned} x^2 - \frac52x + \frac72 &= 0 && \text{divide by } 2 \\ \alpha + \beta &= \frac52 && \text{negative linear coefficient} \\ \alpha\beta &= \frac72 && \text{constant coefficient} \end{aligned}

Why this step? Multiplying an equation by a non-zero constant does not change its roots. The ratios of coefficients carry the root information.

Answer. $\alpha + \beta = \frac52$ and $\alpha\beta = \frac72$ .

Check. $2\{x^2 - \frac52x + \frac72\} = 2x^2 - 5x + 7$ , reconstructing the given polynomial.

The negative sign and the leading coefficient are the two details to retain.

Watch out

Wrong: for $2x^2 - 5x + 7 = 0$ , the sum is $-5$ and the product is $7$ . Right: the sum is $-(-5)/2 = 5/2$ and the product is $7/2$ . Normalise first if you are unsure.

Use the same comparisons with a different leading coefficient.

Sums and products for a cubic

A cubic has three roots, so there are three different symmetric combinations before you start using powers: the sum, the sum of pair products, and the product of all three.

Expand the factors in a way that shows where each coefficient comes from:

\begin{aligned} (x - \alpha)(x - \beta)(x - \gamma) &= x^3 - (\alpha + \beta + \gamma)x^2 \\ &\quad + (\alpha\beta + \beta\gamma + \gamma\alpha)x - \alpha\beta\gamma \end{aligned}

To obtain a term in $x$ , choose a root term from two brackets and $x$ from the third. There are three such choices. Two negative root terms make a positive product. Choosing all three gives a negative product.

Cubic root relationships

For $ax^3 + bx^2 + cx + d = 0$ , where $a \ne 0$ , with roots $\alpha$ , $\beta$ , $\gamma$ , counted with multiplicity:

\begin{aligned} \alpha + \beta + \gamma &= -\frac{b}{a} \\ \alpha\beta + \beta\gamma + \gamma\alpha &= \frac{c}{a} \\ \alpha\beta\gamma &= -\frac{d}{a} \end{aligned}

Not in the formula booklet: learn it.

Notice the alternating signs. They come from the minus signs in the root factors, rather than from a separate mnemonic.

Worked example 3: reading a cubic

Question. The equation $x^3 - 4x^2 + 7x - 6 = 0$ has roots $\alpha$ , $\beta$ , $\gamma$ . Find the sum, the sum of pair products, and the product of the roots.

Plan. Match the monic coefficients with the expanded root factors.

\begin{aligned} \alpha + \beta + \gamma &= -(-4) = 4 && \text{coefficient of } x^2 \\ \alpha\beta + \beta\gamma + \gamma\alpha &= 7 && \text{coefficient of } x \\ \alpha\beta\gamma &= -(-6) = 6 && \text{constant coefficient} \end{aligned}

Why this step? The constant has the opposite sign from the product because three negative root terms were multiplied.

Answer. The three values are $4$ , $7$ and $6$ , respectively.

Check. $(x - 2)(x^2 - 2x + 3) = x^3 - 4x^2 + 7x - 6$ . The root $2$ , together with a pair of sum $2$ and product $3$ , gives total sum $4$ , pair sum $2 \times 2 + 3 = 7$ and product $2 \times 3 = 6$ .

Change only the leading coefficient. All three ratios now change, including the constant ratio.

Worked example 4: keeping the cubic scale

Question. Find the same three symmetric combinations for the roots of $2x^3 - 4x^2 + 7x - 6 = 0$ .

Plan. Identify $a = 2$ , $b = -4$ , $c = 7$ , $d = -6$ , then apply the coefficient ratios.

\begin{aligned} \alpha + \beta + \gamma &= -\frac{-4}{2} = 2 && \text{sum} \\ \alpha\beta + \beta\gamma + \gamma\alpha &= \frac72 && \text{pair products} \\ \alpha\beta\gamma &= -\frac{-6}{2} = 3 && \text{triple product} \end{aligned}

Why this step? The coefficient of the highest power controls the scale of every coefficient comparison.

Answer. $2$ , $7/2$ and $3$ , respectively.

Check. Rebuilding gives $2\{x^3 - 2x^2 + \frac72x - 3\}$ , which is the given polynomial.

You will often see the sum of pair products written using a summation sign. Its meaning is each unordered pair once.

Watch out

Wrong: $\alpha\beta + \alpha\gamma + \beta\alpha + \beta\gamma + \gamma\alpha + \gamma\beta = c/a$ . This counts each pair twice. Right: $\alpha\beta + \alpha\gamma + \beta\gamma = c/a$ .

Write the combinations explicitly while learning the pattern; that makes missing and duplicated terms visible.

The six pairs in a quartic

A quartic continues exactly the same selection pattern: choose one root term, two root terms, three root terms or all four from its four factors.

For roots $\alpha$ , $\beta$ , $\gamma$ , $\delta$ , define the following abbreviations in this lesson:

\begin{aligned} S_1 &= \alpha + \beta + \gamma + \delta \\ S_2 &= \alpha\beta + \alpha\gamma + \alpha\delta + \beta\gamma + \beta\delta + \gamma\delta \\ S_3 &= \alpha\beta\gamma + \alpha\beta\delta + \alpha\gamma\delta + \beta\gamma\delta \\ S_4 &= \alpha\beta\gamma\delta \end{aligned}

There are $4$ single roots, $6$ unordered pairs, $4$ unordered triples and $1$ quadruple. Expanding the root factors gives $x^4 - S_1x^3 + S_2x^2 - S_3x + S_4$ .

Quartic root relationships

For $ax^4 + bx^3 + cx^2 + dx + e = 0$ , with $a \ne 0$ , using the symmetric sums defined above:

S_1 = -\frac ba \qquad S_2 = \frac ca \qquad S_3 = -\frac da \qquad S_4 = \frac ea

Roots are counted with multiplicity. Conversely, these sums give the monic equation

x^4 - S_1x^3 + S_2x^2 - S_3x + S_4 = 0

Not in the formula booklet: learn it.

Building an equation from roots is a useful way to make every sign in that pattern earn its place.

Worked example 5: building a quartic

Question. Find the monic quartic equation with roots $1$ , $2$ , $3$ , $4$ .

Plan. Calculate the four symmetric sums and put them into the quartic pattern.

\begin{aligned} S_1 &= 1 + 2 + 3 + 4 = 10 && \text{four roots} \\ S_2 &= 2 + 3 + 4 + 6 + 8 + 12 = 35 && \text{six pairs} \\ S_3 &= 6 + 8 + 12 + 24 = 50 && \text{four triples} \\ S_4 &= 1 \times 2 \times 3 \times 4 = 24 && \text{all four} \end{aligned}

Why this step? Calculating the sums first separates the root arithmetic from the alternating signs in the equation.

Answer. $x^4 - 10x^3 + 35x^2 - 50x + 24 = 0$ .

Check. $(x - 1)(x - 4)(x - 2)(x - 3) = (x^2 - 5x + 4)(x^2 - 5x + 6)$ , which expands to the answer.

Now change only the root $4$ to $-4$ . Every product containing that root changes sign; the others do not.

Worked example 6: changing one root

Question. Find the monic quartic equation with roots $1$ , $2$ , $3$ , $-4$ .

Plan. Recalculate the sums, keeping the negative root inside each product.

\begin{aligned} S_1 &= 1 + 2 + 3 - 4 = 2 && \text{sum} \\ S_2 &= 2 + 3 - 4 + 6 - 8 - 12 = -13 && \text{six pairs} \\ S_3 &= 6 - 8 - 12 - 24 = -38 && \text{four triples} \\ S_4 &= -24 && \text{product} \end{aligned}

Why this step? The coefficient of $x$ is $-S_3$ , so a negative triple sum produces a positive coefficient.

Answer. $x^4 - 2x^3 - 13x^2 + 38x - 24 = 0$ .

Check. Multiplying $(x - 1)(x + 4) = x^2 + 3x - 4$ by $(x - 2)(x - 3) = x^2 - 5x + 6$ gives this quartic.

The pattern is valid even if roots coincide or are non-real; the factor expansion does not require distinct real numbers.

Watch out

Wrong: using only $\alpha\beta + \beta\gamma + \gamma\delta + \delta\alpha$ as the quartic pair sum. The missing pairs are $\alpha\gamma$ and $\beta\delta$ . Right: list partners of $\alpha$ , then the remaining partners of $\beta$ , then $\gamma\delta$ .

A direct factor expansion is also a valid route to an equation; the root sums explain the coefficients that it produces.

Finding powers and reciprocals without roots

Coefficient relationships become much more powerful when you expand an expression that contains the combination you want.

For any list of roots, squaring their sum gives the sum of their squares plus twice the sum of all pair products. For a cubic, squaring the pair sum gives another useful identity:

\begin{aligned} (\alpha\beta + \beta\gamma + \gamma\alpha)^2 &= \alpha^2\beta^2 + \beta^2\gamma^2 + \gamma^2\alpha^2 \\ &\quad + 2\alpha\beta\gamma(\alpha + \beta + \gamma) \end{aligned}

The mixed products share $\alpha\beta\gamma$ . Factor it out rather than trying to calculate individual roots. Reciprocals have a similar shortcut: put them over the common denominator that is the product of every root.

Derived symmetric expressions

For a cubic, write $S_1 = \alpha + \beta + \gamma$ , $S_2 = \alpha\beta + \beta\gamma + \gamma\alpha$ , $S_3 = \alpha\beta\gamma$ . Then

\begin{aligned} \alpha^2 + \beta^2 + \gamma^2 &= S_1^2 - 2S_2 \\ \alpha^2\beta^2 + \beta^2\gamma^2 + \gamma^2\alpha^2 &= S_2^2 - 2S_1S_3 \\ \frac1\alpha + \frac1\beta + \frac1\gamma &= \frac{S_2}{S_3} \qquad (S_3 \ne 0) \end{aligned}

For a quartic, the square sum is again $S_1^2 - 2S_2$ , and the reciprocal sum is $S_3/S_4$ , provided $S_4 \ne 0$ .
Not in the formula booklet: learn it.

These are consequences of expansions, so you can rebuild them if memory fails. Show the expansion if the question asks you to obtain a result that way.

Worked example 7: three hidden combinations

Question. The roots of $x^3 - 2x^2 - 3x + 5 = 0$ are $\alpha$ , $\beta$ , $\gamma$ . Find the sum of their squares, the sum of squared pair products, and the sum of their reciprocals.

Plan. Read $S_1$ , $S_2$ , $S_3$ , then use expansions and a common denominator.

\begin{aligned} S_1 &= 2 \qquad S_2 = -3 \qquad S_3 = -5 && \text{coefficient relationships} \\ \alpha^2 + \beta^2 + \gamma^2 &= 2^2 - 2(-3) = 10 && \text{square the sum} \\ \alpha^2\beta^2 + \beta^2\gamma^2 + \gamma^2\alpha^2 &= (-3)^2 - 2(2)(-5) = 29 && \text{square the pair sum} \\ \frac1\alpha + \frac1\beta + \frac1\gamma &= \frac{\beta\gamma + \gamma\alpha + \alpha\beta}{\alpha\beta\gamma} = \frac35 && \text{common denominator} \end{aligned}

Why this step? The cross terms are the part you must subtract from the square of a sum; omitting them would confuse a sum of squares with a square of a sum.

Answer. $10$ , $29$ , $3/5$ , respectively.

Check. Reinsert the square sums: $10 + 2(-3) = 4 = S_1^2$ and $29 + 2(2)(-5) = 9 = S_2^2$ .

Change only the constant coefficient from $5$ to $-5$ . The sum and pair sum stay fixed, but the product changes sign. That affects two of the answers.

Worked example 8: a changed constant

Question. Find the same three combinations for the roots of $x^3 - 2x^2 - 3x - 5 = 0$ .

Plan. Keep $S_1 = 2$ , $S_2 = -3$ , and replace $S_3$ by $5$ .

\begin{aligned} \alpha^2 + \beta^2 + \gamma^2 &= 2^2 - 2(-3) = 10 && \text{unchanged first two sums} \\ \alpha^2\beta^2 + \beta^2\gamma^2 + \gamma^2\alpha^2 &= (-3)^2 - 2(2)(5) = -11 && \text{changed product} \\ \frac1\alpha + \frac1\beta + \frac1\gamma &= \frac{-3}{5} && \text{product is non-zero} \end{aligned}

Why this step? A constant coefficient changes the product of the roots, but not their sum or pair sum; identify the affected information before recalculating.

Answer. $10$ , $-11$ , $-3/5$ , respectively.

Check. $-11 + 2(2)(5) = 9 = S_2^2$ . A negative sum of squared products is possible because the roots need not all be real.

The word “square” guarantees a non-negative result for a real number, but not for a complex number: $i^2 = -1$ .

Watch out

Wrong: $\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2$ . Right: subtract $2(\alpha\beta + \beta\gamma + \gamma\alpha)$ . Also check that no root is zero before using reciprocals.

The opening quartic has $S_1 = 0$ , $S_2 = 2$ . Therefore the sum of its four squared roots is $0^2 - 2(2) = -4$ . Its real polynomial is positive for every real $x$ , so none of its roots is real. There is no conflict with the rule for real squares.

Quick Recap

Expand root factors to explain the alternating signs in the coefficient relationships.
Divide by the leading coefficient; multiplying an equation by a non-zero constant does not change its roots.
A quartic has six pair products and four triple products, with repeated roots counted each time.
Obtain new symmetric expressions by expanding a suitable square or using a common denominator.

The four squared roots of $x^4 + 2x^2 + 5 = 0$ sum to $-4$ ; complex squares can be negative.

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4.05a - Roots and coefficients of polynomials

Why coefficients contain root information

Sums and products for a cubic

The six pairs in a quartic

Finding powers and reciprocals without roots

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