NSWMaths Extension 1Syllabus dot point

How do the coefficients of a polynomial already know the sum, the product and every symmetric combination of its roots, so that you can answer questions about the roots without ever finding them?

Relate the coefficients of a polynomial to the elementary symmetric functions of its zeroes (with alternating signs) for quadratics, cubics and quartics, and use these relations to find a missing zero, to evaluate symmetric expressions of the zeroes such as the sum of squares and the sum of reciprocals, to find unknown coefficients from a condition on the zeroes, and to handle zeroes in arithmetic or geometric progression or of a special form

First-contact treatment of sums and products of zeroes for HSC Maths Extension 1. The coefficients are the elementary symmetric functions of the roots with alternating signs, for quadratics, cubics and quartics. Use them to find a missing zero and evaluate symmetric expressions without solving the polynomial.

Generated by Claude Opus 4.819 min answerUpdated 2026-06-21

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Quick answer

The coefficients of a polynomial are the elementary symmetric functions of its zeroes, with the sign alternating. For a quadratic $ax^2 + bx + c$ the sum of the roots is $-\dfrac{b}{a}$ and the product is $\dfrac{c}{a}$ . For a cubic $ax^3 + bx^2 + cx + d$ the sum is $-\dfrac{b}{a}$ , the sum of products in pairs is $\dfrac{c}{a}$ , and the product is $-\dfrac{d}{a}$ . For a quartic the four relations are sum $= -\dfrac{b}{a}$ , pair-sum $= \dfrac{c}{a}$ , triple-sum $= -\dfrac{d}{a}$ and product $= \dfrac{e}{a}$ . These come from expanding the factored form and matching coefficients, so they hold without ever finding the roots. Use them to find a missing zero once one is known (the rest have a known sum and product), to evaluate symmetric expressions such as $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$ and $\dfrac{1}{\alpha} + \dfrac{1}{\beta} = \dfrac{\alpha + \beta}{\alpha\beta}$ by converting them to the sum and product, and to handle roots in arithmetic progression $m - d, m, m + d$ (the sum gives the middle term) or geometric progression $\dfrac{a}{r}, a, ar$ (the product gives the middle term) or of a special form such as $a, -a, b$ (the opposite pair cancels in the sum). Always make the polynomial monic first and keep the alternating sign straight.

What this dot point is asking

Up to now, finding out anything about a polynomial's zeroes meant finding the zeroes themselves: test divisors, divide, factor. This dot point reveals a shortcut that feels almost unfair. The coefficients of a polynomial already contain the sum of its zeroes, the product of its zeroes, and every symmetric combination in between, so a whole class of questions can be answered by reading numbers straight off the polynomial, without solving anything.

NESA groups this under ME-F2 for quadratics, cubics and quartics. You should be able to write down the sum and product of the roots from the coefficients, find a missing root once you know one, and evaluate symmetric expressions of the roots such as $\alpha^2 + \beta^2$ or $\tfrac{1}{\alpha} + \tfrac{1}{\beta}$ . This is the Year 11 first-contact treatment: the relations themselves, derived from scratch, and the standard ways they are used. The Year 12 page on roots and coefficients revisits the same relations (often called Vieta's formulas) and pushes them into transformations of roots and applied work; here the goal is to make the relations second nature and to drill the techniques that turn up most in Year 11 papers.

The answer

Where the relations come from

Everything on this page follows from one fact established by the factor theorem: a polynomial with known zeroes is the product of the matching linear factors, times the leading coefficient. Expanding that product and matching it term by term against the standard form is the entire derivation. Do it once for a quadratic and the pattern is clear.

Let $P(x) = ax^2 + bx + c$ have zeroes $\alpha$ and $\beta$ . By the factor theorem $P(x) = a(x - \alpha)(x - \beta)$ . 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.

Now compare this with $ax^2 + bx + c$ . The coefficient of $x$ must match, so $b = -a(\alpha + \beta)$ , and the constant must match, so $c = a\alpha\beta$ . Solving each for the symmetric function,

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

The coefficients are the sum and product of the roots, up to division by $a$ and a sign on the sum. Nothing here used the values of $\alpha$ and $\beta$ , which is exactly why the relations work without solving for them.

The relations for cubics and quartics

The same expand-and-match works for higher degrees; only the bookkeeping grows. For a cubic $P(x) = ax^3 + bx^2 + cx + d$ with zeroes $\alpha$ , $\beta$ , $\gamma$ , factor and expand:

a(x - \alpha)(x - \beta)(x - \gamma) = ax^3 - a(\alpha + \beta + \gamma)x^2 + a(\alpha\beta + \alpha\gamma + \beta\gamma)x - a\alpha\beta\gamma.

Matching against $ax^3 + bx^2 + cx + d$ gives three relations. A brand-new feature appears at the middle coefficient: it is the sum of the products of the zeroes taken in pairs, not anything to do with the individual zeroes alone.

For a quartic $P(x) = ax^4 + bx^3 + cx^2 + dx + e$ with zeroes $\alpha$ , $\beta$ , $\gamma$ , $\delta$ , the same expansion produces a fourth layer, the sum of the products of the zeroes taken in triples, before the full product appears in the constant term.

Key fact

The sum and product relations. Reading the elementary symmetric functions of the zeroes off the coefficients, the sign alternates at every step (minus, plus, minus, plus).

Quadratic $ax^2 + bx + c$ , zeroes $\alpha, \beta$ :

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

Cubic $ax^3 + bx^2 + cx + d$ , zeroes $\alpha, \beta, \gamma$ :

\alpha + \beta + \gamma = -\frac{b}{a}, \qquad \alpha\beta + \alpha\gamma + \beta\gamma = \frac{c}{a}, \qquad \alpha\beta\gamma = -\frac{d}{a}.

Quartic $ax^4 + bx^3 + cx^2 + dx + e$ , zeroes $\alpha, \beta, \gamma, \delta$ :

\alpha + \beta + \gamma + \delta = -\frac{b}{a}, \qquad \sum_{\text{pairs}} = \frac{c}{a}, \qquad \sum_{\text{triples}} = -\frac{d}{a}, \qquad \alpha\beta\gamma\delta = \frac{e}{a}.

Two things examiners exploit: divide by the leading coefficient $a$ first if the polynomial is not monic, and keep the alternating sign straight (the sum and the triple-sum carry a minus, the pair-sum and the full product a plus).

The pattern is worth saying in words, because the symbols for the quartic get unwieldy. Take each coefficient in turn, from the one just below the leading term, divide it by the leading coefficient, and attach signs that alternate starting with minus: that sequence of numbers is the sum of the zeroes, then the sum of products in pairs, then in triples, and so on down to the full product. The full product of all the zeroes always works out to $(-1)^n$ times the constant over the leading coefficient, which is a plus for even degree and a minus for odd.

Why only two of them usually matter

For a cubic there are three relations and for a quartic four, but most questions lean on the first and the last, the sum and the full product. The reason is practical: the sum and product are the two that "collapse" a structured set of roots. If the roots are evenly spaced the sum pins down the middle one; if they are in a ratio the product pins down the middle one; if a root is the negative of another the opposite pair cancels in the sum. The middle relations (pairs, triples) are then used to finish off or to find a remaining coefficient. Knowing which relation to reach for first is most of the skill.

Finding a missing zero once you know one

The most basic use combines the factor theorem with these relations. Suppose you have found one integer zero of a cubic by testing divisors of the constant term, as on the remainder and factor theorems page. Call it $\gamma$ . The remaining two zeroes $\alpha$ and $\beta$ satisfy

\alpha + \beta = (\text{sum of all three}) - \gamma, \qquad \alpha\beta = \frac{\text{product of all three}}{\gamma},

and any two numbers with a known sum and known product are the roots of a quadratic you can write down and solve, $t^2 - (\alpha + \beta)t + \alpha\beta = 0$ . This recovers the other zeroes with no long division at all. It is the fast alternative to dividing $P(x)$ by $(x - \gamma)$ , and it is the technique the next examples lean on.

Evaluating symmetric expressions without solving

The real power is that any expression in the roots that is symmetric (unchanged if you swap the roots around) can be rewritten in terms of the sum, the product and the pair-sum, then evaluated by substitution. You never find the roots. The two identities that cover almost every Year 11 question are:

\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta, \qquad \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta}.

The first comes from expanding $(\alpha + \beta)^2 = \alpha^2 + 2\alpha\beta + \beta^2$ and moving the cross term across; the second from putting the fractions over the common denominator $\alpha\beta$ . The same ideas scale to three roots:

\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \alpha\gamma + \beta\gamma), \qquad \frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} = \frac{\alpha\beta + \alpha\gamma + \beta\gamma}{\alpha\beta\gamma}.

The sum of squares is "square the sum, subtract twice the pair-sum"; the sum of reciprocals is "pair-sum over product". Recognising a target as symmetric, and knowing the identity that converts it, is the whole method.

Roots in arithmetic or geometric progression

A common dressing-up of these relations is to tell you the roots have a structure. The trick is always to name the roots so the structure builds in the relation for free.

Arithmetic progression: write the three roots as $m - d$ , $m$ , $m + d$ . The two outer terms are symmetric about the middle, so the sum collapses to $3m$ and hands you the middle root straight away. The product then gives $d$ .
Geometric progression: write the three roots as $\dfrac{a}{r}$ , $a$ , $ar$ . The outer terms multiply to $a^2$ , so the product collapses to $a^3$ and hands you the middle root. The sum then gives $r$ .

In both cases one relation isolates the middle root immediately, and a second relation finds the spacing or ratio. This is why the sum and product, the first and last relations, are the ones that matter most.

Roots of a special form

The same naming idea handles a stated relationship between roots. If you are told one root is the negative of another, name the roots $a$ , $-a$ , $b$ : the opposite pair cancels in the sum, so the sum gives $b$ , and the pair-sum reduces to $-a^2$ , giving $a$ . If you are told two roots are equal, name them $a$ , $a$ , $b$ , and the relations become equations in $a$ and $b$ . The rule is the same every time: choose names that encode the condition, then apply the sum, product and pair-sum relations and solve the resulting system.

Exam technique

These questions are quick marks once you name the right relation. To bank them:

Make the polynomial monic in your head first. For $2x^3 + \dots$ the sum is $-\dfrac{b}{2}$ , not $-b$ . Divide every reading by the leading coefficient or you will be out by a factor.
State which relation gives which symmetric function before substituting. Writing "sum $= -\dfrac{b}{a}$ , pair-sum $= \dfrac{c}{a}$ , product $= -\dfrac{d}{a}$ " first catches the alternating-sign slip that loses easy marks.
For a symmetric target, convert with an identity; never solve for the roots. Turn $\alpha^2 + \beta^2$ into $(\alpha + \beta)^2 - 2\alpha\beta$ , or $\dfrac{1}{\alpha} + \dfrac{1}{\beta}$ into $\dfrac{\alpha + \beta}{\alpha\beta}$ , then substitute. Markers reward the identity, not a decimal.
Choose progression names that collapse a relation: $m - d, m, m + d$ for arithmetic (the sum gives $m$ ), or $\dfrac{a}{r}, a, ar$ for geometric (the product gives $a$ ). Find the middle term first, then the spacing or ratio.
Count roots with multiplicity. A repeated root contributes to every relation as many times as its multiplicity; a double root counts twice in the sum.
Finish with a check. Substitute your found roots back into the one relation you did not use; agreement is strong evidence the work is right.

How exam questions ask about sums and products of zeroes

The wordings map cleanly onto the relation to reach for.

"Write down the sum / product of the roots of ..." Read it straight off: $-\dfrac{b}{a}$ for the sum, the constant over $a$ (with the $(-1)^n$ sign) for the product. Make it monic first if needed.
"One root is [given]; find the others." Use sum and product: the remaining roots have a known sum and known product, so they solve a quadratic $t^2 - (\text{sum})t + (\text{product}) = 0$ .
"Find $\alpha^2 + \beta^2$ (or $\tfrac{1}{\alpha} + \tfrac{1}{\beta}$ , or $\tfrac{\alpha}{\beta} + \tfrac{\beta}{\alpha}$ )." The target is symmetric; convert it to the sum, product and pair-sum with an identity, then substitute. Do not find the roots.
"The roots are in arithmetic / geometric progression." Name them $m - d, m, m + d$ or $\dfrac{a}{r}, a, ar$ ; the sum (arithmetic) or product (geometric) gives the middle term, then the other relation gives the spacing or ratio.
"One root is the negative of another" / "two roots are equal" / "the roots are $\alpha, -\alpha, \beta$ ." Name the roots to encode the condition, then solve the system from the three relations.
"Find the value of the unknown coefficient given [a condition on the roots]." Express the condition with the relations, getting an equation in the unknown coefficient, and solve.

Worked example

Find the sum, product and a second root of a quadratic

The quadratic $2x^2 - 5x - 3 = 0$ has roots $\alpha$ and $\beta$ . Find $\alpha + \beta$ and $\alpha\beta$ , and given that $\alpha = 3$ find $\beta$ two ways.

Read off the sum and product. With $a = 2$ , $b = -5$ , $c = -3$ ,

\alpha + \beta = -\frac{b}{a} = -\frac{-5}{2} = \frac{5}{2}, \qquad \alpha\beta = \frac{c}{a} = \frac{-3}{2} = -\frac{3}{2}.

Find the second root from the sum: If $\alpha = 3$ then $3 + \beta = \dfrac{5}{2}$ , so $\beta = \dfrac{5}{2} - 3 = -\dfrac{1}{2}$ .
Find the second root from the product: The product gives $3\beta = -\dfrac{3}{2}$ , so $\beta = -\dfrac{1}{2}$ again. Both relations agree.
Check: The roots $3$ and $-\dfrac{1}{2}$ give $(x - 3)\!\left(x + \tfrac{1}{2}\right) = x^2 - \tfrac{5}{2}x - \tfrac{3}{2}$ , which is $\tfrac{1}{2}(2x^2 - 5x - 3)$ , the original quadratic up to the leading factor. Correct.

Evaluate the sum of squares and reciprocals of a quadratic's roots

For $2x^2 - 5x - 3 = 0$ with roots $\alpha$ and $\beta$ , find $\alpha^2 + \beta^2$ and $\dfrac{1}{\alpha} + \dfrac{1}{\beta}$ without solving the equation.

Carry over the sum and product. From the previous example, $\alpha + \beta = \dfrac{5}{2}$ and $\alpha\beta = -\dfrac{3}{2}$ .

Use the squaring identity for the sum of squares. Since $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$ ,

\alpha^2 + \beta^2 = \left(\frac{5}{2}\right)^2 - 2\!\left(-\frac{3}{2}\right) = \frac{25}{4} + 3 = \frac{37}{4}.

Use the common-denominator identity for the reciprocals. Since $\dfrac{1}{\alpha} + \dfrac{1}{\beta} = \dfrac{\alpha + \beta}{\alpha\beta}$ ,

\frac{1}{\alpha} + \frac{1}{\beta} = \frac{5/2}{-3/2} = -\frac{5}{3}.

State the answers. $\alpha^2 + \beta^2 = \dfrac{37}{4}$ and $\dfrac{1}{\alpha} + \dfrac{1}{\beta} = -\dfrac{5}{3}$ . Neither calculation needed the actual roots, which is the whole point.

State the three relations for a cubic and find the other two zeroes

Given that $x = 2$ is a zero of $P(x) = x^3 - 4x^2 + x + 6$ , write down the three sum-and-product relations and use them to find the other two zeroes.

Write the three relations. With $a = 1$ , $b = -4$ , $c = 1$ , $d = 6$ ,

\alpha + \beta + \gamma = -\frac{b}{a} = 4, \qquad \alpha\beta + \alpha\gamma + \beta\gamma = \frac{c}{a} = 1, \qquad \alpha\beta\gamma = -\frac{d}{a} = -6.

Use the known zero to reduce to a pair: Let the known zero be $\gamma = 2$ and the unknown ones be $\alpha$ and $\beta$ . From the sum, $\alpha + \beta + 2 = 4$ , so $\alpha + \beta = 2$ . From the product, $2\alpha\beta = -6$ , so $\alpha\beta = -3$ .
Solve the quadratic with that sum and product: Two numbers with sum $2$ and product $-3$ are the roots of $t^2 - 2t - 3 = 0$ , that is $(t - 3)(t + 1) = 0$ , so $t = 3$ or $t = -1$ .
State and check: The three zeroes are $2$ , $3$ and $-1$ . Check the middle relation: $\alpha\beta + \alpha\gamma + \beta\gamma = (3)(-1) + (3)(2) + (-1)(2) = -3 + 6 - 2 = 1$ , matching $\dfrac{c}{a}$ . No long division was needed.

Evaluate symmetric expressions of a cubic's roots

The equation $2x^3 + 3x^2 - 11x - 6 = 0$ has roots $\alpha$ , $\beta$ , $\gamma$ . Find $\alpha^2 + \beta^2 + \gamma^2$ and $\dfrac{1}{\alpha} + \dfrac{1}{\beta} + \dfrac{1}{\gamma}$ without solving the equation.

Make it monic and read the relations. Dividing by $2$ , the symmetric functions are

\alpha + \beta + \gamma = -\frac{3}{2}, \qquad \alpha\beta + \alpha\gamma + \beta\gamma = \frac{-11}{2} = -\frac{11}{2}, \qquad \alpha\beta\gamma = -\frac{-6}{2} = 3.

Sum of squares by the identity. Using $\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \alpha\gamma + \beta\gamma)$ ,

\alpha^2 + \beta^2 + \gamma^2 = \left(-\frac{3}{2}\right)^2 - 2\!\left(-\frac{11}{2}\right) = \frac{9}{4} + 11 = \frac{53}{4}.

Sum of reciprocals by the identity. Using $\dfrac{1}{\alpha} + \dfrac{1}{\beta} + \dfrac{1}{\gamma} = \dfrac{\alpha\beta + \alpha\gamma + \beta\gamma}{\alpha\beta\gamma}$ ,

\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} = \frac{-11/2}{3} = -\frac{11}{6}.

State the answers. $\alpha^2 + \beta^2 + \gamma^2 = \dfrac{53}{4}$ and $\dfrac{1}{\alpha} + \dfrac{1}{\beta} + \dfrac{1}{\gamma} = -\dfrac{11}{6}$ . (The roots happen to be $2$ , $-3$ and $-\tfrac{1}{2}$ , but the method never used them.)

State the four relations for a quartic

Write down, in expanded form, the monic quartic whose zeroes are $2$ , $3$ , $5$ and $7$ , and confirm the four sum-and-product relations.

Expand the product of factors. Pairing the factors, $(x - 2)(x - 3) = x^2 - 5x + 6$ and $(x - 5)(x - 7) = x^2 - 12x + 35$ , then

(x^2 - 5x + 6)(x^2 - 12x + 35) = x^4 - 17x^3 + 101x^2 - 247x + 210.

Read the four relations off the coefficients. With $a = 1$ , $b = -17$ , $c = 101$ , $d = -247$ , $e = 210$ ,

\text{sum} = -\frac{b}{a} = 17, \quad \sum_{\text{pairs}} = \frac{c}{a} = 101, \quad \sum_{\text{triples}} = -\frac{d}{a} = 247, \quad \text{product} = \frac{e}{a} = 210.

Confirm each directly from the zeroes. The sum is $2 + 3 + 5 + 7 = 17$ . The product is $2 \times 3 \times 5 \times 7 = 210$ . The sum of products in pairs is $2\cdot3 + 2\cdot5 + 2\cdot7 + 3\cdot5 + 3\cdot7 + 5\cdot7 = 6 + 10 + 14 + 15 + 21 + 35 = 101$ . The sum of products in triples is $2\cdot3\cdot5 + 2\cdot3\cdot7 + 2\cdot5\cdot7 + 3\cdot5\cdot7 = 30 + 42 + 70 + 105 = 247$ .

Conclude. All four match, with the signs alternating minus, plus, minus, plus. This is the four-zero version of the same expand-and-match derivation.

Find roots in arithmetic progression

The zeroes of $P(x) = x^3 - 15x^2 + 66x - 80$ are in arithmetic progression. Find them.

Name the zeroes to use the progression: Write the three zeroes as $m - d$ , $m$ , $m + d$ , so the sum collapses to $3m$ .
Use the sum to find the middle zero: The sum is $-\dfrac{-15}{1} = 15$ , so $3m = 15$ and $m = 5$ . Thus $5$ is a zero.
Use the product to find the spacing: The product is $-\dfrac{-80}{1} = 80$ , and $(m - d)\,m\,(m + d) = m(m^2 - d^2)$ , so

5(25 - d^2) = 80 \implies 25 - d^2 = 16 \implies d^2 = 9 \implies d = 3.

State and check. The zeroes are $5 - 3 = 2$ , $5$ and $5 + 3 = 8$ . Check the pair-sum: $(2)(5) + (2)(8) + (5)(8) = 10 + 16 + 40 = 66$ , matching the coefficient of $x$ .

Find roots of the special form a, -a, b

One zero of $P(x) = x^3 - 2x^2 - 9x + 18$ is the negative of another. Find all three zeroes.

Name the zeroes to encode the condition: Write the zeroes as $a$ , $-a$ , $b$ , so the opposite pair will cancel in the sum.
Use the sum to find $b$: The sum is $-\dfrac{-2}{1} = 2$ , and $a + (-a) + b = b$ , so $b = 2$ .
Use the pair-sum to find $a$: The pairwise products are $a(-a) = -a^2$ , $ab$ and $-ab$ , summing to $-a^2$ . This equals $\dfrac{c}{a_{\text{lead}}} = \dfrac{-9}{1} = -9$ , so

-a^2 = -9 \implies a^2 = 9 \implies a = 3.

State and check. The zeroes are $3$ , $-3$ and $2$ . Check the product: $(3)(-3)(2) = -18$ , which matches $-\dfrac{d}{a_{\text{lead}}} = -18$ . Notice the condition forces $ad = bc$ on the coefficients here too: $a_{\text{lead}}\,d = (1)(18) = 18$ and $b\,c = (-2)(-9) = 18$ , a neat consequence of one zero being the negative of another.

Common traps

Forgetting to divide by the leading coefficient: For $3x^2 + bx + c$ the sum of the roots is $-\dfrac{b}{3}$ and the product is $\dfrac{c}{3}$ , not $-b$ and $c$ . Make the polynomial monic, or divide each reading by $a$ , before doing anything else.
Losing the alternating sign: The sum and the triple-sum carry a minus; the pair-sum and the full product carry a plus. The product of all the roots is $(-1)^n$ times the constant over the leading coefficient: a minus for a cubic, a plus for a quartic. Track the signs explicitly.
Solving for the roots when the target is symmetric: If a question asks for $\alpha^2 + \beta^2$ or $\dfrac{1}{\alpha} + \dfrac{1}{\beta}$ , the intended method converts it to the sum and product with an identity. Finding the roots and substituting wastes time and forfeits the reasoning marks, and is often impossible when the roots are irrational.
Mishandling the pair-sum for a special form: With roots $a$ , $-a$ , $b$ the sum of products in pairs is $-a^2 + ab - ab = -a^2$ , not $-a^2 + ab + ab$ . Write out all three pairwise products before adding; the two mixed terms cancel.
Treating the middle relation as a sum of the roots: For a quartic, the second relation adds the products of zeroes taken two at a time, six terms, and the third adds the products taken three at a time, four terms. They are sums of products, not the roots themselves.

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation3 marksThe quadratic equation

x^2 - 9x + 20 = 0

has roots

\alpha

and

\beta

. Write down the values of

\alpha + \beta

and

\alpha\beta

. Given that one root is

4

, use each of these in turn to find the other root.

Show worked solution →

Read off the sum and product. For $ax^2 + bx + c$ the sum of the roots is $-\dfrac{b}{a}$ and the product is $\dfrac{c}{a}$ . Here $a = 1$ , $b = -9$ , $c = 20$ , so

\alpha + \beta = -\frac{-9}{1} = 9, \qquad \alpha\beta = \frac{20}{1} = 20.

Find the other root from the sum: If $\alpha = 4$ then $4 + \beta = 9$ , so $\beta = 5$ .
Confirm with the product: The product gives $4\beta = 20$ , so $\beta = 5$ again. Both routes agree.
State and check: The other root is $5$ . As a check, $(x - 4)(x - 5) = x^2 - 9x + 20$ , which is the original quadratic, so the roots $4$ and $5$ are correct.

foundation3 marksFor the cubic

P(x) = 2x^3 - 3x^2 - 11x + 6

with zeroes

\alpha

\beta

and

\gamma

, write down the sum of the zeroes, the sum of the products of pairs, and the product of the zeroes.

Show worked solution →

Match to the standard cubic. For $ax^3 + bx^2 + cx + d$ the three relations are

\alpha + \beta + \gamma = -\frac{b}{a}, \qquad \alpha\beta + \alpha\gamma + \beta\gamma = \frac{c}{a}, \qquad \alpha\beta\gamma = -\frac{d}{a}.

Here $a = 2$ , $b = -3$ , $c = -11$ , $d = 6$ .

Substitute the coefficients.

\alpha + \beta + \gamma = -\frac{-3}{2} = \frac{3}{2}, \qquad \alpha\beta + \alpha\gamma + \beta\gamma = \frac{-11}{2} = -\frac{11}{2}, \qquad \alpha\beta\gamma = -\frac{6}{2} = -3.

State the answers. Sum $= \dfrac{3}{2}$ , sum of pairs $= -\dfrac{11}{2}$ , product $= -3$ . (The actual zeroes are $3$ , $-2$ and $\tfrac{1}{2}$ , and indeed $3 - 2 + \tfrac{1}{2} = \tfrac{3}{2}$ and $3 \times (-2) \times \tfrac{1}{2} = -3$ , but no question asked for them.)

core4 marksThe equation

3x^2 - 7x + 2 = 0

has roots

\alpha

and

\beta

. Without solving the equation, find the value of

\alpha^2 + \beta^2

and the value of

\dfrac{\alpha}{\beta} + \dfrac{\beta}{\alpha}

Show worked solution →

Write down the sum and product. With $a = 3$ , $b = -7$ , $c = 2$ ,

\alpha + \beta = -\frac{-7}{3} = \frac{7}{3}, \qquad \alpha\beta = \frac{2}{3}.

Evaluate the sum of squares with the squaring identity. Since $(\alpha + \beta)^2 = \alpha^2 + 2\alpha\beta + \beta^2$ , rearranging gives $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$ :

\alpha^2 + \beta^2 = \left(\frac{7}{3}\right)^2 - 2 \cdot \frac{2}{3} = \frac{49}{9} - \frac{12}{9} = \frac{37}{9}.

Evaluate the symmetric ratio. Over the common denominator $\alpha\beta$ ,

\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha\beta} = \frac{37/9}{2/3} = \frac{37}{9} \cdot \frac{3}{2} = \frac{37}{6}.

State the answers. $\alpha^2 + \beta^2 = \dfrac{37}{9}$ and $\dfrac{\alpha}{\beta} + \dfrac{\beta}{\alpha} = \dfrac{37}{6}$ . (The roots are $2$ and $\tfrac{1}{3}$ ; directly $4 + \tfrac{1}{9} = \tfrac{37}{9}$ , confirming the first value.)

core4 marksThe zeroes of

P(x) = x^3 - 12x^2 + kx - 28

are in arithmetic progression. Find the value of

k

and the three zeroes.

Show worked solution →

Name the zeroes to use the progression. Three numbers in arithmetic progression can be written $m - d$ , $m$ , $m + d$ with middle term $m$ and common difference $d$ . This makes the sum collapse to $3m$ .

Use the sum of the zeroes to find the middle term. For $x^3 - 12x^2 + kx - 28$ the sum is $-\dfrac{-12}{1} = 12$ , so

(m - d) + m + (m + d) = 3m = 12 \implies m = 4.

So $4$ is a zero. (Whenever the zeroes are in arithmetic progression, the middle one is the average, which the sum hands you immediately.)

Use the product to find the common difference. The product is $-\dfrac{-28}{1} = 28$ , and $(m - d)\,m\,(m + d) = m(m^2 - d^2)$ :

4(16 - d^2) = 28 \implies 16 - d^2 = 7 \implies d^2 = 9 \implies d = 3.

The zeroes are $4 - 3 = 1$ , $4$ and $4 + 3 = 7$ .

Find $k$ from the sum of products of pairs. The middle coefficient gives $k = \alpha\beta + \alpha\gamma + \beta\gamma$ :

k = (1)(4) + (1)(7) + (4)(7) = 4 + 7 + 28 = 39.

State and check. $k = 39$ and the zeroes are $1$ , $4$ , $7$ . Check the product $1 \times 4 \times 7 = 28$ , matching $-\dfrac{d}{a}$ .

exam5 marksOne zero of the cubic

P(x) = x^3 - 5x^2 - 4x + 20

is the negative of another. Find all three zeroes, and verify the product-of-zeroes relation.

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Encode the condition in the names of the zeroes. "One zero is the negative of another" means the zeroes can be written $a$ , $-a$ and $b$ . Choosing these names builds the relationship in from the start.

Use the sum of the zeroes to find $b$ . The sum is $-\dfrac{-5}{1} = 5$ , and $a + (-a) + b = b$ , so

b = 5.

The opposite pair cancels in the sum, leaving the third zero directly.

Use the sum of products of pairs to find $a$ . The three pairwise products are $a(-a) = -a^2$ , $a\,b = ab$ and $(-a)b = -ab$ , which sum to $-a^2$ . This equals $\dfrac{c}{a_{\text{lead}}} = \dfrac{-4}{1} = -4$ :

-a^2 = -4 \implies a^2 = 4 \implies a = 2.

So the zeroes are $2$ , $-2$ and $5$ .

Verify the product relation. The product of the zeroes should be $-\dfrac{d}{a_{\text{lead}}} = -\dfrac{20}{1} = -20$ . Directly, $2 \times (-2) \times 5 = -20$ . The relation holds.

State the answer. The zeroes are $2$ , $-2$ and $5$ .

exam5 marksThe zeroes of

P(x) = x^3 - 13x^2 + 39x - 27

are in geometric progression. Find the three zeroes.

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Name the zeroes to use the progression. Three numbers in geometric progression can be written $\dfrac{a}{r}$ , $a$ , $ar$ with middle term $a$ and common ratio $r$ . This makes the product collapse to $a^3$ .

Use the product of the zeroes to find the middle term. The product is $-\dfrac{-27}{1} = 27$ , and $\dfrac{a}{r} \cdot a \cdot ar = a^3$ , so

a^3 = 27 \implies a = 3.

So $3$ is a zero. (For a geometric progression the middle term is the cube root of the product, which the product relation gives at once.)

Use the sum of the zeroes to find the ratio. The sum is $-\dfrac{-13}{1} = 13$ , and $\dfrac{a}{r} + a + ar = a\!\left(\dfrac{1}{r} + 1 + r\right)$ :

3\!\left(\frac{1}{r} + 1 + r\right) = 13 \implies \frac{1}{r} + r = \frac{10}{3}.

Solve for the ratio. Multiplying through by $3r$ gives $3r^2 - 10r + 3 = 0$ , so $(3r - 1)(r - 3) = 0$ and $r = 3$ or $r = \tfrac{1}{3}$ . Either ratio produces the same set of zeroes.

State and check. Taking $a = 3$ , $r = 3$ gives zeroes $1$ , $3$ , $9$ . Check the sum $1 + 3 + 9 = 13$ and the sum of pairs $(1)(3) + (1)(9) + (3)(9) = 3 + 9 + 27 = 39$ , both matching the coefficients.

exam3 marksThe polynomial

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

has zeroes

\alpha

\beta

\gamma

satisfying

\alpha^2 + \beta^2 + \gamma^2 = 14

. Find the value of

k

Show worked solution →

Read off the symmetric functions in terms of $k$ . For $x^3 - 6x^2 + kx - 6$ ,

\alpha + \beta + \gamma = -\frac{-6}{1} = 6, \qquad \alpha\beta + \alpha\gamma + \beta\gamma = \frac{k}{1} = k.

Apply the sum-of-squares identity. Squaring the sum gives $(\alpha + \beta + \gamma)^2 = \alpha^2 + \beta^2 + \gamma^2 + 2(\alpha\beta + \alpha\gamma + \beta\gamma)$ , so

\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \alpha\gamma + \beta\gamma) = 6^2 - 2k = 36 - 2k.

Solve for $k$ . Setting this equal to $14$ ,

36 - 2k = 14 \implies 2k = 22 \implies k = 11.

State and check. $k = 11$ . With $k = 11$ the cubic is $x^3 - 6x^2 + 11x - 6$ , whose zeroes are $1$ , $2$ , $3$ ; directly $1 + 4 + 9 = 14$ , confirming the condition.

What this dot point is asking

The answer

Where the relations come from

The relations for cubics and quartics

Why only two of them usually matter

Finding a missing zero once you know one

Evaluating symmetric expressions without solving

Roots in arithmetic or geometric progression

Roots of a special form

How exam questions ask about sums and products of zeroes

Practice questions

Related dot points