MathsCBSE Class 10 · Polynomials

Relationship Between Zeros and Coefficients of a Quadratic

Q: Why is there a minus sign in the sum of zeros but not in the product?

When you expand a times (x minus alpha) times (x minus beta), the x term becomes minus a times the sum of zeros, while the constant becomes plus a times the product. Comparing with ax squared plus bx plus c gives the minus sign only in the sum.

Q: What does verify the relationship mean in a board answer?

It means you find the actual zeros, compute their sum and product, and show these equal minus b over a and c over a. Write both sides for each relation; just saying hence verified is not enough.

Q: Do these relations work when the zeros are surds or fractions?

Yes. The rules hold for any real zeros of a quadratic, including surds and fractions. Surd zeros often come as a pair whose sum and product are rational.

Q: Is the relationship between zeros and coefficients of a cubic polynomial in the syllabus?

No. For the 2026-27 syllabus only the quadratic relationship is assessed.

Updated 14 October 2026 · Fact-checked

For a quadratic ax squared plus bx plus c, the sum of its zeros is minus b over a and the product is c over a. The board asks you to find the zeros, verify these two relations, find unknown coefficients from a condition on the zeros, and evaluate expressions such as the sum of squares of zeros.

For x² − 5x + 6, the zeros 2 and 3 add to 5, which is −b/a, and multiply to 6, which is c/a.

Illustration drawn for QuestionBuddy.

Understand Relationship Between Zeros and Coefficients

Take a quadratic polynomial $p (x) = a x^{2} + b x + c$ with $a \neq = 0$ , and call its zeros $α$ and $β$ . The two zeros are not random numbers. They are tied to the coefficients $a$ , $b$ and $c$ by two simple rules: $α + β = - \frac{b}{a}$ and $α β = \frac{c}{a}$ .

Why it works. If $α$ is a zero, then $(x - α)$ is a factor, and the same holds for $β$ . So $a x^{2} + b x + c = a (x - α) (x - β)$ . Expanding the right side gives $a x^{2} - a (α + β) x + a α β$ . Two polynomials are equal only when matching coefficients are equal, so $b = - a (α + β)$ and $c = a α β$ . Divide by $a$ and you get both rules. The minus sign in the sum comes straight from the minus signs inside the brackets.

How to verify. A typical board question says: find the zeros and verify the relationship. You factorise, state the two zeros, then compute their sum and product and show each one equals $- \frac{b}{a}$ and $\frac{c}{a}$ respectively. Both sides must be written out. This acts as a built-in check: if your sum does not match, your factorisation is wrong.

Using the rules without finding zeros. Because the sum and product come directly from the coefficients, you can answer many questions without solving. You can find an unknown $k$ from a condition such as the difference of zeros, decide the signs of the zeros, or find $α^{2} + β^{2} = (α + β)^{2} - 2 α β$ . The reverse use, building a polynomial from a given sum and product, is covered on the forming-polynomials page.

Conditions. The rules need the polynomial in standard form, with terms in the order $x^{2}$ , $x$ , constant, and $a \neq = 0$ . Always rearrange first: in $5 - 3 x + 2 x^{2}$ , $a = 2$ , $b = - 3$ , $c = 5$ . Read the signs with the coefficients. If a term is missing, its coefficient is $0$ , as in $x^{2} - 9$ where $b = 0$ . Only the quadratic relations are in your syllabus. The rules describe the two zeros together, and equal (repeated) zeros count twice. Arguments about the signs of the zeros apply only when the zeros are real, that is when $b^{2} - 4 a c \geq 0$ ; Class 10 questions on these relations always use polynomials with real zeros.

Relationship Between Zeros and Coefficients: formulas to remember

Sum of zeros: $α + β = - \frac{b}{a} = - \frac{coefficient of x}{coefficient of x ^{2}}$; Keep the minus sign. Read b with its own sign from the standard form.
Product of zeros: $α β = \frac{c}{a} = \frac{constant term}{coefficient of x ^{2}}$; No minus sign. A negative product means the zeros have opposite signs.
Factored form (source of the rules): $a x^{2} + b x + c = a (x - α) (x - β)$; Expand and compare coefficients to prove both relations.
Sum of squares: $α^{2} + β^{2} = (α + β)^{2} - 2 α β$; Never equal to just the square of the sum.
Square of the difference: $(α - β)^{2} = (α + β)^{2} - 4 α β$; Use when the difference of zeros is given or asked.
Sum of reciprocals: $\frac{1}{α} + \frac{1}{β} = \frac{α + β}{α β}$; Needs only the sum and product.
Sum of cubes: $α^{3} + β^{3} = (α + β)^{3} - 3 α β (α + β)$; Use only when the question asks for a cube expression; it needs just the sum and product.

How to solve Relationship Between Zeros and Coefficients questions

Use this method for any question that says find the zeros and verify, or that gives a condition on the zeros.

1Rearrange the polynomial into standard form $a x^{2} + b x + c$ and write $a$ , $b$ , $c$ with their signs.
2Write $α + β = - \frac{b}{a}$ and $α β = \frac{c}{a}$ straight away.
3If the zeros are asked, factorise by splitting the middle term (two numbers with product $a c$ and sum $b$ ), and state the zeros as values of $x$ .
4To verify, compute the sum of the zeros on one line and compare it with $- \frac{b}{a}$ ; do the same for the product and $\frac{c}{a}$ .
5If an unknown $k$ appears, translate the given condition into an equation using the sum and product, then solve for $k$ .
6If an expression like $α^{2} + β^{2}$ is asked, rewrite it using only $α + β$ and $α β$ before substituting.
7Check: substitute your zeros or your $k$ back and confirm the condition holds.

Relationship Between Zeros and Coefficients in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Relationship between zeros and coefficients of a quadratic

Suppose a puzzle on your school's maths notice board says: the ages of two siblings, in years, are the zeros of $p (x) = x^{2} - 18 x + 80$ . You are asked to learn about their ages using only the coefficients first.

Without finding the ages, find the sum of their ages, the product of their ages and the difference between their ages. Then find the ages and verify the sum and product.

Try it first, then show the solutionSolution

Here $a = 1$ , $b = - 18$ , $c = 80$ .
Sum of ages $= - \frac{b}{a} = 18$ years.
Product of ages $= \frac{c}{a} = 80$ .
$(α - β)^{2} = (α + β)^{2} - 4 α β = 324 - 320 = 4$ , so the difference is $2$ years (ages are positive, take the positive root).
Sum $18$ and difference $2$ give ages $\frac{18 + 2}{2} = 10$ and $\frac{18 - 2}{2} = 8$ .
Check: $10 + 8 = 18$ and $10 \times 8 = 80$ , matching the coefficients.

Answer: Sum 18 years, product 80, difference 2 years; the ages are 10 years and 8 years.

Memory hook: Sibling ages puzzle: the middle coefficient tells the total, the constant tells the product.

Tricks to remember Relationship Between Zeros and Coefficients

Short memory aids. Say each one aloud once, then use it on a question the same day.

Sum is shy, product is plain
The sum hides behind a minus sign: $- \frac{b}{a}$ . The product shows up as it is: $\frac{c}{a}$ .
Use it when: Write both down before any working so the sign never slips.
Both over a
Both rules divide by $a$ , the coefficient of $x^{2}$ . Sum uses $b$ (the middle), product uses $c$ (the end).
Use it when: Stops you dividing by $b$ or forgetting $a$ when $a \neq = 1$ .
Sign check
Product negative: zeros of opposite signs. Product positive and sum negative: both negative. Product positive and sum positive: both positive.
Use it when: Quick check on factorised zeros and on assertion-reason items.

Relationship Between Zeros and Coefficients: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Find zeros and verify (standard) · 3 marks · Short answerBoard level

Find the zeros of $p (x) = 3 x^{2} - x - 4$ and verify the relationship between the zeros and the coefficients.

Try it first, then show the solutionSolution

$a c = - 12$ ; numbers with product $- 12$ and sum $- 1$ are $- 4$ and $3$ .
$3 x^{2} - 4 x + 3 x - 4 = x (3 x - 4) + 1 (3 x - 4) = (3 x - 4) (x + 1)$ (1 mark)
Zeros: $x = \frac{4}{3}$ and $x = - 1$ (½ mark)
Sum $= \frac{4}{3} - 1 = \frac{1}{3}$ and $- \frac{b}{a} = - \frac{- 1}{3} = \frac{1}{3}$ , verified (1 mark)
Product $= \frac{4}{3} \times (- 1) = - \frac{4}{3}$ and $\frac{c}{a} = - \frac{4}{3}$ , verified (½ mark)

Answer: Zeros are $\frac{4}{3}$ and $- 1$ ; sum $\frac{1}{3}$ and product $- \frac{4}{3}$ match $- \frac{b}{a}$ and $\frac{c}{a}$ .

Example 2: Surd coefficients · 3 marks · Short answerBoard level

Find the zeros of $p (x) = x^{2} - 22 x - 6$ and verify the relationship between the zeros and the coefficients.

Try it first, then show the solutionSolution

$a c = - 6$ ; we need two terms with product $- 6$ and sum $- 22$ : $- 32$ and $2$ , since $(- 32) (2) = - 6$ .
$x^{2} - 32 x + 2 x - 6 = x (x - 32) + 2 (x - 32) = (x + 2) (x - 32)$ (1 mark)
Zeros: $32$ and $- 2$ (½ mark)
Sum $= 32 - 2 = 22$ ; $- \frac{b}{a} = - \frac{- 2 2}{1} = 22$ , verified (1 mark)
Product $= 32 \times (- 2) = - 6$ ; $\frac{c}{a} = - 6$ , verified (½ mark)

Answer: Zeros are $32$ and $- 2$ ; sum $22$ and product $- 6$ verify the relations.

Example 3: Unknown from the difference of zeros · 3 marks · Short answerHarder board level

The difference of the zeros of $p (x) = x^{2} - 5 x + k$ is $1$ . Find $k$ , find the zeros and verify the relationship between the zeros and the coefficients.

Try it first, then show the solutionSolution

$α + β = 5$ and $α β = k$ (½ mark)
$(α - β)^{2} = (α + β)^{2} - 4 α β$ , so $1 = 25 - 4 k$ (½ mark)
$4 k = 24$ , so $k = 6$ (½ mark)
$p (x) = x^{2} - 5 x + 6 = (x - 2) (x - 3)$ , so zeros are $2$ and $3$ (½ mark)
Sum $= 5 = - \frac{- 5}{1}$ , verified (½ mark)
Product $= 6 = \frac{6}{1}$ , verified; difference $3 - 2 = 1$ also holds (½ mark)

Answer: $k = 6$ ; zeros $2$ and $3$ , with sum $5$ and product $6$ as required.

Example 4: Expressions in the zeros (long answer) · 5 marks · Long answerHarder board level

Let $α$ and $β$ be the zeros of $p (x) = 2 x^{2} - 5 x + 2$ .
(i) Find the zeros and verify the relationship between the zeros and the coefficients.
(ii) Without using the zeros, find $α^{2} + β^{2}$ .
(iii) Find $α^{3} + β^{3}$ .
(iv) Find $\frac{1}{α} + \frac{1}{β}$ .

Try it first, then show the solutionSolution

(i) $2 x^{2} - 4 x - x + 2 = 2 x (x - 2) - 1 (x - 2) = (2 x - 1) (x - 2)$ (½ mark)
Zeros: $\frac{1}{2}$ and $2$ (½ mark)
Sum $= \frac{5}{2} = - \frac{- 5}{2}$ ; product $= 1 = \frac{2}{2}$ , both verified (1 mark)
(ii) $α + β = \frac{5}{2}$ , $α β = 1$ ; $α^{2} + β^{2} = \frac{25}{4} - 2 = \frac{17}{4}$ (1 mark)
(iii) $α^{3} + β^{3} = (\frac{5}{2})^{3} - 3 (1) (\frac{5}{2}) = \frac{125}{8} - \frac{60}{8} = \frac{65}{8}$ (1 mark)
(iv) $\frac{1}{α} + \frac{1}{β} = \frac{α + β}{α β} = \frac{5}{2}$ (½ mark)
Check with the zeros: $\frac{1}{4} + 4 = \frac{17}{4}$ and $\frac{1}{8} + 8 = \frac{65}{8}$ , both agree (½ mark)

Answer: (i) Zeros $\frac{1}{2}$ and $2$ , relations verified; (ii) $\frac{17}{4}$ ; (iii) $\frac{65}{8}$ ; (iv) $\frac{5}{2}$ .

Example 5: Signs of zeros by reasoning · 3 marks · Short answerHOTS / competency

Without finding the zeros, show that both zeros of $p (x) = 3 x^{2} + 11 x + 6$ are negative. Then find the zeros and verify the relationship between the zeros and the coefficients.

Try it first, then show the solutionSolution

The zeros are real, because $b^{2} - 4 a c = 121 - 72 = 49 > 0$ ; the sign argument below works only for real zeros (½ mark)
Product $= \frac{c}{a} = 2 > 0$ , so the two real zeros have the same sign (½ mark)
Sum $= - \frac{b}{a} = - \frac{11}{3} < 0$ ; two numbers of the same sign with a negative sum are both negative (½ mark)
$3 x^{2} + 9 x + 2 x + 6 = 3 x (x + 3) + 2 (x + 3) = (3 x + 2) (x + 3)$ , so zeros are $- \frac{2}{3}$ and $- 3$ (1 mark)
Sum $= - \frac{2}{3} - 3 = - \frac{11}{3}$ and product $= (- \frac{2}{3}) (- 3) = 2$ , both verified (½ mark)

Answer: The zeros are real (discriminant 49 > 0); positive product and negative sum then force both to be negative. Zeros are $- \frac{2}{3}$ and $- 3$ , relations verified.

Where students lose marks in Relationship Between Zeros and Coefficients

The slips examiners see most often, and what to write instead.

Mistake: Dropping the minus sign in the sum
Students remember b over a and forget the sign comes from expanding the brackets.
Wrong working
For $x^{2} - 7 x + 12$ : $α + β = \frac{b}{a} = - 7$
Right working
$α + β = - \frac{b}{a} = - \frac{- 7}{1} = 7$ (zeros $3$ and $4$ )
Do this: Write the formula with its minus sign before substituting.
Usually costs: The verification mark, and every later part that uses the sum.
Mistake: Reading coefficients before rearranging
The polynomial is given out of order, so the wrong number is taken as a.
Wrong working
For $5 - 3 x + 2 x^{2}$ : $a = 5$ , $b = - 3$ , $c = 2$ , product $= \frac{2}{5}$
Right working
Standard form $2 x^{2} - 3 x + 5$ : $a = 2$ , $b = - 3$ , $c = 5$ , product $= \frac{5}{2}$
Do this: Always rewrite in the order $x^{2}$ , $x$ , constant first.
Usually costs: All marks for the sum and product.
Mistake: Taking the square of the sum as the sum of squares
Confusing $(α + β)^{2}$ with $α^{2} + β^{2}$ .
Wrong working
For $x^{2} - 5 x + 6$ : $α^{2} + β^{2} = 5^{2} = 25$
Right working
$α^{2} + β^{2} = 25 - 2 (6) = 13$ (check: $4 + 9 = 13$ )
Do this: Remember the $- 2 α β$ correction.
Usually costs: 1 to 2 marks in expression questions.
Mistake: Verifying only one side
Students compute the sum of zeros but do not show $- \frac{b}{a}$ separately.
Wrong working
Sum $= \frac{1}{3}$ , product $= - \frac{4}{3}$ . Hence verified.
Right working
Sum $= \frac{1}{3}$ and $- \frac{b}{a} = \frac{1}{3}$ ; product $= - \frac{4}{3}$ and $\frac{c}{a} = - \frac{4}{3}$ . Hence verified.
Do this: Show both the zero side and the coefficient side for each relation.
Usually costs: Up to 1 mark in a 3-mark verify question.

Exam tips

In verify questions, keep the sum and product on separate lines, each comparing two computed values.
With fraction coefficients, factor out a constant to factorise, but verify using the original $a$ , $b$ , $c$ .
When a question says without finding the zeros, use only identities in $α + β$ and $α β$ ; solving instead may lose method marks.
After finding an unknown $k$ , put it back and factorise to confirm the condition.
For assertion-reason items on signs of zeros, check the sign of the product first, then the sum.

Practice questions on Relationship Between Zeros and Coefficients

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQHarder board level
If $α$ and $β$ are the zeros of $x^{2} - 6 x + k$ and $3 α + 2 β = 20$ , then the value of $k$ is
(a) $- 16$
(b) $16$
(c) $- 12$
(d) $12$
Hint
Write $3 α + 2 β = 2 (α + β) + α$ and use the sum of zeros.
Answer
(a) $- 16$ . $α + β = 6$ , so $12 + α = 20$ , giving $α = 8$ , $β = - 2$ and $k = α β = - 16$ . Check: $x^{2} - 6 x - 16 = (x - 8) (x + 2)$ .
Q2 · 1 mark · MCQBoard level
If the zeros of $(k + 1) x^{2} - 13 x + 6$ are reciprocals of each other, then $k$ equals
(a) $6$
(b) $5$
(c) $- 5$
(d) $7$
Hint
If the zeros are $α$ and $\frac{1}{α}$ , their product is $1$ .
Answer
(b) $5$ . Product $= \frac{6}{k + 1} = 1$ , so $k = 5$ . Check: $6 x^{2} - 13 x + 6 = (2 x - 3) (3 x - 2)$ , zeros $\frac{3}{2}$ and $\frac{2}{3}$ .
Q3 · 2 marks · Very short answerBoard level
One zero of the polynomial $2 x^{2} - 7 x + k$ is 3. Find $k$ and the other zero, and verify the sum of the zeros.
Hint
A zero makes the polynomial equal to 0. Then use the product (or sum) of zeros.
Answer
$p (3) = 18 - 21 + k = 0$ , so $k = 3$ . (1) Product of zeros $= \frac{3}{2}$ , so the other zero is $\frac{3/2}{3} = \frac{1}{2}$ . Check: sum $= 3 + \frac{1}{2} = \frac{7}{2} = - \frac{- 7}{2}$ . (1)
Q4 · 1 mark · Assertion-reasonBoard level
Assertion (A): Both zeros of $x^{2} + 7 x + 12$ are negative.
Reason (R): If the zeros of a quadratic polynomial are real, their product is positive and their sum is negative, then both zeros are negative.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Check that the zeros are real, then look at the signs of $\frac{c}{a}$ and $- \frac{b}{a}$ .
Answer
(a). Discriminant $49 - 48 = 1 > 0$ , so the zeros are real; product $12 > 0$ and sum $- 7 < 0$ , so both are negative (they are $- 3$ and $- 4$ ). R is exactly the reason.
Q5 · 1 mark · MCQBoard level
The graph of $y = x^{2} + b x + c$ cuts the x-axis at the points $(- 2, 0)$ and $(5, 0)$ . The value of $b + c$ is
(a) $- 13$
(b) $7$
(c) $- 7$
(d) $13$
Hint
The x-coordinates where the graph cuts the x-axis are the zeros.
Answer
(a) $- 13$ . Zeros $- 2$ and $5$ : sum $3 = - b$ , so $b = - 3$ ; product $- 10 = c$ . So $b + c = - 13$ .
Q6 · 3 marks · Short answerHarder board level
Find the zeros of $p (y) = 7 y^{2} - \frac{11}{3} y - \frac{2}{3}$ and verify the relationship between the zeros and the coefficients.
Hint
Write $p (y) = \frac{1}{3} (21 y^{2} - 11 y - 2)$ and factorise the bracket. Verify using the original $a$ , $b$ , $c$ .
Answer
$21 y^{2} - 14 y + 3 y - 2 = (3 y - 2) (7 y + 1)$ , zeros $\frac{2}{3}$ and $- \frac{1}{7}$ . Sum $= \frac{11}{21} = \frac{11/3}{7}$ ; product $= - \frac{2}{21} = \frac{- 2/3}{7}$ . Verified.
Q7 · 3 marks · Short answerHarder board level
The sum of the zeros of $p (x) = 2 x^{2} + k x - 6$ equals their product. Find $k$ , and then find $α^{2} + β^{2}$ for this polynomial.
Hint
Set minus k over 2 equal to minus 3.
Answer
$- \frac{k}{2} = - 3$ gives $k = 6$ . Then $α + β = - 3$ , $α β = - 3$ , so $α^{2} + β^{2} = 9 + 6 = 15$ .
Q8 · 5 marks · Long answerHarder board level
Let $α$ and $β$ be the zeros of $x^{2} - 4 x + 1$ . Find (i) $α^{2} + β^{2}$ , (ii) $\frac{1}{α ^{2}} + \frac{1}{β ^{2}}$ , (iii) $α^{3} + β^{3}$ , (iv) the positive value of $α - β$ . Then find the zeros and check (i).
Hint
Sum is 4 and product is 1. Use the identities, then solve using sum and difference.
Answer
(i) $16 - 2 = 14$ ; (ii) $\frac{14}{1} = 14$ ; (iii) $64 - 12 = 52$ ; (iv) $(α - β)^{2} = 16 - 4 = 12$ , so $23$ . Zeros $2 + 3$ and $2 - 3$ ; squares add to $2 (4 + 3) = 14$ , matching.
Q9 · 3 marks · Short answerHOTS / competency
Neha says the zeros of $2 x^{2} - 5 x - 12$ are $4$ and $\frac{3}{2}$ . Without factorising, use the relationship between zeros and coefficients to show she is wrong. Then find the correct zeros and verify.
Hint
Compare her sum and product with minus b over a and c over a.
Answer
Required sum $\frac{5}{2}$ , product $- 6$ ; her sum is $\frac{11}{2}$ and product $6$ , so she is wrong. $(2 x + 3) (x - 4)$ gives zeros $4$ and $- \frac{3}{2}$ : sum $\frac{5}{2}$ , product $- 6$ , verified.
Q10 · 4 marks · Case-basedHOTS / competency
Case study: The curved edge of a garden arch is drawn on a grid as the graph of $p (x) = - x^{2} + 6 x + k$ , where x is measured in metres along the ground. The arch meets the ground (the x-axis) at its two feet, whose x-coordinates are the zeros $α$ and $β$ of $p (x)$ .
(i) Write $α + β$ and $α β$ in terms of $k$ . (1 mark)
(ii) The feet of the arch are 8 m apart. Find $k$ . (2 marks)
(iii) Find the zeros, and the height of the arch above the point $x = 3$ . (1 mark)
Hint
The distance between the feet is $∣ α - β ∣$ , and $(α - β)^{2} = (α + β)^{2} - 4 α β$ .
Answer
(i) Here $a = - 1$ , $b = 6$ , $c = k$ : $α + β = - \frac{6}{- 1} = 6$ and $α β = \frac{k}{- 1} = - k$ . (1)
(ii) $(α - β)^{2} = 36 - 4 (- k) = 36 + 4 k = 8^{2} = 64$ , so $k = 7$ . (2)
(iii) $- x^{2} + 6 x + 7 = - (x - 7) (x + 1)$ , so the zeros are $7$ and $- 1$ (8 m apart, as required). Height at $x = 3$ : $p (3) = - 9 + 18 + 7 = 16$ m. (1)

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Sum of zeros of $a x^{2} + b x + c$ ?Show answer
$- \frac{b}{a}$
2.Product of zeros of $a x^{2} + b x + c$ ?Show answer
$\frac{c}{a}$
3.Sum and product of zeros of $x^{2} - 9$ ?Show answer
Sum $0$ , product $- 9$ (zeros $3$ and $- 3$ ).
4. $α^{2} + β^{2}$ in terms of sum and product?Show answer
$(α + β)^{2} - 2 α β$
5.If the product of zeros is negative, what about the signs?Show answer
The zeros have opposite signs.

Relationship Between Zeros and Coefficients: frequently asked questions

Why is there a minus sign in the sum of zeros but not in the product?

When you expand a times (x minus alpha) times (x minus beta), the x term becomes minus a times the sum of zeros, while the constant becomes plus a times the product. Comparing with ax squared plus bx plus c gives the minus sign only in the sum.

What does verify the relationship mean in a board answer?

It means you find the actual zeros, compute their sum and product, and show these equal minus b over a and c over a. Write both sides for each relation; just saying hence verified is not enough.

Do these relations work when the zeros are surds or fractions?

Yes. The rules hold for any real zeros of a quadratic, including surds and fractions. Surd zeros often come as a pair whose sum and product are rational.

Is the relationship between zeros and coefficients of a cubic polynomial in the syllabus?

No. For the 2026-27 syllabus only the quadratic relationship is assessed.