CBSE • Class X • Mathematics
Chapter 2 — Polynomials
Unit Test / Examination Paper
PAPER CODE
MTH-P02
Reprint 2025–26
Name
Roll No.
Class & Section
Date
General Instructions
- This paper contains four sections: A, B, C and D.
- All questions are compulsory.
- Section A: 10 MCQs of 1 mark each (10 marks).
- Section B: 6 Very Short Answer questions of 2 marks each (12 marks).
- Section C: 6 Short Answer questions of 3 marks each (18 marks).
- Section D: 5 Long Answer questions of 4 marks each (20 marks).
- Case-Based: 2 questions of 10 marks each (20 marks).
- No marks will be deducted for wrong answers in Section A.
- Draw neat diagrams wherever required. Label clearly.
- Calculators are not permitted.
Marks Distribution
A
MCQ • 1×10
10
B
VSAQ • 2×6
12
C
SAQ • 3×6
18
D
LAQ • 4×5
20
CBQ
Case • 10×2
20
Section A — Multiple Choice Questions
10 Questions
1 Mark Each • 10 Marks
1
[1 Mark]
The degree of the polynomial p(x) = 7u6 − (3/2)u4 + 4u2 − 8 is:
2
[1 Mark]
Which of the following is NOT a polynomial?
3
[1 Mark]
The zeroes of the polynomial p(x) = x2 − 3x − 4 are:
4
[1 Mark]
If α and β are the zeroes of 2x2 − 8x + 6, then αβ is equal to:
5
[1 Mark]
A quadratic polynomial whose graph (parabola) does not intersect the x-axis has:
6
[1 Mark]
The zero of the linear polynomial p(x) = ax + b (a ≠ 0) is:
7
[1 Mark]
If α, β, γ are zeroes of ax3 + bx2 + cx + d, then αβγ equals:
8
[1 Mark]
The graph of a quadratic polynomial ax2 + bx + c opens downward when:
9
[1 Mark]
A cubic polynomial can have at most how many real zeroes?
10
[1 Mark]
The sum of zeroes of the polynomial 3x2 + 5x − 2 is:
Section B — Very Short Answer
6 Questions
2 Marks Each • 12 Marks
11
[2 Marks]
Find the zero of the linear polynomial p(x) = 5x − 10. Verify your answer by substitution.
12
[2 Marks]
Find the zeroes of p(x) = x2 − 3 using the difference-of-squares identity.
13
[2 Marks]
The graph of y = p(x) is a parabola that intersects the x-axis at exactly one point (tangent). State how many zeroes the polynomial has and what special name is given to such a zero.
14
[2 Marks]
For the polynomial p(x) = x2 + 7x + 10, find the sum and product of its zeroes using Vieta’s formulas (without finding the zeroes individually).
15
[2 Marks]
Determine whether x = 3 is a zero of p(x) = 3x3 − 5x2 − 11x − 3. Show your working.
16
[2 Marks]
Write the most general form of a cubic polynomial. State the condition on the leading coefficient. Give one concrete example.
Section C — Short Answer
6 Questions
3 Marks Each • 18 Marks
17
[3 Marks]
Find the zeroes of the quadratic polynomial p(x) = 2x2 − 8x + 6 by factorisation. Verify the relationship between the zeroes and the coefficients.
18
[3 Marks]
Find the zeroes of 3x2 + 5x − 2. Verify that the sum of zeroes = −b/a and product of zeroes = c/a.
19
[3 Marks]
Find a quadratic polynomial, the sum and product of whose zeroes are −3 and 2, respectively. Find a general family of such polynomials.
20
[3 Marks]
The graph of y = p(x) for a polynomial p(x) is shown below (described in words). It is a curve that crosses the x-axis at x = −2, x = 0, and x = 2, and no other point. State the minimum degree of p(x), name its type, and write one possible polynomial for p(x).
21
[3 Marks]
Verify that 3, −1, and −1/3 are the zeroes of p(x) = 3x3 − 5x2 − 11x − 3 by substitution.
22
[3 Marks]
If α and β are the zeroes of the polynomial f(x) = x2 − 5x + k such that α − β = 1, find the value of k.
Section D — Long Answer
5 Questions
4 Marks Each • 20 Marks
23
[4 Marks]
Find the zeroes of the quadratic polynomial p(x) = 4s2 − 4s + 1. Verify all relationships between zeroes and coefficients. Also state what type of zero it is geometrically.
24
[4 Marks]
For the cubic polynomial p(x) = 2x3 − 5x2 − 14x + 8 with zeroes 4, −2, and 1/2:
(i)
Verify the sum of zeroes using α + β + γ = −b/a.
(ii)
Verify the sum of products taken two at a time using αβ + βγ + γα = c/a.
(iii)
Verify the product of all zeroes using αβγ = −d/a.
25
[4 Marks]
The sum of zeroes of a quadratic polynomial p(x) = kx2 + 2x + 3k is equal to the product of its zeroes. Find the value of k. Also find the polynomial and its zeroes.
26
[4 Marks]
One zero of the polynomial p(x) = 3x3 − 5x2 − 11x − 3 is given as 3. Using this, find the remaining two zeroes by the following method:
(i)
Use the sum of all three zeroes to set up an equation for the remaining two.
(ii)
Use the sum of pairwise products to find a second equation.
(iii)
Solve the two equations to find the remaining zeroes.
Case-Based Questions (CBQ)
2 Questions
10 Marks Each • 20 Marks
Case Study I — The Roller Coaster Track
A theme park engineer is designing a roller coaster track. The height of the track (in metres) above the ground at a horizontal distance of x metres from the start can be modelled by the polynomial:
h(x) = −x3 + 4x (for −2 ≤ x ≤ 2)
The engineer notes that the track meets ground level (h = 0) at certain points. Based on this information, answer the following:
C1
[2 Marks]
Find all values of x where the track meets the ground (h(x) = 0). Factorise h(x) completely.
C2
[2 Marks]
What is the degree of h(x)? What is the maximum number of times the track can cross ground level according to this model?
C3
[3 Marks]
Using Vieta’s formulas, verify the sum of zeroes, sum of pairwise products, and product of all zeroes of h(x) = −x3 + 4x (rewrite as −x3 + 0x2 + 4x + 0, so a=−1, b=0, c=4, d=0).
C4
[3 Marks]
The engineer modifies the track so that only ONE section of the track touches the ground and bounces back (a tangent point). Describe what this looks like geometrically, and write a quadratic polynomial whose graph would model a single tangent touch at x = 2.
Case Study II — The Farming Plot
A farmer has a rectangular field. The length is 3 metres more than twice its width. The area of the field (in m2) is modelled by the quadratic polynomial:
A(w) = 2w2 + 3w where w = width in metres
The farmer needs to fence the field and also wants to divide it into sections. Answer the following questions based on this context:
C5
[2 Marks]
Find the zeroes of A(w) = 2w2 + 3w by factorisation. Interpret what each zero means in the context of the farming plot.
C6
[2 Marks]
Using the coefficient formula, verify the sum and product of zeroes of A(w) = 2w2 + 3w.
C7
[3 Marks]
The farmer wants the area to be exactly 45 m2. Set up the equation 2w2 + 3w = 45 and solve it by splitting the middle term. Find w. Is the negative root valid in this context? Explain.
C8
[3 Marks]
Write a quadratic polynomial whose zeroes are √2 and −√2. Verify using Vieta’s formulas. State what the graph of this polynomial looks like (open up/down, x-intercepts).
Examiner / Teacher View
Answer Key & Marking Scheme
Chapter 2 — Polynomials • Class X Mathematics • 80 Marks
For Examiners Only. Award marks as indicated. Accept equivalent correct methods. Where steps are shown, partial credit may be awarded as marked.
Section A — MCQ Answers (1 mark each)
Q.1
(C)
Degree 6
Q.2
(B)
√x + 2
Q.3
(B)
−1 and 4
Q.4
(A)
αβ = 3
Q.5
(C)
No real zeroes
Q.6
(C)
−b/a
Q.7
(C)
−d/a
Q.8
(C)
a < 0
Q.9
(C)
3 zeroes
Q.10
(B)
−5/3
Section B — Very Short Answer (2 marks each)
11
2 Marks
Find zero of p(x) = 5x − 10.
Set p(x) = 0: 5x − 10 = 0
5x = 10
x = 2 ← Zero [1 mark]
Verification: p(2) = 5(2) − 10 = 10 − 10 = 0 ✓ [1 mark]
5x = 10
x = 2 ← Zero [1 mark]
Verification: p(2) = 5(2) − 10 = 10 − 10 = 0 ✓ [1 mark]
Answer: x = 2
12
2 Marks
Find zeroes of p(x) = x2 − 3 using difference of squares.
a2 − b2 = (a−b)(a+b) [identity stated: ½ mark]
x2 − 3 = x2 − (√3)2 = (x − √3)(x + √3) [1 mark]
x − √3 = 0 ⇒ x = √3 x + √3 = 0 ⇒ x = −√3 [½ mark]
x2 − 3 = x2 − (√3)2 = (x − √3)(x + √3) [1 mark]
x − √3 = 0 ⇒ x = √3 x + √3 = 0 ⇒ x = −√3 [½ mark]
Zeroes: x = √3 and x = −√3
13
2 Marks
Parabola tangent to x-axis at exactly one point.
Number of zeroes: 1 zero (the two zeroes coincide) [1 mark]
Special name: Repeated zero (or double root / coincident zeroes) [1 mark]
Algebraically: discriminant b2 − 4ac = 0 for this case.
Special name: Repeated zero (or double root / coincident zeroes) [1 mark]
Algebraically: discriminant b2 − 4ac = 0 for this case.
14
2 Marks
Sum and product of zeroes of x2 + 7x + 10 using Vieta’s formulas.
a = 1, b = 7, c = 10
Sum = −b/a = −7/1 = −7 [1 mark]
Product = c/a = 10/1 = 10 [1 mark]
Sum = −b/a = −7/1 = −7 [1 mark]
Product = c/a = 10/1 = 10 [1 mark]
15
2 Marks
Verify x = 3 is a zero of p(x) = 3x3 − 5x2 − 11x − 3.
p(3) = 3(3)3 − 5(3)2 − 11(3) − 3 [substitution: 1 mark]
= 3(27) − 5(9) − 33 − 3
= 81 − 45 − 33 − 3
= 81 − 81 = 0 [correct computation: 1 mark]
Since p(3) = 0, x = 3 IS a zero. ✓
= 3(27) − 5(9) − 33 − 3
= 81 − 45 − 33 − 3
= 81 − 81 = 0 [correct computation: 1 mark]
Since p(3) = 0, x = 3 IS a zero. ✓
16
2 Marks
General form of a cubic polynomial.
General form: ax3 + bx2 + cx + d [1 mark]
Condition: a, b, c, d ∈ ℜ and a ≠ 0 (otherwise not cubic) [½ mark]
Example: 2x3 − 5x2 + 3x − 1 (any valid cubic) [½ mark]
Condition: a, b, c, d ∈ ℜ and a ≠ 0 (otherwise not cubic) [½ mark]
Example: 2x3 − 5x2 + 3x − 1 (any valid cubic) [½ mark]
Section C — Short Answer (3 marks each)
17
3 Marks
Zeroes of p(x) = 2x2 − 8x + 6 and verification.
Factorisation: 2x2 − 8x + 6 = 2(x2 − 4x + 3) = 2(x−1)(x−3) [1 mark]
Zeroes: α = 1, β = 3 [½ mark]
Verification (a=2, b=−8, c=6):
Sum: α+β = 1+3 = 4 = −(−8)/2 = 4 ✓ [½ mark]
Product: αβ = 1×3 = 3 = 6/2 = 3 ✓ [1 mark]
Zeroes: α = 1, β = 3 [½ mark]
Verification (a=2, b=−8, c=6):
Sum: α+β = 1+3 = 4 = −(−8)/2 = 4 ✓ [½ mark]
Product: αβ = 1×3 = 3 = 6/2 = 3 ✓ [1 mark]
18
3 Marks
Zeroes of 3x2 + 5x − 2 and verification.
Splitting middle term: 3x2 + 6x − x − 2 = 3x(x+2) − 1(x+2) = (3x−1)(x+2) [1 mark]
Zeroes: x = 1/3 and x = −2 [½ mark]
Verification (a=3, b=5, c=−2):
Sum: 1/3 + (−2) = 1/3 − 6/3 = −5/3 = −b/a = −5/3 ✓ [½ mark]
Product: (1/3)(−2) = −2/3 = c/a = −2/3 ✓ [1 mark]
Zeroes: x = 1/3 and x = −2 [½ mark]
Verification (a=3, b=5, c=−2):
Sum: 1/3 + (−2) = 1/3 − 6/3 = −5/3 = −b/a = −5/3 ✓ [½ mark]
Product: (1/3)(−2) = −2/3 = c/a = −2/3 ✓ [1 mark]
19
3 Marks
Find quadratic polynomial with sum = −3, product = 2.
Template: x2 − (sum)x + (product) [1 mark]
= x2 − (−3)x + 2 [substitution: 1 mark]
= x2 + 3x + 2 [1 mark]
General family: k(x2 + 3x + 2), k ≠ 0 [bonus/partial credit]
= x2 − (−3)x + 2 [substitution: 1 mark]
= x2 + 3x + 2 [1 mark]
General family: k(x2 + 3x + 2), k ≠ 0 [bonus/partial credit]
Polynomial: x2 + 3x + 2 (or any scalar multiple)
20
3 Marks
Curve crosses x-axis at −2, 0, 2 only.
Min. degree: 3 (at least 3 zeroes ⇒ degree ≥ 3) [1 mark]
Type: Cubic polynomial [½ mark]
Polynomial: p(x) = (x+2)(x)(x−2) = x(x2−4) = x3 − 4x [1½ marks]
Note: any scalar multiple also accepted.
Type: Cubic polynomial [½ mark]
Polynomial: p(x) = (x+2)(x)(x−2) = x(x2−4) = x3 − 4x [1½ marks]
Note: any scalar multiple also accepted.
One valid polynomial: p(x) = x3 − 4x
21
3 Marks
Verify zeroes 3, −1, −1/3 of p(x) = 3x3 − 5x2 − 11x − 3.
p(3) = 3(27) − 5(9) − 33 − 3 = 81−45−33−3 = 0 ✓ [1 mark]
p(−1) = 3(−1) − 5(1) − 11(−1) − 3 = −3−5+11−3 = 0 ✓ [1 mark]
p(−1/3) = 3(−1/27) − 5(1/9) − 11(−1/3) − 3
= −1/9 − 5/9 + 11/3 − 3 = −6/9 + 33/9 − 27/9 = 0/9 = 0 ✓ [1 mark]
p(−1) = 3(−1) − 5(1) − 11(−1) − 3 = −3−5+11−3 = 0 ✓ [1 mark]
p(−1/3) = 3(−1/27) − 5(1/9) − 11(−1/3) − 3
= −1/9 − 5/9 + 11/3 − 3 = −6/9 + 33/9 − 27/9 = 0/9 = 0 ✓ [1 mark]
22
3 Marks
f(x) = x2 − 5x + k, α − β = 1. Find k.
By Vieta’s: α + β = 5 and αβ = k [1 mark]
Given: α − β = 1 ...(i)
From α+β=5 and α−β=1: adding ⇒ 2α = 6 ⇒ α = 3, β = 2 [1 mark]
αβ = 3×2 = 6 = k [1 mark]
Given: α − β = 1 ...(i)
From α+β=5 and α−β=1: adding ⇒ 2α = 6 ⇒ α = 3, β = 2 [1 mark]
αβ = 3×2 = 6 = k [1 mark]
k = 6
Section D — Long Answer (4 marks each)
23
4 Marks
Zeroes of p(x) = 4s2 − 4s + 1 and full verification.
Factorise: 4s2 − 4s + 1 = (2s−1)2 [1 mark]
(2s−1)2 = 0 ⇒ s = 1/2 (repeated zero) [1 mark]
Verification (a=4, b=−4, c=1):
Sum: 1/2 + 1/2 = 1 = −(−4)/4 = 4/4 = 1 ✓ [1 mark]
Product: (1/2)(1/2) = 1/4 = c/a = 1/4 ✓ [½ mark]
Geometric interpretation: parabola is tangent to x-axis at s = 1/2 (touches, does not cross) [½ mark]
(2s−1)2 = 0 ⇒ s = 1/2 (repeated zero) [1 mark]
Verification (a=4, b=−4, c=1):
Sum: 1/2 + 1/2 = 1 = −(−4)/4 = 4/4 = 1 ✓ [1 mark]
Product: (1/2)(1/2) = 1/4 = c/a = 1/4 ✓ [½ mark]
Geometric interpretation: parabola is tangent to x-axis at s = 1/2 (touches, does not cross) [½ mark]
Zero: s = 1/2 (repeated). Geometric: tangent touch.
24
4 Marks
Verify all three Vieta’s relations for 2x3 − 5x2 − 14x + 8 (zeroes: 4, −2, 1/2).
a=2, b=−5, c=−14, d=8
(i) Sum: 4+(−2)+1/2 = 2+1/2 = 5/2 = −(−5)/2 = 5/2 ✓ [1⅓ mark]
(ii) Pairwise: (4)(−2)+(−2)(1/2)+(1/2)(4) = −8−1+2 = −7 = −14/2 = c/a ✓ [1⅓ mark]
(iii) Product: 4×(−2)×(1/2) = −4 = −8/2 = −d/a ✓ [1⅓ mark]
(i) Sum: 4+(−2)+1/2 = 2+1/2 = 5/2 = −(−5)/2 = 5/2 ✓ [1⅓ mark]
(ii) Pairwise: (4)(−2)+(−2)(1/2)+(1/2)(4) = −8−1+2 = −7 = −14/2 = c/a ✓ [1⅓ mark]
(iii) Product: 4×(−2)×(1/2) = −4 = −8/2 = −d/a ✓ [1⅓ mark]
All three Vieta’s relations verified.
25
4 Marks
Sum of zeroes = Product of zeroes for p(x) = kx2 + 2x + 3k.
Sum of zeroes = −2/k Product of zeroes = 3k/k = 3 [1 mark]
Given: Sum = Product ⇒ −2/k = 3 [1 mark]
k = −2/3 [1 mark]
Polynomial: (−2/3)x2 + 2x + 3(−2/3) = (−2/3)x2 + 2x − 2
Multiply by −3/2: x2 − 3x + 3 (or leave in original form)
Zeroes: use quadratic formula on x2 − 3x + 3: disc = 9−12 = −3 < 0 → no real zeroes [1 mark for correct working]
Given: Sum = Product ⇒ −2/k = 3 [1 mark]
k = −2/3 [1 mark]
Polynomial: (−2/3)x2 + 2x + 3(−2/3) = (−2/3)x2 + 2x − 2
Multiply by −3/2: x2 − 3x + 3 (or leave in original form)
Zeroes: use quadratic formula on x2 − 3x + 3: disc = 9−12 = −3 < 0 → no real zeroes [1 mark for correct working]
k = −2/3. The resulting polynomial has no real zeroes.
26
4 Marks
Find remaining zeroes of 3x3−5x2−11x−3 given one zero is 3.
Let other zeroes be β, γ. a=3, b=−5, c=−11, d=−3.
(i) Sum: 3+β+γ = −(−5)/3 = 5/3
⇒ β+γ = 5/3−3 = 5/3−9/3 = −4/3 [1 mark]
(ii) Pairwise: 3β+3γ+βγ = c/a = −11/3
3(β+γ)+βγ = −11/3
3(−4/3)+βγ = −11/3
−4+βγ = −11/3
βγ = −11/3+4 = −11/3+12/3 = 1/3 [1 mark]
(iii) β,γ are zeroes of t2−(β+γ)t+βγ = t2+(4/3)t+1/3
Multiply by 3: 3t2+4t+1 = (3t+1)(t+1) [1 mark]
⇒ t = −1/3 or t = −1 [1 mark]
(i) Sum: 3+β+γ = −(−5)/3 = 5/3
⇒ β+γ = 5/3−3 = 5/3−9/3 = −4/3 [1 mark]
(ii) Pairwise: 3β+3γ+βγ = c/a = −11/3
3(β+γ)+βγ = −11/3
3(−4/3)+βγ = −11/3
−4+βγ = −11/3
βγ = −11/3+4 = −11/3+12/3 = 1/3 [1 mark]
(iii) β,γ are zeroes of t2−(β+γ)t+βγ = t2+(4/3)t+1/3
Multiply by 3: 3t2+4t+1 = (3t+1)(t+1) [1 mark]
⇒ t = −1/3 or t = −1 [1 mark]
All three zeroes: 3, −1, −1/3
Case-Based Questions — Answer Key
C1
2 Marks
h(x) = −x3 + 4x = 0. Factorise.
−x(x2−4) = 0 ⇒ −x(x−2)(x+2) = 0 [1 mark]
x = 0, x = 2, x = −2 [1 mark]
Track meets ground at x = −2, 0, and 2.
x = 0, x = 2, x = −2 [1 mark]
Track meets ground at x = −2, 0, and 2.
C2
2 Marks
Degree of h(x) and maximum crossings.
Degree = 3 (highest power of x is x3) [1 mark]
Maximum crossings = 3 (a degree-n polynomial crosses x-axis at MOST n times) [1 mark]
Maximum crossings = 3 (a degree-n polynomial crosses x-axis at MOST n times) [1 mark]
C3
3 Marks
Verify Vieta’s for h(x) = −x3+4x (a=−1, b=0, c=4, d=0). Zeroes: 0, 2, −2.
Sum: 0+2+(−2) = 0 = −b/a = −0/(−1) = 0 ✓ [1 mark]
Pairwise: (0)(2)+(2)(−2)+(−2)(0) = 0−4+0 = −4 = c/a = 4/(−1) = −4 ✓ [1 mark]
Product: 0×2×(−2) = 0 = −d/a = −0/(−1) = 0 ✓ [1 mark]
Pairwise: (0)(2)+(2)(−2)+(−2)(0) = 0−4+0 = −4 = c/a = 4/(−1) = −4 ✓ [1 mark]
Product: 0×2×(−2) = 0 = −d/a = −0/(−1) = 0 ✓ [1 mark]
C4
3 Marks
Quadratic with single tangent touch at x = 2.
Geometric description: parabola touches x-axis at exactly one point (vertex on x-axis), does not cross — this is Case (ii): two coincident zeroes [1 mark]
For a tangent touch at x = 2: zero of multiplicity 2
Polynomial: (x−2)2 = x2 − 4x + 4 [1 mark]
Verify: opens upward (a=1 > 0), vertex at (2,0), touches x-axis at x=2 only [1 mark]
For a tangent touch at x = 2: zero of multiplicity 2
Polynomial: (x−2)2 = x2 − 4x + 4 [1 mark]
Verify: opens upward (a=1 > 0), vertex at (2,0), touches x-axis at x=2 only [1 mark]
Polynomial: (x−2)2 = x2 − 4x + 4
C5
2 Marks
Zeroes of A(w) = 2w2 + 3w.
A(w) = w(2w + 3) = 0 [1 mark]
w = 0 or w = −3/2 [½ mark]
Context: w = 0 means zero width (no field). w = −3/2 is negative (invalid for width) [½ mark]
w = 0 or w = −3/2 [½ mark]
Context: w = 0 means zero width (no field). w = −3/2 is negative (invalid for width) [½ mark]
C6
2 Marks
Verify sum and product for 2w2 + 3w (a=2, b=3, c=0).
Sum: 0 + (−3/2) = −3/2 = −b/a = −3/2 ✓ [1 mark]
Product: 0 × (−3/2) = 0 = c/a = 0/2 = 0 ✓ [1 mark]
Product: 0 × (−3/2) = 0 = c/a = 0/2 = 0 ✓ [1 mark]
C7
3 Marks
Solve 2w2 + 3w = 45.
2w2 + 3w − 45 = 0 [1 mark]
Split middle term: product = 2×(−45) = −90, sum = 3
Factors: 15 and −6 ⇒ 2w2+15w−12w−45 = w(2w+15)−3(2w+15) = (w−3)(2w+15) [1 mark]
w = 3 or w = −15/2
Negative root w = −15/2 is not valid (width cannot be negative) [1 mark]
Split middle term: product = 2×(−45) = −90, sum = 3
Factors: 15 and −6 ⇒ 2w2+15w−12w−45 = w(2w+15)−3(2w+15) = (w−3)(2w+15) [1 mark]
w = 3 or w = −15/2
Negative root w = −15/2 is not valid (width cannot be negative) [1 mark]
w = 3 metres (width of field)
C8
3 Marks
Quadratic with zeroes √2 and −√2.
Sum of zeroes: √2 + (−√2) = 0 Product: (√2)(−√2) = −2 [1 mark]
Polynomial: x2 − 0·x + (−2) = x2 − 2 [1 mark]
Verification: −b/a = 0 ✓, c/a = −2 ✓ [½ mark]
Graph description: parabola opens upward (a=1 > 0), crosses x-axis at √2 and −√2 [½ mark]
Polynomial: x2 − 0·x + (−2) = x2 − 2 [1 mark]
Verification: −b/a = 0 ✓, c/a = −2 ✓ [½ mark]
Graph description: parabola opens upward (a=1 > 0), crosses x-axis at √2 and −√2 [½ mark]
Polynomial: x2 − 2
Marks Summary
SEC A
10
MCQ × 1
SEC B
12
VSAQ × 2
SEC C
18
SAQ × 3
SEC D
20
LAQ × 4
TOTAL
80
CBQ = 20
