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CHAPTER 2 — INTRODUCTION TO LINEAR POLYNOMIALS
Class IX Mathematics — Polynomials
Preeti Kushwah Classes — Unit Test (Set 4: HOTS & Mixed Application)
General Instructions:
1. All questions are compulsory.
2. Section A has 6 questions of 1 mark each.
3. Section B has 5 questions of 2 marks each.
4. Section C has 4 questions of 3 marks each.
5. Section D has 2 questions of 5 marks each.
6. Bonus question is optional (2 marks).
7. Draw neat figures wherever required. Show all working steps.
1. All questions are compulsory.
2. Section A has 6 questions of 1 mark each.
3. Section B has 5 questions of 2 marks each.
4. Section C has 4 questions of 3 marks each.
5. Section D has 2 questions of 5 marks each.
6. Bonus question is optional (2 marks).
7. Draw neat figures wherever required. Show all working steps.
Section A — (1 Mark Each) [6 × 1 = 6]
Q1.1
If p(x) = x² − 3x + 2 and p(1) = 0, which factor of p(x) does this give?
Q2.1
The remainder when any polynomial p(x) is divided by (x − 0) is p(___).
Q3.1
A shopkeeper earns P(x) = 8x − 400. The break-even point (P = 0) is at x = ___ items.
Q4.1
The graph below shows a linear polynomial crossing the x-axis at x = −2 and the y-axis at y = 4. The polynomial is p(x) = ___ x + ___.
Q5.1
If both (x − 1) and (x − 2) are factors of p(x), then p(1) = ___ and p(2) = ___.
Q6.1
The zero of p(x) = −(3/4)x + 6 is ___.
Section B — (2 Marks Each) [5 × 2 = 10]
Q7.2
Find the remainder when p(x) = 4x³ − 12x² + 14x − 3 is divided by (2x − 1). (Hint: set 2x − 1 = 0)
Q8.2
If x = −1 is a zero of p(x) = x² + kx − 6, find k.
Q9.2
Two linear polynomials A and B are graphed below.
(i) Which polynomial has a positive slope?
(ii) Write the zeros of A and B.
(i) Which polynomial has a positive slope?
(ii) Write the zeros of A and B.
Q10.2
A student's marks are modelled by M(h) = 5h + 30, where h = hours studied. Find h when M = 80. What does the zero of M(h) represent?
Q11.2
Check whether x = ½ is a zero of p(x) = 2x² − 3x + 1.
Section C — (3 Marks Each) [4 × 3 = 12]
Q12.3
If (x − 2) is a factor of p(x) = x² − kx + 2, find k. Hence find all zeros of p(x).
Q13.3
Find the remainder when p(x) = x³ + 3x² + 3x + 1 is divided by:
(i) x + 1 (ii) x − ½ (iii) x
What do the results tell you?
(i) x + 1 (ii) x − ½ (iii) x
What do the results tell you?
Q14.3
The graph of p(x) = 2x − 6 is shown below. From the graph:
(i) State the zero.
(ii) State the y-intercept.
(iii) Verify: does −b/a match the zero on the graph?
(i) State the zero.
(ii) State the y-intercept.
(iii) Verify: does −b/a match the zero on the graph?
Q15.3
A taxi company charges: Fare = 12d + 50, where d = distance in km.
(i) Is this a polynomial? If so, what type?
(ii) For what distance is the fare ₹290?
(iii) What does the constant 50 represent in real life?
(i) Is this a polynomial? If so, what type?
(ii) For what distance is the fare ₹290?
(iii) What does the constant 50 represent in real life?
Section D — (5 Marks Each) [2 × 5 = 10]
Q16.5
Mixed problem: p(x) = x³ − 4x² + x + 6.
(a) Find p(−1), p(2), p(3). [2 marks]
(b) Which values give p = 0? Write the corresponding factors. [1 mark]
(c) Hence factorise p(x) completely. [1 mark]
(d) Verify your answer by expanding the factors. [1 mark]
(a) Find p(−1), p(2), p(3). [2 marks]
(b) Which values give p = 0? Write the corresponding factors. [1 mark]
(c) Hence factorise p(x) completely. [1 mark]
(d) Verify your answer by expanding the factors. [1 mark]
Q17.5
A factory produces x items daily. Revenue: R(x) = 20x. Cost: C(x) = 5x + 750.
(a) Write the profit polynomial P(x) = R(x) − C(x). Simplify. [1 mark]
(b) What type of polynomial is P(x)? What is its degree? [1 mark]
(c) Find the break-even point (P(x) = 0). [1 mark]
(d) From the graph below, read off: for what x values is the factory making a loss? [1 mark]
(e) The factory wants a profit of ₹1500. How many items must it produce? [1 mark]
(a) Write the profit polynomial P(x) = R(x) − C(x). Simplify. [1 mark]
(b) What type of polynomial is P(x)? What is its degree? [1 mark]
(c) Find the break-even point (P(x) = 0). [1 mark]
(d) From the graph below, read off: for what x values is the factory making a loss? [1 mark]
Bonus Question (Optional) [2 Marks]
Q18.2
★ Two friends argue: Arun says “every polynomial of degree n has exactly n zeros” and Bina says “every polynomial of degree n has AT MOST n zeros.” Who is correct? Give an example of a degree-2 polynomial with:
(i) exactly 2 zeros
(ii) exactly 1 zero
(iii) no real zeros
to support your answer.
(i) exactly 2 zeros
(ii) exactly 1 zero
(iii) no real zeros
to support your answer.
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Answer Key & Detailed Solutions
Q1. [1 Mark]
Since p(1) = 0, by the Factor Theorem, (x − 1) is a factor of p(x).
Verification: p(x) = x² − 3x + 2 = (x − 1)(x − 2).
Verification: p(x) = x² − 3x + 2 = (x − 1)(x − 2).
Q2. [1 Mark]
By the Remainder Theorem, the remainder when p(x) is divided by (x − a) is p(a).
Here a = 0, so the remainder is p(0).
Here a = 0, so the remainder is p(0).
Q3. [1 Mark]
Set P(x) = 0:
8x − 400 = 0
8x = 400
x = 50 items.
8x − 400 = 0
8x = 400
x = 50 items.
Q4. [1 Mark]
The line crosses the x-axis at (−2, 0) and the y-axis at (0, 4).
y-intercept = 4, so constant term = 4.
Slope = (4 − 0) / (0 − (−2)) = 4/2 = 2.
So p(x) = 2x + 4.
y-intercept = 4, so constant term = 4.
Slope = (4 − 0) / (0 − (−2)) = 4/2 = 2.
So p(x) = 2x + 4.
Q5. [1 Mark]
By the Factor Theorem, if (x − a) is a factor of p(x), then p(a) = 0.
So p(1) = 0 and p(2) = 0.
So p(1) = 0 and p(2) = 0.
Q6. [1 Mark]
Set p(x) = 0:
−(3/4)x + 6 = 0
(3/4)x = 6
x = 6 × (4/3) = 8.
−(3/4)x + 6 = 0
(3/4)x = 6
x = 6 × (4/3) = 8.
Q7. [2 Marks]
Divisor: (2x − 1). Set 2x − 1 = 0 ⇒ x = 1/2. [½ mark]
By Remainder Theorem, remainder = p(1/2):
p(1/2) = 4(1/2)³ − 12(1/2)² + 14(1/2) − 3 [½ mark]
= 4(1/8) − 12(1/4) + 7 − 3
= 1/2 − 3 + 7 − 3 [½ mark]
= 3/2 (or 1.5). [½ mark]
By Remainder Theorem, remainder = p(1/2):
p(1/2) = 4(1/2)³ − 12(1/2)² + 14(1/2) − 3 [½ mark]
= 4(1/8) − 12(1/4) + 7 − 3
= 1/2 − 3 + 7 − 3 [½ mark]
= 3/2 (or 1.5). [½ mark]
Q8. [2 Marks]
Since x = −1 is a zero, p(−1) = 0. [½ mark]
(−1)² + k(−1) − 6 = 0 [½ mark]
1 − k − 6 = 0
−k − 5 = 0
k = −5. [1 mark]
(−1)² + k(−1) − 6 = 0 [½ mark]
1 − k − 6 = 0
−k − 5 = 0
k = −5. [1 mark]
Q9. [2 Marks]
(i) Polynomial A has a positive slope (line goes upward from left to right). [1 mark]
(ii) From the graph:
Zero of A = 1 (line A crosses x-axis at x = 1).
Zero of B = −3 (line B crosses x-axis at x = −3). [1 mark]
(ii) From the graph:
Zero of A = 1 (line A crosses x-axis at x = 1).
Zero of B = −3 (line B crosses x-axis at x = −3). [1 mark]
Q10. [2 Marks]
Finding h when M = 80:
5h + 30 = 80
5h = 50
h = 10 hours. [1 mark]
Zero of M(h): Set M(h) = 0:
5h + 30 = 0 ⇒ h = −6.
The zero is h = −6. Since negative hours are not meaningful, the zero tells us that the model predicts 0 marks at h = −6, which indicates the model is only valid for h ≥ 0. In context, it means the student starts with 30 base marks (at h = 0), and the zero has no practical meaning in this scenario. [1 mark]
5h + 30 = 80
5h = 50
h = 10 hours. [1 mark]
Zero of M(h): Set M(h) = 0:
5h + 30 = 0 ⇒ h = −6.
The zero is h = −6. Since negative hours are not meaningful, the zero tells us that the model predicts 0 marks at h = −6, which indicates the model is only valid for h ≥ 0. In context, it means the student starts with 30 base marks (at h = 0), and the zero has no practical meaning in this scenario. [1 mark]
Q11. [2 Marks]
Substitute x = 1/2 in p(x) = 2x² − 3x + 1: [½ mark]
p(1/2) = 2(1/2)² − 3(1/2) + 1 [½ mark]
= 2(1/4) − 3/2 + 1
= 1/2 − 3/2 + 1
= 1/2 − 3/2 + 2/2
= 0. [½ mark]
Since p(1/2) = 0, yes, x = 1/2 is a zero of p(x). [½ mark]
p(1/2) = 2(1/2)² − 3(1/2) + 1 [½ mark]
= 2(1/4) − 3/2 + 1
= 1/2 − 3/2 + 1
= 1/2 − 3/2 + 2/2
= 0. [½ mark]
Since p(1/2) = 0, yes, x = 1/2 is a zero of p(x). [½ mark]
Q12. [3 Marks]
Finding k: Since (x − 2) is a factor, p(2) = 0. [½ mark]
(2)² − k(2) + 2 = 0
4 − 2k + 2 = 0
6 − 2k = 0
k = 3. [1 mark]
Finding zeros: p(x) = x² − 3x + 2 = (x − 1)(x − 2). [1 mark]
Zeros: x = 1 and x = 2. [½ mark]
(2)² − k(2) + 2 = 0
4 − 2k + 2 = 0
6 − 2k = 0
k = 3. [1 mark]
Finding zeros: p(x) = x² − 3x + 2 = (x − 1)(x − 2). [1 mark]
Zeros: x = 1 and x = 2. [½ mark]
Q13. [3 Marks]
p(x) = x³ + 3x² + 3x + 1.
(i) Divisor: (x + 1). Set x + 1 = 0 ⇒ x = −1.
p(−1) = (−1)³ + 3(−1)² + 3(−1) + 1 = −1 + 3 − 3 + 1 = 0.
Remainder = 0. This means (x + 1) is a factor. [1 mark]
(ii) Divisor: (x − ½). Set x = 1/2.
p(1/2) = (1/2)³ + 3(1/2)² + 3(1/2) + 1 = 1/8 + 3/4 + 3/2 + 1 = 1/8 + 6/8 + 12/8 + 8/8 = 27/8.
Remainder = 27/8. [1 mark]
(iii) Divisor: x. Set x = 0.
p(0) = 0 + 0 + 0 + 1 = 1.
Remainder = 1. This equals the constant term. [½ mark]
Observation: Result (i) shows (x + 1) is a factor of p(x). In fact, p(x) = (x + 1)³. Result (iii) shows that the remainder on dividing by x is always the constant term of the polynomial. [½ mark]
(i) Divisor: (x + 1). Set x + 1 = 0 ⇒ x = −1.
p(−1) = (−1)³ + 3(−1)² + 3(−1) + 1 = −1 + 3 − 3 + 1 = 0.
Remainder = 0. This means (x + 1) is a factor. [1 mark]
(ii) Divisor: (x − ½). Set x = 1/2.
p(1/2) = (1/2)³ + 3(1/2)² + 3(1/2) + 1 = 1/8 + 3/4 + 3/2 + 1 = 1/8 + 6/8 + 12/8 + 8/8 = 27/8.
Remainder = 27/8. [1 mark]
(iii) Divisor: x. Set x = 0.
p(0) = 0 + 0 + 0 + 1 = 1.
Remainder = 1. This equals the constant term. [½ mark]
Observation: Result (i) shows (x + 1) is a factor of p(x). In fact, p(x) = (x + 1)³. Result (iii) shows that the remainder on dividing by x is always the constant term of the polynomial. [½ mark]
Q14. [3 Marks]
p(x) = 2x − 6. Here a = 2, b = −6.
(i) From the graph, the line crosses the x-axis at x = 3. Zero = 3. [1 mark]
(ii) From the graph, the line crosses the y-axis at y = −6. Y-intercept = (0, −6). [1 mark]
(iii) Verification: −b/a = −(−6)/2 = 6/2 = 3.
Yes, −b/a = 3 matches the zero read from the graph. ✓ [1 mark]
(i) From the graph, the line crosses the x-axis at x = 3. Zero = 3. [1 mark]
(ii) From the graph, the line crosses the y-axis at y = −6. Y-intercept = (0, −6). [1 mark]
(iii) Verification: −b/a = −(−6)/2 = 6/2 = 3.
Yes, −b/a = 3 matches the zero read from the graph. ✓ [1 mark]
Q15. [3 Marks]
Fare(d) = 12d + 50.
(i) Yes, it is a polynomial in d. It is a linear polynomial (degree 1) with coefficient 12 and constant 50. [1 mark]
(ii) Set Fare = 290:
12d + 50 = 290
12d = 240
d = 20 km. [1 mark]
(iii) The constant 50 represents the base/minimum fare (the fixed charge before any distance is covered). It is the fare at d = 0, i.e., the amount charged the moment you board the taxi. [1 mark]
(i) Yes, it is a polynomial in d. It is a linear polynomial (degree 1) with coefficient 12 and constant 50. [1 mark]
(ii) Set Fare = 290:
12d + 50 = 290
12d = 240
d = 20 km. [1 mark]
(iii) The constant 50 represents the base/minimum fare (the fixed charge before any distance is covered). It is the fare at d = 0, i.e., the amount charged the moment you board the taxi. [1 mark]
Q16. [5 Marks]
p(x) = x³ − 4x² + x + 6.
(a) p(−1) = (−1)³ − 4(−1)² + (−1) + 6 = −1 − 4 − 1 + 6 = 0. [⅔ mark]
p(2) = (2)³ − 4(2)² + (2) + 6 = 8 − 16 + 2 + 6 = 0. [⅔ mark]
p(3) = (3)³ − 4(3)² + (3) + 6 = 27 − 36 + 3 + 6 = 0. [⅔ mark]
(b) p(−1) = 0, p(2) = 0, p(3) = 0.
All three give p = 0.
Corresponding factors: (x + 1), (x − 2), (x − 3). [1 mark]
(c) Since p(x) is degree 3 and has three linear factors:
p(x) = (x + 1)(x − 2)(x − 3). [1 mark]
(d) Verification:
(x + 1)(x − 2) = x² − x − 2.
(x² − x − 2)(x − 3) = x³ − 3x² − x² + 3x − 2x + 6
= x³ − 4x² + x + 6 = p(x). ✓ [1 mark]
(a) p(−1) = (−1)³ − 4(−1)² + (−1) + 6 = −1 − 4 − 1 + 6 = 0. [⅔ mark]
p(2) = (2)³ − 4(2)² + (2) + 6 = 8 − 16 + 2 + 6 = 0. [⅔ mark]
p(3) = (3)³ − 4(3)² + (3) + 6 = 27 − 36 + 3 + 6 = 0. [⅔ mark]
(b) p(−1) = 0, p(2) = 0, p(3) = 0.
All three give p = 0.
Corresponding factors: (x + 1), (x − 2), (x − 3). [1 mark]
(c) Since p(x) is degree 3 and has three linear factors:
p(x) = (x + 1)(x − 2)(x − 3). [1 mark]
(d) Verification:
(x + 1)(x − 2) = x² − x − 2.
(x² − x − 2)(x − 3) = x³ − 3x² − x² + 3x − 2x + 6
= x³ − 4x² + x + 6 = p(x). ✓ [1 mark]
Q17. [5 Marks]
R(x) = 20x, C(x) = 5x + 750.
(a) P(x) = R(x) − C(x) = 20x − (5x + 750) = 20x − 5x − 750 = 15x − 750. [1 mark]
(b) P(x) = 15x − 750 is a linear polynomial (degree 1). [1 mark]
(c) Break-even: P(x) = 0.
15x − 750 = 0
15x = 750
x = 50 items. [1 mark]
(d) From the graph, P(x) < 0 (below x-axis) when x < 50.
The factory makes a loss when it produces fewer than 50 items per day. [1 mark]
(e) Set P(x) = 1500:
15x − 750 = 1500
15x = 2250
x = 150 items. [1 mark]
(a) P(x) = R(x) − C(x) = 20x − (5x + 750) = 20x − 5x − 750 = 15x − 750. [1 mark]
(b) P(x) = 15x − 750 is a linear polynomial (degree 1). [1 mark]
(c) Break-even: P(x) = 0.
15x − 750 = 0
15x = 750
x = 50 items. [1 mark]
(d) From the graph, P(x) < 0 (below x-axis) when x < 50.
The factory makes a loss when it produces fewer than 50 items per day. [1 mark]
(e) Set P(x) = 1500:
15x − 750 = 1500
15x = 2250
x = 150 items. [1 mark]
Q18. Bonus [2 Marks]
Bina is correct. A polynomial of degree n has at most n real zeros, not necessarily exactly n. [½ mark]
Examples (degree 2):
(i) Exactly 2 zeros: p(x) = x² − 1 = (x − 1)(x + 1).
Zeros: x = 1 and x = −1. [½ mark]
(ii) Exactly 1 zero: p(x) = x² = (x)(x).
Zero: x = 0 only (repeated root). [½ mark]
(iii) No real zeros: p(x) = x² + 1.
x² + 1 = 0 ⇒ x² = −1, which has no real solution.
This degree-2 polynomial has 0 real zeros. [½ mark]
This proves Arun is wrong — not every polynomial of degree n has exactly n zeros. Bina’s statement (“at most n”) is the correct one.
Examples (degree 2):
(i) Exactly 2 zeros: p(x) = x² − 1 = (x − 1)(x + 1).
Zeros: x = 1 and x = −1. [½ mark]
(ii) Exactly 1 zero: p(x) = x² = (x)(x).
Zero: x = 0 only (repeated root). [½ mark]
(iii) No real zeros: p(x) = x² + 1.
x² + 1 = 0 ⇒ x² = −1, which has no real solution.
This degree-2 polynomial has 0 real zeros. [½ mark]
This proves Arun is wrong — not every polynomial of degree n has exactly n zeros. Bina’s statement (“at most n”) is the correct one.