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CHAPTER 2 — INTRODUCTION TO LINEAR POLYNOMIALS
Class IX Mathematics — Linear Polynomials
Preeti Kushwah Classes — Unit Test (Set 2: Zeros & Remainder Theorem)
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. Show all working steps clearly. Marks are awarded for correct method even if the final answer is wrong.
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. Show all working steps clearly. Marks are awarded for correct method even if the final answer is wrong.
Section A — (1 Mark Each) [6 × 1 = 6]
Q1.1
The zero of p(x) = 3x − 6 is:
(a) −2 (b) 2 (c) 6 (d) 3
(a) −2 (b) 2 (c) 6 (d) 3
Q2.1
If p(x) = x² − 5x + 6, then p(2) = ___.
Q3.1
The remainder when p(x) is divided by (x − a) is ___ (state the theorem).
Q4.1
How many zeros can a linear polynomial have?
Q5.1
The zero of p(x) = −4x + 12 is ___.
Q6.1
True or False: A non-zero constant polynomial has no zero.
Section B — (2 Marks Each) [5 × 2 = 10]
Q7.2
Find the zero of p(x) = 7x − 21 and verify by substitution.
Q8.2
If p(x) = x² + 3x + 2, find p(0) and p(−1).
Q9.2
Find the remainder when p(x) = x² − 5x + 6 is divided by (x − 2).
Q10.2
The zero of p(x) = kx − 8 is x = 2. Find the value of k.
Q11.2
Find the zero of p(x) = ⅔x + 4.
Section C — (3 Marks Each) [4 × 3 = 12]
Q12.3
Find the remainder when p(x) = x³ + 3x² − 5x + 2 is divided by (x − 1). Is (x − 1) a factor of p(x)?
Q13.3
Find the remainder when p(x) = 2x² + 3x − 1 is divided by (x + 2). Show all steps.
Q14.3
If x = 3 is a zero of p(x) = 2x² − kx + 3, find the value of k.
Q15.3
A taxi charges ₹25 per km plus a base fare of ₹50. Express the total fare as a linear polynomial in distance d. Find the distance for which the fare is ₹300.
Section D — (5 Marks Each) [2 × 5 = 10]
Q16.5
(a) State the Remainder Theorem. [1]
(b) Prove it using the division algorithm. [1]
(c) Use it to find the remainder when p(x) = x³ − 6x² + 11x − 6 is divided by (x − 2). [1]
(d) What does the result tell you about (x − 2)? [1]
(e) Can you guess another factor of p(x)? [1]
(b) Prove it using the division algorithm. [1]
(c) Use it to find the remainder when p(x) = x³ − 6x² + 11x − 6 is divided by (x − 2). [1]
(d) What does the result tell you about (x − 2)? [1]
(e) Can you guess another factor of p(x)? [1]
Q17.5
Find the zeros of each polynomial and verify:
(i) p(x) = 4x − 20 [1]
(ii) p(x) = −½x + 3 [1]
(iii) p(x) = 3x [1]
(iv) If the zero of p(x) = ax + 6 is x = −3, find a. [1]
(v) For what value of x is 2x + 7 equal to 15? [1]
(i) p(x) = 4x − 20 [1]
(ii) p(x) = −½x + 3 [1]
(iii) p(x) = 3x [1]
(iv) If the zero of p(x) = ax + 6 is x = −3, find a. [1]
(v) For what value of x is 2x + 7 equal to 15? [1]
Bonus Question (Optional) [2 Marks]
Q18.2
★ p(x) = x² − 5x + 6. Without finding zeros, determine whether x = 2 AND x = 3 are both zeros. What can you conclude about the factorisation of p(x)?
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Answer Key & Detailed Solutions
Q1. [1 Mark]
(b) 2
Set p(x) = 0: 3x − 6 = 0 → 3x = 6 → x = 2.
Set p(x) = 0: 3x − 6 = 0 → 3x = 6 → x = 2.
Q2. [1 Mark]
p(2) = (2)² − 5(2) + 6 = 4 − 10 + 6 = 0.
Q3. [1 Mark]
Remainder Theorem: The remainder when p(x) is divided by (x − a) is p(a).
Q4. [1 Mark]
A linear polynomial has exactly one zero.
(A polynomial of degree 1 has at most 1 zero, and it always exists.)
(A polynomial of degree 1 has at most 1 zero, and it always exists.)
Q5. [1 Mark]
Set p(x) = 0: −4x + 12 = 0 → −4x = −12 → x = 3.
Q6. [1 Mark]
True. A non-zero constant polynomial (e.g., p(x) = 5) never equals zero for any value of x, so it has no zero.
Q7. [2 Marks]
Finding zero: Set p(x) = 0:
7x − 21 = 0 → 7x = 21 → x = 3. [1 mark]
Verification: p(3) = 7(3) − 21 = 21 − 21 = 0 ✔ [1 mark]
7x − 21 = 0 → 7x = 21 → x = 3. [1 mark]
Verification: p(3) = 7(3) − 21 = 21 − 21 = 0 ✔ [1 mark]
Q8. [2 Marks]
p(0) = (0)² + 3(0) + 2 = 0 + 0 + 2 = 2. [1 mark]
p(−1) = (−1)² + 3(−1) + 2 = 1 − 3 + 2 = 0. [1 mark]
(This means x = −1 is a zero of p(x).)
p(−1) = (−1)² + 3(−1) + 2 = 1 − 3 + 2 = 0. [1 mark]
(This means x = −1 is a zero of p(x).)
Q9. [2 Marks]
By the Remainder Theorem, remainder = p(2). [1 mark]
p(2) = (2)² − 5(2) + 6 = 4 − 10 + 6 = 0. [1 mark]
Since the remainder is 0, (x − 2) is a factor of p(x).
p(2) = (2)² − 5(2) + 6 = 4 − 10 + 6 = 0. [1 mark]
Since the remainder is 0, (x − 2) is a factor of p(x).
Q10. [2 Marks]
If x = 2 is a zero, then p(2) = 0. [1 mark]
k(2) − 8 = 0 → 2k = 8 → k = 4. [1 mark]
k(2) − 8 = 0 → 2k = 8 → k = 4. [1 mark]
Q11. [2 Marks]
Set p(x) = 0: ⅔x + 4 = 0 [1 mark]
⅔x = −4
x = −4 × (3/2) = −6. [1 mark]
Verification: p(−6) = ⅔(−6) + 4 = −4 + 4 = 0 ✔
⅔x = −4
x = −4 × (3/2) = −6. [1 mark]
Verification: p(−6) = ⅔(−6) + 4 = −4 + 4 = 0 ✔
Q12. [3 Marks]
By the Remainder Theorem, remainder = p(1). [1 mark]
p(1) = (1)³ + 3(1)² − 5(1) + 2
= 1 + 3 − 5 + 2 = 1. [1 mark]
Since the remainder is 1 ≠ 0, (x − 1) is NOT a factor of p(x). [1 mark]
p(1) = (1)³ + 3(1)² − 5(1) + 2
= 1 + 3 − 5 + 2 = 1. [1 mark]
Since the remainder is 1 ≠ 0, (x − 1) is NOT a factor of p(x). [1 mark]
Q13. [3 Marks]
Dividing by (x + 2) means we use the Remainder Theorem with a = −2. [1 mark]
Remainder = p(−2):
p(−2) = 2(−2)² + 3(−2) − 1
= 2(4) + (−6) − 1
= 8 − 6 − 1 = 1. [1 mark]
The remainder when p(x) is divided by (x + 2) is 1. [1 mark]
Remainder = p(−2):
p(−2) = 2(−2)² + 3(−2) − 1
= 2(4) + (−6) − 1
= 8 − 6 − 1 = 1. [1 mark]
The remainder when p(x) is divided by (x + 2) is 1. [1 mark]
Q14. [3 Marks]
If x = 3 is a zero of p(x), then p(3) = 0. [1 mark]
p(3) = 2(3)² − k(3) + 3 = 0
= 2(9) − 3k + 3 = 0
= 18 − 3k + 3 = 0
= 21 − 3k = 0 [1 mark]
3k = 21 → k = 7. [1 mark]
p(3) = 2(3)² − k(3) + 3 = 0
= 2(9) − 3k + 3 = 0
= 18 − 3k + 3 = 0
= 21 − 3k = 0 [1 mark]
3k = 21 → k = 7. [1 mark]
Q15. [3 Marks]
Total fare = Base fare + (Rate per km × distance)
f(d) = 25d + 50 (a linear polynomial in d). [1 mark]
For fare = ₹300:
25d + 50 = 300
25d = 250 [1 mark]
d = 10 km. [1 mark]
f(d) = 25d + 50 (a linear polynomial in d). [1 mark]
For fare = ₹300:
25d + 50 = 300
25d = 250 [1 mark]
d = 10 km. [1 mark]
Q16. [5 Marks]
(a) Remainder Theorem: If p(x) is a polynomial of degree ≥ 1 and a is any real number, then when p(x) is divided by (x − a), the remainder is p(a). [1 mark]
(b) Proof using Division Algorithm:
By the division algorithm: p(x) = (x − a) · q(x) + r
where q(x) is the quotient and r is the remainder (a constant, since divisor is degree 1).
Substituting x = a: p(a) = (a − a) · q(a) + r = 0 · q(a) + r = r.
Therefore, remainder r = p(a). ◼ [1 mark]
(c) Remainder = p(2):
p(2) = (2)³ − 6(2)² + 11(2) − 6
= 8 − 24 + 22 − 6 = 0. [1 mark]
(d) Since the remainder is 0, by the Factor Theorem, (x − 2) is a factor of p(x). [1 mark]
(e) Try x = 1: p(1) = 1 − 6 + 11 − 6 = 0. So (x − 1) is also a factor.
Try x = 3: p(3) = 27 − 54 + 33 − 6 = 0. So (x − 3) is also a factor.
Thus p(x) = (x − 1)(x − 2)(x − 3). [1 mark]
(b) Proof using Division Algorithm:
By the division algorithm: p(x) = (x − a) · q(x) + r
where q(x) is the quotient and r is the remainder (a constant, since divisor is degree 1).
Substituting x = a: p(a) = (a − a) · q(a) + r = 0 · q(a) + r = r.
Therefore, remainder r = p(a). ◼ [1 mark]
(c) Remainder = p(2):
p(2) = (2)³ − 6(2)² + 11(2) − 6
= 8 − 24 + 22 − 6 = 0. [1 mark]
(d) Since the remainder is 0, by the Factor Theorem, (x − 2) is a factor of p(x). [1 mark]
(e) Try x = 1: p(1) = 1 − 6 + 11 − 6 = 0. So (x − 1) is also a factor.
Try x = 3: p(3) = 27 − 54 + 33 − 6 = 0. So (x − 3) is also a factor.
Thus p(x) = (x − 1)(x − 2)(x − 3). [1 mark]
Q17. [5 Marks]
(i) p(x) = 4x − 20 = 0 → 4x = 20 → x = 5.
Verification: p(5) = 4(5) − 20 = 20 − 20 = 0 ✔ [1 mark]
(ii) p(x) = −½x + 3 = 0 → −½x = −3 → x = 6.
Verification: p(6) = −½(6) + 3 = −3 + 3 = 0 ✔ [1 mark]
(iii) p(x) = 3x = 0 → x = 0.
Verification: p(0) = 3(0) = 0 ✔ [1 mark]
(iv) Zero is x = −3, so p(−3) = 0:
a(−3) + 6 = 0 → −3a = −6 → a = 2. [1 mark]
(v) 2x + 7 = 15 → 2x = 8 → x = 4.
Verification: 2(4) + 7 = 8 + 7 = 15 ✔ [1 mark]
Verification: p(5) = 4(5) − 20 = 20 − 20 = 0 ✔ [1 mark]
(ii) p(x) = −½x + 3 = 0 → −½x = −3 → x = 6.
Verification: p(6) = −½(6) + 3 = −3 + 3 = 0 ✔ [1 mark]
(iii) p(x) = 3x = 0 → x = 0.
Verification: p(0) = 3(0) = 0 ✔ [1 mark]
(iv) Zero is x = −3, so p(−3) = 0:
a(−3) + 6 = 0 → −3a = −6 → a = 2. [1 mark]
(v) 2x + 7 = 15 → 2x = 8 → x = 4.
Verification: 2(4) + 7 = 8 + 7 = 15 ✔ [1 mark]
Q18. Bonus [2 Marks]
Check x = 2: p(2) = (2)² − 5(2) + 6 = 4 − 10 + 6 = 0.
So x = 2 is a zero. ✔ [½ mark]
Check x = 3: p(3) = (3)² − 5(3) + 6 = 9 − 15 + 6 = 0.
So x = 3 is a zero. ✔ [½ mark]
Conclusion: Since both x = 2 and x = 3 are zeros, by the Factor Theorem, both (x − 2) and (x − 3) are factors of p(x). [½ mark]
Since p(x) is a quadratic (degree 2), it can have at most 2 factors of degree 1. Therefore:
p(x) = (x − 2)(x − 3). [½ mark]
Verification: (x − 2)(x − 3) = x² − 5x + 6 ✔
So x = 2 is a zero. ✔ [½ mark]
Check x = 3: p(3) = (3)² − 5(3) + 6 = 9 − 15 + 6 = 0.
So x = 3 is a zero. ✔ [½ mark]
Conclusion: Since both x = 2 and x = 3 are zeros, by the Factor Theorem, both (x − 2) and (x − 3) are factors of p(x). [½ mark]
Since p(x) is a quadratic (degree 2), it can have at most 2 factors of degree 1. Therefore:
p(x) = (x − 2)(x − 3). [½ mark]
Verification: (x − 2)(x − 3) = x² − 5x + 6 ✔