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CHAPTER 2 — INTRODUCTION TO LINEAR POLYNOMIALS
Class IX Mathematics — Factor Theorem & Factorisation
Preeti Kushwah Classes — Unit Test (Set 3: Factor Theorem & Factorisation)
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. Draw neat figures wherever required. Show all steps clearly.
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. Draw neat figures wherever required. Show all steps clearly.
Section A — (1 Mark Each) [6 × 1 = 6]
Q1.1
If p(a) = 0, then (x − a) is a ___ of p(x).
Q2.1
Is (x − 1) a factor of p(x) = x² − 1? (Yes/No with reason in one line)
Q3.1
The remainder when p(x) = x² + x + 1 is divided by (x + 1) is ___.
Q4.1
The graph below shows a linear polynomial p(x). From the graph, the zero of the polynomial is ___.
Q5.1
If (x − 2) is a factor of p(x), then p(___) = 0.
Q6.1
True or False: The Factor Theorem is a special case of the Remainder Theorem.
Section B — (2 Marks Each) [5 × 2 = 10]
Q7.2
Check whether (x − 3) is a factor of p(x) = x² − 5x + 6.
Q8.2
Check whether (x + 1) is a factor of p(x) = x³ + x² + x + 1.
Q9.2
Find the value of k if (x − 2) is a factor of p(x) = x² − kx + 4.
Q10.2
The graph below shows two linear polynomials p(x) and q(x). Write the zero of each polynomial from the graph.
Q11.2
Find the remainder when p(x) = 2x³ − 5x² + 3x + 7 is divided by (x − 1). Is (x − 1) a factor?
Section C — (3 Marks Each) [4 × 3 = 12]
Q12.3
Using the Factor Theorem, factorise x² − 7x + 12 completely.
Q13.3
Find the value of k if (x + 2) is a factor of p(x) = x³ + kx² + 7x + 10. Then find the other factors.
Q14.3
The graph of a linear polynomial p(x) = ax + b is shown below.
From the graph: (i) Find the zero of p(x). (ii) Find the y-intercept. (iii) Find the values of a and b.
From the graph: (i) Find the zero of p(x). (ii) Find the y-intercept. (iii) Find the values of a and b.
Q15.3
Check if (x − 1), (x − 2), and (x − 3) are factors of p(x) = x³ − 6x² + 11x − 6. Hence factorise p(x).
Section D — (5 Marks Each) [2 × 5 = 10]
Q16.5
(a) State the Factor Theorem. [1]
(b) Explain how it is related to the Remainder Theorem. [1]
(c) Check if (x + 1) is a factor of x³ + 1. [1]
(d) If (x − 1) is a factor of p(x) = x³ − 2x² + kx + 4, find k. [1]
(e) Using the Factor Theorem, factorise x³ − 3x² + 2x completely. (Hint: first take x common.) [1]
(b) Explain how it is related to the Remainder Theorem. [1]
(c) Check if (x + 1) is a factor of x³ + 1. [1]
(d) If (x − 1) is a factor of p(x) = x³ − 2x² + kx + 4, find k. [1]
(e) Using the Factor Theorem, factorise x³ − 3x² + 2x completely. (Hint: first take x common.) [1]
Q17.5
Factorise x³ − 2x² − x + 2 completely using the Factor Theorem. Show each step:
(a) Find one zero by trial. [1]
(b) Write one factor. [1]
(c) Divide to find the quotient. [1]
(d) Factorise the quotient. [1]
(e) Write the complete factorisation and verify by expanding. [1]
(a) Find one zero by trial. [1]
(b) Write one factor. [1]
(c) Divide to find the quotient. [1]
(d) Factorise the quotient. [1]
(e) Write the complete factorisation and verify by expanding. [1]
Bonus Question (Optional) [2 Marks]
Q18.2
★ A linear polynomial p(x) = ax + b passes through (0, 6) and has its zero at x = −3.
(i) Find the values of a and b from this information. [1]
(ii) Using the diagram below, verify that the line passes through the y-intercept (0, 6) and the zero (−3, 0). [1]
(i) Find the values of a and b from this information. [1]
(ii) Using the diagram below, verify that the line passes through the y-intercept (0, 6) and the zero (−3, 0). [1]
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Answer Key & Detailed Solutions
Q1. [1 Mark]
(x − a) is a factor of p(x).
By the Factor Theorem, if p(a) = 0, then (x − a) is a factor of p(x).
By the Factor Theorem, if p(a) = 0, then (x − a) is a factor of p(x).
Q2. [1 Mark]
Yes. p(1) = 1² − 1 = 1 − 1 = 0.
Since p(1) = 0, by the Factor Theorem, (x − 1) is a factor of x² − 1.
Since p(1) = 0, by the Factor Theorem, (x − 1) is a factor of x² − 1.
Q3. [1 Mark]
By the Remainder Theorem, remainder = p(−1).
p(−1) = (−1)² + (−1) + 1 = 1 − 1 + 1 = 1.
p(−1) = (−1)² + (−1) + 1 = 1 − 1 + 1 = 1.
Q4. [1 Mark]
From the graph, the line crosses the x-axis at the point (2, 0).
Therefore, the zero of the polynomial is x = 2.
Therefore, the zero of the polynomial is x = 2.
Q5. [1 Mark]
p(2) = 0.
If (x − 2) is a factor of p(x), then by the Factor Theorem, p(2) = 0.
If (x − 2) is a factor of p(x), then by the Factor Theorem, p(2) = 0.
Q6. [1 Mark]
True.
The Remainder Theorem states: when p(x) is divided by (x − a), the remainder is p(a).
The Factor Theorem says: if p(a) = 0 (i.e., remainder = 0), then (x − a) is a factor.
So the Factor Theorem is indeed a special case of the Remainder Theorem (when remainder = 0).
The Remainder Theorem states: when p(x) is divided by (x − a), the remainder is p(a).
The Factor Theorem says: if p(a) = 0 (i.e., remainder = 0), then (x − a) is a factor.
So the Factor Theorem is indeed a special case of the Remainder Theorem (when remainder = 0).
Q7. [2 Marks]
To check if (x − 3) is a factor, we find p(3).
p(3) = (3)² − 5(3) + 6 = 9 − 15 + 6 = 0. [1 mark]
Since p(3) = 0, by the Factor Theorem, (x − 3) is a factor of p(x) = x² − 5x + 6. [1 mark]
p(3) = (3)² − 5(3) + 6 = 9 − 15 + 6 = 0. [1 mark]
Since p(3) = 0, by the Factor Theorem, (x − 3) is a factor of p(x) = x² − 5x + 6. [1 mark]
Q8. [2 Marks]
To check if (x + 1) is a factor, we find p(−1).
p(−1) = (−1)³ + (−1)² + (−1) + 1 = −1 + 1 − 1 + 1 = 0. [1 mark]
Since p(−1) = 0, by the Factor Theorem, (x + 1) is a factor of p(x) = x³ + x² + x + 1. [1 mark]
p(−1) = (−1)³ + (−1)² + (−1) + 1 = −1 + 1 − 1 + 1 = 0. [1 mark]
Since p(−1) = 0, by the Factor Theorem, (x + 1) is a factor of p(x) = x³ + x² + x + 1. [1 mark]
Q9. [2 Marks]
If (x − 2) is a factor, then p(2) = 0. [1 mark]
p(2) = (2)² − k(2) + 4 = 4 − 2k + 4 = 8 − 2k
Setting p(2) = 0:
8 − 2k = 0
2k = 8
k = 4. [1 mark]
p(2) = (2)² − k(2) + 4 = 4 − 2k + 4 = 8 − 2k
Setting p(2) = 0:
8 − 2k = 0
2k = 8
k = 4. [1 mark]
Q10. [2 Marks]
From the graph:
p(x) crosses the x-axis at (−1, 0). Therefore, the zero of p(x) is x = −1. [1 mark]
q(x) crosses the x-axis at (3, 0). Therefore, the zero of q(x) is x = 3. [1 mark]
p(x) crosses the x-axis at (−1, 0). Therefore, the zero of p(x) is x = −1. [1 mark]
q(x) crosses the x-axis at (3, 0). Therefore, the zero of q(x) is x = 3. [1 mark]
Q11. [2 Marks]
By the Remainder Theorem, the remainder when p(x) is divided by (x − 1) is p(1).
p(1) = 2(1)³ − 5(1)² + 3(1) + 7 = 2 − 5 + 3 + 7 = 7. [1 mark]
Since the remainder is 7 ≠ 0, (x − 1) is NOT a factor of p(x). [1 mark]
p(1) = 2(1)³ − 5(1)² + 3(1) + 7 = 2 − 5 + 3 + 7 = 7. [1 mark]
Since the remainder is 7 ≠ 0, (x − 1) is NOT a factor of p(x). [1 mark]
Q12. [3 Marks]
Let p(x) = x² − 7x + 12.
Step 1: Try x = 3: p(3) = 9 − 21 + 12 = 0. So (x − 3) is a factor. [1 mark]
Step 2: Try x = 4: p(4) = 16 − 28 + 12 = 0. So (x − 4) is a factor. [1 mark]
Step 3: Since p(x) is quadratic and both (x − 3) and (x − 4) are factors:
x² − 7x + 12 = (x − 3)(x − 4). [1 mark]
Step 1: Try x = 3: p(3) = 9 − 21 + 12 = 0. So (x − 3) is a factor. [1 mark]
Step 2: Try x = 4: p(4) = 16 − 28 + 12 = 0. So (x − 4) is a factor. [1 mark]
Step 3: Since p(x) is quadratic and both (x − 3) and (x − 4) are factors:
x² − 7x + 12 = (x − 3)(x − 4). [1 mark]
Q13. [3 Marks]
If (x + 2) is a factor, then p(−2) = 0. [1 mark]
p(−2) = (−2)³ + k(−2)² + 7(−2) + 10
= −8 + 4k − 14 + 10 = 4k − 12
Setting p(−2) = 0:
4k − 12 = 0 ⇒ k = 3. [1 mark]
So p(x) = x³ + 3x² + 7x + 10.
Dividing by (x + 2):
x³ + 3x² + 7x + 10 = (x + 2)(x² + x + 5).
The quotient x² + x + 5 has discriminant = 1 − 20 = −19 < 0, so it cannot be factorised further over reals.
p(x) = (x + 2)(x² + x + 5). [1 mark]
p(−2) = (−2)³ + k(−2)² + 7(−2) + 10
= −8 + 4k − 14 + 10 = 4k − 12
Setting p(−2) = 0:
4k − 12 = 0 ⇒ k = 3. [1 mark]
So p(x) = x³ + 3x² + 7x + 10.
Dividing by (x + 2):
x³ + 3x² + 7x + 10 = (x + 2)(x² + x + 5).
The quotient x² + x + 5 has discriminant = 1 − 20 = −19 < 0, so it cannot be factorised further over reals.
p(x) = (x + 2)(x² + x + 5). [1 mark]
Q14. [3 Marks]
(i) From the graph, the line crosses the x-axis at (4, 0).
The zero of p(x) is x = 4. [1 mark]
(ii) From the graph, the line crosses the y-axis at (0, −8).
The y-intercept is −8. [1 mark]
(iii) Since p(x) = ax + b:
y-intercept: p(0) = b = −8, so b = −8.
Zero: p(4) = 0 ⇒ 4a + (−8) = 0 ⇒ 4a = 8 ⇒ a = 2.
So p(x) = 2x − 8. [1 mark]
The zero of p(x) is x = 4. [1 mark]
(ii) From the graph, the line crosses the y-axis at (0, −8).
The y-intercept is −8. [1 mark]
(iii) Since p(x) = ax + b:
y-intercept: p(0) = b = −8, so b = −8.
Zero: p(4) = 0 ⇒ 4a + (−8) = 0 ⇒ 4a = 8 ⇒ a = 2.
So p(x) = 2x − 8. [1 mark]
Q15. [3 Marks]
Let p(x) = x³ − 6x² + 11x − 6.
p(1) = 1 − 6 + 11 − 6 = 0. So (x − 1) is a factor. [1 mark]
p(2) = 8 − 24 + 22 − 6 = 0. So (x − 2) is a factor. [1 mark]
p(3) = 27 − 54 + 33 − 6 = 0. So (x − 3) is a factor.
Since p(x) is cubic and has three linear factors:
x³ − 6x² + 11x − 6 = (x − 1)(x − 2)(x − 3). [1 mark]
p(1) = 1 − 6 + 11 − 6 = 0. So (x − 1) is a factor. [1 mark]
p(2) = 8 − 24 + 22 − 6 = 0. So (x − 2) is a factor. [1 mark]
p(3) = 27 − 54 + 33 − 6 = 0. So (x − 3) is a factor.
Since p(x) is cubic and has three linear factors:
x³ − 6x² + 11x − 6 = (x − 1)(x − 2)(x − 3). [1 mark]
Q16. [5 Marks]
(a) Factor Theorem: If p(x) is a polynomial of degree ≥ 1 and a is any real number, then (x − a) is a factor of p(x) if and only if p(a) = 0. [1 mark]
(b) Relation to Remainder Theorem: The Remainder Theorem states that when p(x) is divided by (x − a), the remainder is p(a). The Factor Theorem is a special case: when this remainder p(a) = 0, it means (x − a) divides p(x) exactly, i.e., (x − a) is a factor. [1 mark]
(c) Check if (x + 1) is a factor of x³ + 1.
p(−1) = (−1)³ + 1 = −1 + 1 = 0.
Since p(−1) = 0, (x + 1) is a factor of x³ + 1. [1 mark]
(d) If (x − 1) is a factor, then p(1) = 0.
p(1) = (1)³ − 2(1)² + k(1) + 4 = 1 − 2 + k + 4 = 3 + k
Setting p(1) = 0: 3 + k = 0 ⇒ k = −3. [1 mark]
(e) Factorise x³ − 3x² + 2x.
Take x common: x(x² − 3x + 2).
Now factorise x² − 3x + 2 using Factor Theorem:
Let q(x) = x² − 3x + 2. q(1) = 1 − 3 + 2 = 0. So (x − 1) is a factor.
q(2) = 4 − 6 + 2 = 0. So (x − 2) is a factor.
Therefore: x³ − 3x² + 2x = x(x − 1)(x − 2). [1 mark]
(b) Relation to Remainder Theorem: The Remainder Theorem states that when p(x) is divided by (x − a), the remainder is p(a). The Factor Theorem is a special case: when this remainder p(a) = 0, it means (x − a) divides p(x) exactly, i.e., (x − a) is a factor. [1 mark]
(c) Check if (x + 1) is a factor of x³ + 1.
p(−1) = (−1)³ + 1 = −1 + 1 = 0.
Since p(−1) = 0, (x + 1) is a factor of x³ + 1. [1 mark]
(d) If (x − 1) is a factor, then p(1) = 0.
p(1) = (1)³ − 2(1)² + k(1) + 4 = 1 − 2 + k + 4 = 3 + k
Setting p(1) = 0: 3 + k = 0 ⇒ k = −3. [1 mark]
(e) Factorise x³ − 3x² + 2x.
Take x common: x(x² − 3x + 2).
Now factorise x² − 3x + 2 using Factor Theorem:
Let q(x) = x² − 3x + 2. q(1) = 1 − 3 + 2 = 0. So (x − 1) is a factor.
q(2) = 4 − 6 + 2 = 0. So (x − 2) is a factor.
Therefore: x³ − 3x² + 2x = x(x − 1)(x − 2). [1 mark]
Q17. [5 Marks]
Let p(x) = x³ − 2x² − x + 2.
(a) Find one zero by trial:
p(1) = 1 − 2 − 1 + 2 = 0. So x = 1 is a zero. [1 mark]
(b) Write one factor:
Since p(1) = 0, by Factor Theorem, (x − 1) is a factor of p(x). [1 mark]
(c) Divide to find quotient:
Dividing x³ − 2x² − x + 2 by (x − 1):
x³ − 2x² − x + 2 = (x − 1)(x² − x − 2)
[We can verify: (x − 1)(x² − x − 2) = x³ − x² − 2x − x² + x + 2 = x³ − 2x² − x + 2 ✓]
Quotient = x² − x − 2. [1 mark]
(d) Factorise the quotient:
x² − x − 2: try x = 2: 4 − 2 − 2 = 0. So (x − 2) is a factor.
try x = −1: 1 + 1 − 2 = 0. So (x + 1) is a factor.
x² − x − 2 = (x − 2)(x + 1). [1 mark]
(e) Complete factorisation:
x³ − 2x² − x + 2 = (x − 1)(x − 2)(x + 1)
Verification:
(x − 1)(x − 2)(x + 1)
= (x − 1)(x + 1)(x − 2)
= (x² − 1)(x − 2)
= x³ − 2x² − x + 2 ✓ [1 mark]
(a) Find one zero by trial:
p(1) = 1 − 2 − 1 + 2 = 0. So x = 1 is a zero. [1 mark]
(b) Write one factor:
Since p(1) = 0, by Factor Theorem, (x − 1) is a factor of p(x). [1 mark]
(c) Divide to find quotient:
Dividing x³ − 2x² − x + 2 by (x − 1):
x³ − 2x² − x + 2 = (x − 1)(x² − x − 2)
[We can verify: (x − 1)(x² − x − 2) = x³ − x² − 2x − x² + x + 2 = x³ − 2x² − x + 2 ✓]
Quotient = x² − x − 2. [1 mark]
(d) Factorise the quotient:
x² − x − 2: try x = 2: 4 − 2 − 2 = 0. So (x − 2) is a factor.
try x = −1: 1 + 1 − 2 = 0. So (x + 1) is a factor.
x² − x − 2 = (x − 2)(x + 1). [1 mark]
(e) Complete factorisation:
x³ − 2x² − x + 2 = (x − 1)(x − 2)(x + 1)
Verification:
(x − 1)(x − 2)(x + 1)
= (x − 1)(x + 1)(x − 2)
= (x² − 1)(x − 2)
= x³ − 2x² − x + 2 ✓ [1 mark]
Q18. Bonus [2 Marks]
(i) p(x) = ax + b passes through (0, 6):
p(0) = a(0) + b = b = 6. So b = 6.
Zero at x = −3 means p(−3) = 0:
a(−3) + 6 = 0
−3a = −6
a = 2.
So p(x) = 2x + 6. [1 mark]
(ii) Verification from the graph:
The graph shows the line passing through (−3, 0) on the x-axis (the zero) and (0, 6) on the y-axis (the y-intercept).
p(−3) = 2(−3) + 6 = −6 + 6 = 0 ✓
p(0) = 2(0) + 6 = 6 ✓
Both points are confirmed from the graph and the equation. [1 mark]
p(0) = a(0) + b = b = 6. So b = 6.
Zero at x = −3 means p(−3) = 0:
a(−3) + 6 = 0
−3a = −6
a = 2.
So p(x) = 2x + 6. [1 mark]
(ii) Verification from the graph:
The graph shows the line passing through (−3, 0) on the x-axis (the zero) and (0, 6) on the y-axis (the y-intercept).
p(−3) = 2(−3) + 6 = −6 + 6 = 0 ✓
p(0) = 2(0) + 6 = 6 ✓
Both points are confirmed from the graph and the equation. [1 mark]