Factor Theorem

Q: What is the Factor Theorem in Class 9 maths?

The Factor Theorem states that (x − a) is a factor of p(x) if and only if p(a) = 0. Substituting x = a into the polynomial and getting zero means (x − a) divides it exactly.

Q: How is the Factor Theorem different from the Remainder Theorem?

The Remainder Theorem gives the remainder p(a) when dividing by (x − a). The Factor Theorem is the special case where p(a) = 0 (remainder is zero), meaning (x − a) is a factor.

Q: How do you use the Factor Theorem to factorise a cubic polynomial?

Step 1: Try values like 0, ±1, ±2, ±3 (factors of the constant term) to find a where p(a) = 0. Step 2: (x − a) is a factor. Step 3: Divide to get a quadratic quotient. Step 4: Factorise the quadratic by splitting the middle term or using the quadratic formula.

Q: What values should you try when using the Factor Theorem?

Try factors of the constant term (±1, ±2, ±3, etc.). If the leading coefficient is not 1, also try fractions: factors of the constant term divided by factors of the leading coefficient.

Q: Can the Factor Theorem be used for quadratic polynomials?

Yes, it works for any degree. However, for quadratics, splitting the middle term or the quadratic formula is usually faster. The Factor Theorem is most useful for cubic and higher-degree polynomials.

Q: What is the connection between factors and zeroes of a polynomial?

If (x − a) is a factor, then a is a zero (p(a) = 0). Conversely, if a is a zero, then (x − a) is a factor. Zeroes are the x-values where the graph crosses the x-axis.

Q: How many factors can a polynomial have?

A polynomial of degree n has at most n linear factors. A cubic can have at most 3. Some factors may be repeated (e.g., (x − 1)² counts as 2), and some polynomials may have fewer real factors.

Q: Is the Factor Theorem important for CBSE board exams?

Yes. It is a key topic in Chapter 2: Polynomials. Questions on checking factors, factorising polynomials, and finding unknown coefficients frequently appear, typically carrying 3–5 marks.

Q: Can you use the Factor Theorem for expressions that are not polynomials?

No. The Factor Theorem applies only to polynomials with non-negative integer exponents and no variables in the denominator. Expressions like 1/x + 3 or √x + 2 are not polynomials.

Q: How do you verify if your factorisation is correct?

Two methods: (1) Multiply all factors and check that you get the original polynomial. (2) Substitute each zero into the original polynomial and verify each gives zero.

Class 9Polynomials

The Factor Theorem establishes a direct connection between the factors of a polynomial and its zeroes (roots).

It builds on the Remainder Theorem: while the Remainder Theorem gives the remainder when dividing by a linear expression, the Factor Theorem identifies when that remainder is exactly zero.

The Factor Theorem is essential for factorising cubic and higher-degree polynomials that cannot be factorised by grouping or splitting the middle term. It transforms the problem of factorisation into finding the right values to substitute.

What is Factor Theorem?

Definition: For a polynomial p(x) and a real number a:

(x − a) is a factor of p(x) if and only if p(a) = 0

Two parts:

Part 1 (If): If p(a) = 0, then (x − a) is a factor of p(x).
Part 2 (Only if): If (x − a) is a factor of p(x), then p(a) = 0.

The Factor Theorem is a special case of the Remainder Theorem where the remainder equals zero.

Extensions:

(ax − b) is a factor of p(x) if and only if p(b/a) = 0
(ax + b) is a factor of p(x) if and only if p(−b/a) = 0

Trial-and-error strategy:

Try values of a that are factors of the constant term.
Check if p(a) = 0.
If yes, (x − a) is a factor.
This is the basic form of the Rational Root Test.

Factor Theorem Formula

Key Formulas:

1. Factor Theorem (basic form):

(x − a) is a factor of p(x) ⇔ p(a) = 0

2. General linear factors:

(ax − b) is a factor ⇔ p(b/a) = 0
(ax + b) is a factor ⇔ p(−b/a) = 0

3. Factorisation process for cubic polynomials:

Find one zero a such that p(a) = 0 (try factors of the constant term).
Then (x − a) is a factor.
Divide p(x) by (x − a) to get the quadratic quotient q(x).
Factorise q(x) further if possible.

4. Rational Root Test (basic version):

Possible rational zeroes = ± (factors of constant term) / (factors of leading coefficient)

5. Factor-zero connection:

If p(x) = (x − r₁)(x − r₂)...(x − rₙ), then r₁, r₂, ..., rₙ are the zeroes.

6. Maximum linear factors:

A polynomial of degree n has at most n linear factors (counting repetitions).

Derivation and Proof

Proof of the Factor Theorem:

Given: p(x) is a polynomial and a is a real number.

To prove: (x − a) is a factor of p(x) if and only if p(a) = 0.

Part 1: If p(a) = 0, then (x − a) is a factor

By the Division Algorithm: p(x) = (x − a) × q(x) + r
By the Remainder Theorem: r = p(a)
Given p(a) = 0, so r = 0
Therefore: p(x) = (x − a) × q(x)
So (x − a) is a factor of p(x).

Part 2: If (x − a) is a factor, then p(a) = 0

If (x − a) is a factor: p(x) = (x − a) × q(x)
Substitute x = a: p(a) = (a − a) × q(a) = 0 × q(a) = 0
Therefore p(a) = 0.

Since both directions are proved, the Factor Theorem is established.

Geometric interpretation:

The zeroes of a polynomial correspond to the x-intercepts of its graph.
Each factor (x − a) corresponds to the graph crossing or touching the x-axis at x = a.
If p(a) = 0, the graph passes through the point (a, 0).

Types and Properties

Types of Factor Theorem Problems:

Type 1: Checking if a linear expression is a factor — Substitute the zero of the expression into the polynomial. If the result is 0, it is a factor.
Type 2: Finding one factor by trial and error — Try x = 0, ±1, ±2, ±3 (factors of the constant term). The first value giving p(a) = 0 gives a factor (x − a).
Type 3: Complete factorisation of cubic polynomials — Find one factor, divide to get a quadratic, then factorise the quadratic.
Type 4: Factorisation using algebraic identities — Recognise special forms like a³ + b³ or a³ − b³.
Type 5: Finding unknown coefficients given factor conditions — If a linear expression is a factor, set p(a) = 0 and solve for unknowns.
Type 6: Finding all zeroes of a polynomial — Use the Factor Theorem with polynomial division to find all zeroes systematically.
Type 7: Constructing polynomials from given zeroes — If zeroes are 1, 2, 3, the polynomial is p(x) = (x − 1)(x − 2)(x − 3).

Solved Examples

Example 1: Example 1: Checking if (x − 1) is a factor

Problem: Determine whether (x − 1) is a factor of p(x) = x³ − 2x² − x + 2.

Solution:

By the Factor Theorem, (x − 1) is a factor if p(1) = 0.

p(1) = (1)³ − 2(1)² − 1 + 2 = 1 − 2 − 1 + 2 = 0

Since p(1) = 0, (x − 1) IS a factor.

Answer: Yes, (x − 1) is a factor of x³ − 2x² − x + 2.

Example 2: Example 2: Checking if (x + 2) is a factor

Problem: Is (x + 2) a factor of p(x) = 2x³ + 5x² − x − 6?

Solution:

The zero of (x + 2) is x = −2. Check if p(−2) = 0.

p(−2) = 2(−8) + 5(4) + 2 − 6
= −16 + 20 + 2 − 6 = 0

Since p(−2) = 0, (x + 2) IS a factor.

Answer: Yes, (x + 2) is a factor of 2x³ + 5x² − x − 6.

Example 3: Example 3: Complete factorisation of a cubic polynomial

Problem: Factorise p(x) = x³ − 6x² + 11x − 6 completely.

Solution:

Step 1: Find one factor by trial.

Try x = 1: p(1) = 1 − 6 + 11 − 6 = 0 ✓
So (x − 1) is a factor.

Step 2: Divide p(x) by (x − 1).

Quotient = x² − 5x + 6

Step 3: Factorise the quadratic.

x² − 5x + 6 = (x − 2)(x − 3)

Step 4: Complete factorisation:

p(x) = (x − 1)(x − 2)(x − 3)

Verification: p(1) = 0, p(2) = 0, p(3) = 0 ✓

Example 4: Example 4: Finding unknown coefficients given factors

Problem: If both (x − 1) and (x + 2) are factors of p(x) = x³ + ax² + bx + 6, find a and b.

Solution:

Condition 1: p(1) = 0

1 + a + b + 6 = 0 ⇒ a + b = −7 ... (i)

Condition 2: p(−2) = 0

−8 + 4a − 2b + 6 = 0 ⇒ 4a − 2b = 2 ⇒ 2a − b = 1 ... (ii)

Solving:

Adding (i) and (ii): 3a = −6 ⇒ a = −2
From (i): −2 + b = −7 ⇒ b = −5

Answer: a = −2 and b = −5.

Example 5: Example 5: Factorising using sum of cubes identity

Problem: Factorise x³ + 125 using the Factor Theorem and verify.

Solution:

Method 1: Factor Theorem

Try x = −5: p(−5) = (−5)³ + 125 = −125 + 125 = 0
So (x + 5) is a factor.

Dividing: x³ + 125 = (x + 5)(x² − 5x + 25)

Method 2: Identity verification

Using a³ + b³ = (a + b)(a² − ab + b²) with a = x, b = 5
x³ + 125 = (x + 5)(x² − 5x + 25) ✓

Note: x² − 5x + 25 cannot be factorised further (discriminant = 25 − 100 = −75 < 0).

Answer: x³ + 125 = (x + 5)(x² − 5x + 25)

Example 6: Example 6: Checking (2x − 1) as a factor

Problem: Determine whether (2x − 1) is a factor of p(x) = 2x³ − x² − 2x + 1.

Solution:

The zero of (2x − 1) is x = 1/2. Check if p(1/2) = 0.

p(1/2) = 2(1/8) − 1/4 − 1 + 1
= 1/4 − 1/4 − 1 + 1 = 0

Since p(1/2) = 0, (2x − 1) IS a factor.

Complete factorisation:

Divide by (2x − 1): quotient = x² − 1
x² − 1 = (x − 1)(x + 1)

Answer: 2x³ − x² − 2x + 1 = (2x − 1)(x − 1)(x + 1)

Example 7: Example 7: Factorisation of a polynomial with repeated roots

Problem: Factorise p(x) = x³ − 3x² + 3x − 1.

Solution:

Try x = 1: p(1) = 1 − 3 + 3 − 1 = 0 ✓
Divide by (x − 1): quotient = x² − 2x + 1
x² − 2x + 1 = (x − 1)²

Therefore: p(x) = (x − 1)(x − 1)² = (x − 1)³

This matches the identity (a − b)³ with a = x, b = 1.

Answer: x³ − 3x² + 3x − 1 = (x − 1)³

Example 8: Example 8: Factorising when the leading coefficient is not 1

Problem: Factorise p(x) = 6x³ + 11x² − 3x − 2.

Solution:

Given: constant term = −2, leading coefficient = 6

Possible rational zeroes: ±1, ±2, ±1/2, ±1/3, ±2/3, ±1/6

Try x = −2: p(−2) = 6(−8) + 11(4) + 6 − 2 = −48 + 44 + 6 − 2 = 0 ✓
So (x + 2) is a factor.
Divide: quotient = 6x² − x − 1
Factorise 6x² − x − 1: need two numbers multiplying to −6 and adding to −1 (−3 and 2)
6x² − 3x + 2x − 1 = 3x(2x − 1) + 1(2x − 1) = (3x + 1)(2x − 1)

Answer: 6x³ + 11x² − 3x − 2 = (x + 2)(3x + 1)(2x − 1)

Example 9: Example 9: Constructing a polynomial from given zeroes

Problem: Find a cubic polynomial whose zeroes are −1, 2, and 4.

Solution:

Given zeroes: −1, 2, 4 ⇒ factors are (x + 1), (x − 2), (x − 4)

p(x) = (x + 1)(x − 2)(x − 4)
(x + 1)(x − 2) = x² − x − 2
(x² − x − 2)(x − 4) = x³ − 4x² − x² + 4x − 2x + 8
= x³ − 5x² + 2x + 8

Verification:

p(−1) = −1 − 5 − 2 + 8 = 0 ✓
p(2) = 8 − 20 + 4 + 8 = 0 ✓
p(4) = 64 − 80 + 8 + 8 = 0 ✓

Answer: p(x) = x³ − 5x² + 2x + 8

Example 10: Example 10: Word problem involving the Factor Theorem

Problem: A rectangular box has dimensions (x + 1) cm, (x − 2) cm, and (x + 3) cm. Its volume is V(x) = x³ + 2x² − 5x − 6. Verify and find dimensions when x = 3.

Solution:

Verification: If (x + 1), (x − 2), (x + 3) are factors, then x = −1, 2, −3 should be zeroes.

V(−1) = −1 + 2 + 5 − 6 = 0 ✓
V(2) = 8 + 8 − 10 − 6 = 0 ✓
V(−3) = −27 + 18 + 15 − 6 = 0 ✓

When x = 3:

Dimensions: (3 + 1) = 4 cm, (3 − 2) = 1 cm, (3 + 3) = 6 cm
Volume = 4 × 1 × 6 = 24 cm³
Check: V(3) = 27 + 18 − 15 − 6 = 24 ✓

Answer: Dimensions are 4 cm, 1 cm, 6 cm; volume = 24 cm³.

Real-World Applications

Applications of the Factor Theorem:

Polynomial factorisation: Breaking down complex polynomials into simpler factors for solving equations and simplifying rational expressions.
Solving equations: To solve x³ − 6x² + 11x − 6 = 0, find one root using the Factor Theorem, then reduce to a quadratic.
Graph sketching: Factors reveal where the graph crosses the x-axis. Each factor (x − a) gives an x-intercept at (a, 0).
Engineering design: Factors of characteristic polynomials determine system stability in control systems engineering.
Cryptography: Polynomial factorisation over finite fields forms the basis of certain encryption algorithms.
Error detection: Cyclic redundancy checks (CRC) use polynomial factors to detect transmission errors.
Curve fitting: Polynomial interpolation ensures the interpolating polynomial passes through each data point exactly.

Key Points to Remember

(x − a) is a factor of p(x) if and only if p(a) = 0.
It is a special case of the Remainder Theorem where the remainder is zero.
To check if (x − a) is a factor, substitute x = a and check if the result is zero.
For (ax + b) type factors, substitute x = −b/a.
To factorise cubic polynomials: find one factor by trial, divide, then factorise the quadratic.
Try factors of the constant term (divided by factors of the leading coefficient) when searching.
A polynomial of degree n has at most n linear factors and n zeroes.
Zeroes r₁, r₂, ..., rₙ give: p(x) = constant × (x − r₁)(x − r₂)...(x − rₙ).
The Factor Theorem connects three concepts: factors, zeroes, and x-intercepts.
Essential topic in the CBSE Class 9 Polynomials chapter.

Practice Problems

Determine whether (x − 3) is a factor of p(x) = x³ − 3x² + 4x − 12.
Factorise x³ + 2x² − x − 2 completely using the Factor Theorem.
If (x + 1) is a factor of p(x) = kx³ + 5x² + x − 2, find the value of k.
Factorise 2x³ + 3x² − 11x − 6 completely.
Find a polynomial of degree 3 whose zeroes are 1, −2, and 3. Verify your answer.
If both (x − 1) and (x − 2) are factors of p(x) = x³ − 6x² + ax + b, find the values of a and b, and the third factor.
Show that (x + 1) is a factor of xⁿ + 1 when n is an odd positive integer.
Factorise x³ − 8 using the Factor Theorem and verify with the difference of cubes identity.

Frequently Asked Questions

Q1. What is the Factor Theorem in Class 9 maths?

The Factor Theorem states that (x − a) is a factor of p(x) if and only if p(a) = 0. Substituting x = a into the polynomial and getting zero means (x − a) divides it exactly.

Q2. How is the Factor Theorem different from the Remainder Theorem?

The Remainder Theorem gives the remainder p(a) when dividing by (x − a). The Factor Theorem is the special case where p(a) = 0 (remainder is zero), meaning (x − a) is a factor.

Q3. How do you use the Factor Theorem to factorise a cubic polynomial?

Step 1: Try values like 0, ±1, ±2, ±3 (factors of the constant term) to find a where p(a) = 0. Step 2: (x − a) is a factor. Step 3: Divide to get a quadratic quotient. Step 4: Factorise the quadratic by splitting the middle term or using the quadratic formula.

Q4. What values should you try when using the Factor Theorem?

Try factors of the constant term (±1, ±2, ±3, etc.). If the leading coefficient is not 1, also try fractions: factors of the constant term divided by factors of the leading coefficient.

Q5. Can the Factor Theorem be used for quadratic polynomials?

Yes, it works for any degree. However, for quadratics, splitting the middle term or the quadratic formula is usually faster. The Factor Theorem is most useful for cubic and higher-degree polynomials.

Q6. What is the connection between factors and zeroes of a polynomial?

If (x − a) is a factor, then a is a zero (p(a) = 0). Conversely, if a is a zero, then (x − a) is a factor. Zeroes are the x-values where the graph crosses the x-axis.

Q7. How many factors can a polynomial have?

A polynomial of degree n has at most n linear factors. A cubic can have at most 3. Some factors may be repeated (e.g., (x − 1)² counts as 2), and some polynomials may have fewer real factors.

Q8. Is the Factor Theorem important for CBSE board exams?

Yes. It is a key topic in Chapter 2: Polynomials. Questions on checking factors, factorising polynomials, and finding unknown coefficients frequently appear, typically carrying 3–5 marks.

Q9. Can you use the Factor Theorem for expressions that are not polynomials?

No. The Factor Theorem applies only to polynomials with non-negative integer exponents and no variables in the denominator. Expressions like 1/x + 3 or √x + 2 are not polynomials.

Q10. How do you verify if your factorisation is correct?

Two methods: (1) Multiply all factors and check that you get the original polynomial. (2) Substitute each zero into the original polynomial and verify each gives zero.

Factor Theorem

What is Factor Theorem?

Factor Theorem Formula

Derivation and Proof

Types and Properties

Solved Examples

Example 1: Example 1: Checking if (x − 1) is a factor

Example 2: Example 2: Checking if (x + 2) is a factor

Example 3: Example 3: Complete factorisation of a cubic polynomial

Example 4: Example 4: Finding unknown coefficients given factors

Example 5: Example 5: Factorising using sum of cubes identity

Example 6: Example 6: Checking (2x − 1) as a factor

Example 7: Example 7: Factorisation of a polynomial with repeated roots

Example 8: Example 8: Factorising when the leading coefficient is not 1

Example 9: Example 9: Constructing a polynomial from given zeroes

Example 10: Example 10: Word problem involving the Factor Theorem

Real-World Applications

Key Points to Remember

Practice Problems

Frequently Asked Questions

Q1. What is the Factor Theorem in Class 9 maths?

Q2. How is the Factor Theorem different from the Remainder Theorem?

Q3. How do you use the Factor Theorem to factorise a cubic polynomial?

Q4. What values should you try when using the Factor Theorem?

Q5. Can the Factor Theorem be used for quadratic polynomials?

Q6. What is the connection between factors and zeroes of a polynomial?

Q7. How many factors can a polynomial have?

Q8. Is the Factor Theorem important for CBSE board exams?

Q9. Can you use the Factor Theorem for expressions that are not polynomials?

Q10. How do you verify if your factorisation is correct?

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