Welcome, Class 9 students! In this guide, we will explore step-by-step solutions for Exercise 2.4 of Chapter 2 (Polynomials), fully aligned with the MP Board (NCERT) syllabus. This exercise focuses on the application of the Factor Theorem and techniques for factorizing quadratic and cubic polynomials. Mastering these concepts is essential for scoring high in your school exams and building a strong foundation for higher classes.
Core Concepts: The Factor Theorem
The Factor Theorem states that if p(x) is a polynomial of degree n ≥ 1 and a is any real number, then:
- (x - a) is a factor of p(x) if p(a) = 0.
- Conversely, if (x - a) is a factor of p(x), then p(a) = 0.
Step-by-Step Solutions
Question 1: Determine which of the following polynomials has (x + 1) as a factor:
(i) x3 + x2 + x + 1
Solution:
Let p(x) = x3 + x2 + x + 1
To check if (x + 1) is a factor, we find the value of p(-1) [since x + 1 = 0 ⇒ x = -1]:
p(-1) = (-1)3 + (-1)2 + (-1) + 1
p(-1) = -1 + 1 - 1 + 1
p(-1) = 0
Since the remainder p(-1) is 0, by the Factor Theorem, (x + 1) is a factor of x3 + x2 + x + 1.
Question 2: Use the Factor Theorem to determine whether g(x) is a factor of p(x) in the following case:
p(x) = 2x3 + x2 - 2x - 1, g(x) = x + 1
Solution:
To find if g(x) = x + 1 is a factor of p(x), we substitute x = -1 into p(x):
p(-1) = 2(-1)3 + (-1)2 - 2(-1) - 1
p(-1) = 2(-1) + 1 + 2 - 1
p(-1) = -2 + 1 + 2 - 1
p(-1) = 0
Since p(-1) = 0, g(x) is a factor of p(x).
Question 3: Find the value of k, if x - 1 is a factor of p(x) = x2 + x + k.
Solution:
Since (x - 1) is a factor of p(x), by the Factor Theorem, p(1) must be equal to 0.
p(x) = x2 + x + k
p(1) = (1)2 + (1) + k = 0
1 + 1 + k = 0
2 + k = 0
k = -2
Thus, the value of k is -2.
Question 4: Factorise: 12x2 - 7x + 1
Solution:
We will factorise this quadratic polynomial by splitting the middle term.
We need to find two numbers whose sum is -7 (coefficient of x) and whose product is 12 × 1 = 12 (coefficient of x2 × constant term).
Those two numbers are -4 and -3 (since -4 + -3 = -7 and -4 × -3 = 12).
12x2 - 7x + 1 = 12x2 - 4x - 3x + 1
= 4x(3x - 1) - 1(3x - 1)
= (3x - 1)(4x - 1)
Therefore, the factors are (3x - 1)(4x - 1).
Question 5: Factorise the cubic polynomial: x3 - 23x2 + 142x - 120
Solution:
Let p(x) = x3 - 23x2 + 142x - 120.
Step 1: Find one factor using Trial and Error.
The constant term is -120. Its factors are ±1, ±2, ±3, etc.
Let's test x = 1:
p(1) = (1)3 - 23(1)2 + 142(1) - 120
p(1) = 1 - 23 + 142 - 120
p(1) = 143 - 143 = 0
Since p(1) = 0, (x - 1) is a factor of p(x).
Step 2: Divide the polynomial to find the quadratic quotient.
Let's rewrite the polynomial to extract (x - 1):
x3 - 23x2 + 142x - 120
= x2(x - 1) - 22x(x - 1) + 120(x - 1)
= (x - 1)(x2 - 22x + 120)
Step 3: Factorise the quadratic expression (x2 - 22x + 120).
We split the middle term. We need two numbers with sum -22 and product 120. These are -12 and -10.
x2 - 22x + 120 = x2 - 12x - 10x + 120
= x(x - 12) - 10(x - 12)
= (x - 12)(x - 10)
Final Answer:
The fully factorised form of the cubic polynomial is: (x - 1)(x - 10)(x - 12).
ЁЯТб Aptitude Corner (NTSE / Olympiad Special)
The Coefficient Sum Rule:
For any polynomial, if the sum of all its coefficients is equal to 0, then (x - 1) is guaranteed to be a factor of that polynomial.
Example: In Q5 above, p(x) = 1x3 - 23x2 + 142x - 120.
Sum of coefficients = 1 - 23 + 142 - 120 = 0.
This instantly tells you that (x - 1) is a factor without calculating anything else!
Understanding the Factor Theorem and mastering the factorization of cubic polynomials are crucial steps toward high algebraic proficiency. Keep practicing these steps to ensure perfect execution in your Class 9 board exams. Happy learning!
Academic Discussion & Doubts