Hello Students! Welcome back. Today, we are going to dive deep into Exercise 2.3 of Chapter 2 (Polynomials) from your MP Board Class 9 Mathematics textbook. This exercise is built around the highly important Factor Theorem and the methods of factorization (splitting the middle term and cubic factorization). Mastering these concepts will not only help you score full marks in your school exams but also lay a rock-solid foundation for higher-level algebra. Let's learn together step-by-step!
Quick Revision: The Factor Theorem
If p(x) is a polynomial of degree n ≥ 1 and a is any real number, then:
- (i) (x - a) is a factor of p(x), if p(a) = 0.
- (ii) p(a) = 0, if (x - a) is a factor of p(x).
Solutions for Exercise 2.3
Q1. Determine which of the following polynomials has (x + 1) a factor:
(i) x3 + x2 + x + 1
Solution:
Let the given polynomial be 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].
Substituting x = -1 in p(x):
p(-1) = (-1)3 + (-1)2 + (-1) + 1
= -1 + 1 - 1 + 1
= 0
Since p(-1) = 0, by the Factor Theorem, (x + 1) is a factor of x3 + x2 + x + 1.
(ii) x4 + x3 + x2 + x + 1
Solution:
Let p(x) = x4 + x3 + x2 + x + 1.
Substituting x = -1:
p(-1) = (-1)4 + (-1)3 + (-1)2 + (-1) + 1
= 1 - 1 + 1 - 1 + 1
= 1 ≠ 0
Since p(-1) ≠ 0, by the Factor Theorem, (x + 1) is not a factor of x4 + x3 + x2 + x + 1.
Q2. Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases:
p(x) = 2x3 + x2 - 2x - 1, g(x) = x + 1
Solution:
Given, g(x) = x + 1. To find its zero, we equate it to 0:
x + 1 = 0 ⇒ x = -1.
Now, substitute x = -1 in p(x):
p(-1) = 2(-1)3 + (-1)2 - 2(-1) - 1
= 2(-1) + 1 + 2 - 1
= -2 + 1 + 2 - 1
= 0
Since p(-1) = 0, therefore, by the Factor Theorem, g(x) is a factor of p(x).
Q3. Find the value of k, if (x - 1) is a factor of p(x) in the following case:
p(x) = x2 + x + k
Solution:
Since (x - 1) is a factor of p(x), by the Factor Theorem, we must have:
p(1) = 0
Substitute x = 1 in the polynomial:
(1)2 + (1) + k = 0
1 + 1 + k = 0
2 + k = 0
k = -2
Thus, the value of k is -2.
Q4. Factorise: 12x2 - 7x + 1
Solution (Splitting the Middle Term):
We need to find two numbers whose product is 12 × 1 = 12 and whose sum is -7.
Those numbers are -3 and -4 (since -3 × -4 = 12 and -3 + (-4) = -7).
12x2 - 7x + 1 = 12x2 - 4x - 3x + 1
= 4x(3x - 1) - 1(3x - 1)
= (3x - 1)(4x - 1)
Hence, the factors are (3x - 1)(4x - 1).
Q5. Factorise: x3 - 2x2 - x + 2
Solution (Cubic Polynomial Factorisation):
Let p(x) = x3 - 2x2 - x + 2.
The constant term is 2. The factors of 2 are ±1, ±2.
Let's try x = 1:
p(1) = (1)3 - 2(1)2 - (1) + 2
= 1 - 2 - 1 + 2
= 0
Since p(1) = 0, (x - 1) is a factor of p(x).
Now, let's group the terms to factorize directly:
x3 - 2x2 - x + 2 = x2(x - 2) - 1(x - 2)
= (x - 2)(x2 - 1)
= (x - 2)(x - 1)(x + 1) [Using identity: a2 - b2 = (a-b)(a+b)]
Thus, the factorized form is (x - 1)(x + 1)(x - 2).
ЁЯЪА Aptitude Corner (For NTSE & Olympiads)
Super Fast Trick to check if (x - 1) is a factor:
If the sum of all the coefficients of a polynomial is 0, then (x - 1) is always a factor!
Example: For x3 - 2x2 - x + 2, the coefficients are: 1, -2, -1, and 2.
Sum = 1 + (-2) + (-1) + 2 = 0. Hence, (x - 1) is instantly proven to be a factor without division!
That's it for Exercise 2.3! With the Factor Theorem and these algebraic tricks up your sleeve, you are ready to tackle any question on polynomial factorization. Keep practicing, and feel free to reach out with your questions in class. Keep learning, keep shining!
Academic Discussion & Doubts