2. Polynomials Mathematics Exercise - 2.4 class 9 Maths in English - CBSE Study
NCERT Solutions for Class 9 Mathematics are carefully prepared according to the latest CBSE syllabus and NCERT textbooks to help students understand every concept clearly. These solutions cover all important 2. Polynomials with detailed explanations and step-by-step answers for better exam preparation. Each Exercise 2.4 is explained in simple language so that students can easily grasp the fundamentals and improve their academic performance. The study material is designed to support daily homework, revision practice, and final exam preparation for Class 9 students. With accurate answers, concept clarity, and structured content, these NCERT solutions help learners build confidence and score higher marks in their examinations. Whether you are revising a specific topic or preparing an entire chapter, this resource provides reliable and syllabus-based guidance for complete success in Mathematics.
Class 9 English Medium Mathematics All Chapters:
2. Polynomials
4. Exercise 2.4
Chapter 2. Polynomials
Exercise 2.4
Q.1. Determine which of the following polynomials has (x+ 1) a factor:
(i) x3 +x2 + x +1 (ii) x4 + x3 + x2 + x +1
(iii) x4 + 3x3 + 3x2 + x +1 
Solution:
(i) If (x + 1) is a factor of p(x) = x3+ x2+ x + 1, then p (−1) must be zero, otherwise (x + 1) is not a factor of p(x).
P(x) = x3 + x2 + x + 1
P(−1)= (−1)3 + (−1)2 + (−1) + 1
= − 1 + 1 − 1 − 1
= 0
∵ P(x) = 0
Hence, x + 1 is a factor of this polynomial.
(ii) If (x + 1) is a factor of p(x) = x4+ x3+ x2+ x + 1, then p (−1) must be zero, Otherwise (x + 1) is not a factor of p(x).
P(x) = x4+ x3+ x2+ x + 1
P(−1) = (−1)4 + (−1)3 + (−1)2 + (−1) + 1
= 1 − 1 + 1 −1 + 1
= 1
As P(x) ≠ 0, (− 1)
Therefore, x + 1 is not a factor of this polynomial.
(iii) If (x + 1) is a factor of polynomial p(x) = x4+ 3x3+ 3x2+ x + 1, then p (−1) must be 0, otherwise (x + 1) is not a factor of this polynomial.
P(x) = x4+ 3x3+ 3x2+ x + 1
P(−1) = (−1)4+ 3(−1)3+ 3(−1)2+ (−1) + 1
= 1 − 3 + 3 − 1 + 1
= 1
As P(x) ≠ 0, (−1)
Therefore (x+1) is not a factor of this polynomial .
Q.2. Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases:
(i) P(x) = 2x3+ x2− 2x − 1, g(x) = x + 1
(ii) P(x) = x3 + 3x2 + 3x + 1, g(x) = x + 2
(iii) P(x) = x3 − 4 x2 + x + 6, g(x) = x − 3
Solution:
(i) If g(x) = x + 1 is a factor of the given polynomial p(x), then p (−1) must be zero.
P (x) = 2x3 + x2 − 2x − 1
P (−1) = 2(−1)3+ (−1)2− 2(−1) − 1
= 2(−1) + 1 + 2 – 1
= 0
∵ P(x) = 0
Hence, g(x) = x + 1 is a factor of the given polynomial.
(ii) If g(x) = x + 2 is a factor of the given polynomial p(x), then p (−2) must be 0.
P (x) = x3+3x2+ 3x + 1
P (−2) = (−2)3+ 3(−2)2+ 3(−2) + 1
= − 8 + 12 − 6 + 1
= −1
As P(x) ≠ 0,
Hence, g(x) = x + 2 is not a factor of the given polynomial.
(iii) If g(x) = x − 3 is a factor of the given polynomial p(x), then P(3) must be 0.
P(x) = x3− 4 x2+ x + 6
P(3) = (3)3 − 4(3)2 + 3 + 6
= 27 −36 + 9
= 0
Hence, g(x) = x − 3 is a factor of the given polynomial.
Solution:
: If x − 1 is a factor of polynomial p(x), then P(1) must be 0.
(i) P(x) = x2+ x + k
P(1) = (1)2+ 1 + k
= 1+1+
= 2+k
k =−2
(iv) P(x) = kx2-3x + k
P(1) = k(1)2-3(1) + k
= k-3+k
2k = 3
K = 3/2
Question 4: Factorise:
(i) 12x2− 7x + 1 (ii) 2x2+ 7x + 3
(iii) 6x2+ 5x – 6 (iv) 3x2− x − 4
Solution:
(i) 12x2− 7x + 1 we can find two numbers,
Such that pq = 12 × 1 = 12 and p + q = −7.
They are p = −4 and q = −3
Here, 12x2− 7x + 1
= 12x2− 4x − 3x + 1
= 4x (3x − 1) − 1 (3x − 1)
= (3x − 1) (4x − 1)
(ii) 2x2+ 7x + 3 we can find two numbers such that pq = 2 × 3= 6 and p + q = 7.
They are p = 6 and q = 1.
Here, 2x2 + 7x + 3
= 2x2+ 6x + x + 3
= 2x (x + 3) + 1 (x + 3)
= (x + 3) (2x+ 1)
(iii) 6x2+ 5x − 6 we can find two numbers such that pq = −36 and p + q = 5.
They are p = 9 and q = −4.
Here, 6x2+ 5x – 6
= 6x2+ 9x − 4x – 6
= 3x (2x + 3) − 2 (2x + 3)
= (2x + 3) (3x − 2)
(iv) 3x2− x − 4 we can find two numbers,
such that pq = 3 × (−4) = −12 and p + q = −1.
They are p = −4 and q = 3
Here, 3x2− x − 4
= 3x2− 4x + 3x – 4
= x (3x − 4) + 1 (3x − 4)
= (3x − 4) (x + 1)
Question 5. Factorize:
(i) x3− 2x2− x + 2 (ii) x3+ 3x2−9x − 5
(iii) x3+ 13x2+ 32x + 20 (iv) 2y3+ y2− 2y – 1
Solution:
(i) Let P(x) = x3− 2x2− x + 2 all the factor are there. These are ± 1, ± 2.
By trial method, P (1) = (1)3− 2(1)2− 1 + 2
= 1 − 2 − 1+ 2
= 0 Therefore, (x − 1) is factor of polynomial p(x)
Let us find the quotient on dividing x3− 2x2− x + 2 by x − 1.
By long division method
Now,
Dividend = Divisor × Quotient + remainder
x3− 2x2− x + 2 = (x – 1) ( X2– x – 2) + 0
= (x – 1) (x2–2x+x–2)
= (x – 1) [x (x–2) + 1(x–2)]
= (x – 1) (x + 1) (x – 2)
(ii) Let p(x) = x3 – 3x2−9x – 5 all the factor are there. These are ± 1, ± 2.
By trial method, p (–1) = (–1)3– 3(1)2− 9(1) – 5
= –1– 3–9–5 =0
Therefore (x+1) is the factor of polynomial p(x).
. Let us find the quotient on dividing x3– 3x2−9x – 5 by x+1.
By long division method
Now,
Dividend = Divisor × Quotient + remainder
x3– 3x2−9x – 5 = (x +1) ( X2–4x – 5) + 0
=(x + 1) (x2–5x+x–5)
=(x + 1) [x (x–5) +1(x–5)]
=(x + 1) (x + 1) (x – 5)
(iii) Let p(x) = x3+ 13x2+ 32x + 20 all the factor are there.
These are ± 1, ± 2, ± 3, ± 4.
By trial method, p (–1) = (–1)3+13(–1)2+ 32(–1) +20
= –1+13–32+20 = 0
Therefore (x+1) is the factor of polynomial p(x).
Let us find the quotient on dividing x3+ 13x2+ 32x + 20 by x+1
By long division method
Now,
Dividend = Divisor × Quotient + remainder
x3 +13x2 + 32x + 20 = (x +1) ( x2 + 12x + 20) + 0
=(x + 1) (x2+10x+2x+20)
=(x + 1) [x (x+10) +2(x+10)]
=(x + 1) (x + 2) (x + 10)
(iv) Let p(y) = 2y3+ y2− 2y – 1 all the factor are there. These are ± 1, ± 2.
By trial method, p (1) =2(1)3 + (1)2 – 2(1) – 1
=2 + 1 – 2 – 1 =0
Therefore (y–1) is the factor of polynomial p(y).
Let us find the quotient on dividing 2y3+ y2− 2y – 1 by y–1.
By long division method
Now,
Dividend = Divisor × Quotient + remainder
2y3+ y2− 2y −1 =(y − 1) (2y2+3y + 1)
= (y − 1) (2y2+2y
= (y − 1) [2y (y+1) + 1 (y + 1)]
= (y − 1) (y + 1) (2y + 1)
Topic Lists: