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2. Polynomials Mathematics class 9 in English Medium ncert book solutions Exercise 2.4
2. Polynomials Exercise 2.4 – Complete NCERT Book Solutions for Class 9 Mathematics (English Medium). Get all chapter explanations, extra questions, solved examples and additional practice questions for 2. Polynomials Exercise 2.4 to help you master concepts and score higher.
2. Polynomials Mathematics class 9 in English Medium ncert book solutions Exercise 2.4
NCERT Solutions for Class 9 Mathematics play an important role in helping students understand the concepts of the chapter 2. Polynomials clearly. This chapter includes the topic Exercise 2.4, which is essential from both academic and examination point of view. The solutions provided here are prepared strictly according to the latest NCERT syllabus and follow the guidelines of CBSE to ensure accuracy and relevance. Each question is explained in a simple and student-friendly manner so that learners can grasp the concepts without confusion. These NCERT Solutions are useful for regular study, homework help, and exam preparation. All textbook questions are solved step by step to improve problem-solving skills and conceptual clarity. Students of Class 9 studying Mathematics can use these solutions to revise important topics, understand difficult questions, and practise effectively before examinations. The chapter 2. Polynomials is explained in a structured way, making it easier for students to connect the theory with the topic Exercise 2.4. By studying these updated NCERT Solutions for Class 9 Mathematics, students can build a strong foundation, boost their confidence, and score better marks in school and board exams.
2. Polynomials
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)
See other sub-topics of this chapter:
1. Exercise 2.1 2. Exercise 2.2 3. Exercise 2.3 4. Exercise 2.4 5. Exercise 2.5