Toppers Study Toppers Study Toppers StudyDetailed NCERT Solutions for 9 Mathematics 2. Polynomials to simplify learning. Understand chapters clearly and practice with free solutions for better results.
Detailed NCERT Solutions for 9 Mathematics 2. Polynomials to simplify learning. Understand chapters clearly and practice with free solutions for better results.
Preparing for exams becomes easier with Exercise 2.4. Whether you are studying for board exams or mid-term exams, 9 Mathematics Chapter 2. Polynomials solutions provide quick revising points, well-structured answers, and additional practice material to help you score better.
ncert_solutionsChapter 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)
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