Class 9 2. Polynomials Exercise 2.4: NCERT Book Solutions
Class 9 chapter 2. Polynomials important extra long questions with solution for board exams and term 1 and term 2 exams.
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Class 9 chapter 2. Polynomials important extra long questions with solution for board exams and term 1 and term 2 exams. - 2. Polynomials - Exercise 2.4: NCERT Book Solutions for class 9th. All solutions and extra or additional solved questions for 2. Polynomials : Exercise 2.4 Mathematics class 9th:English Medium NCERT Book Solutions. Class 9 chapter 2. Polynomials important extra long questions with solution for board exams and term 1 and term 2 exams.
2. Polynomials : Exercise 2.4 Mathematics class 9th:English Medium NCERT Book Solutions
Class 9 chapter 2. Polynomials important extra long questions with solution for board exams and term 1 and term 2 exams. - 2. Polynomials - Exercise 2.4: NCERT Book Solutions for class 9th. All solutions and extra or additional solved questions for 2. Polynomials : Exercise 2.4 Mathematics class 9th:English Medium NCERT Book Solutions.
Class 9 2. Polynomials Exercise 2.4: NCERT Book Solutions
NCERT Books Subjects for class 9th Hindi Medium
2. Polynomials
Class 9 chapter 2. Polynomials important extra long questions with solution for board exams and term 1 and term 2 exams.
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)
Class 9 chapter 2. Polynomials important extra long questions with solution for board exams and term 1 and term 2 exams.
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Class 9 chapter 2. Polynomials important extra long questions with solution for board exams and term 1 and term 2 exams. - 2. Polynomials - Exercise 2.4: NCERT Book Solutions for class 9th. All solutions and extra or additional solved questions for 2. Polynomials : Exercise 2.4 Mathematics class 9th:English Medium NCERT Book Solutions. Class 9 chapter 2. Polynomials important extra long questions with solution for board exams and term 1 and term 2 exams.
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Mathematics Chapter List
1. Number Systems
2. Polynomials
3. Coordinate Geometry
4. Linear Equation In Two Variables
5. Introduction To Euclid’s Geometry
6. Lines and Angles
7. Triangles
8. Quadrilaterals
9. Area Parallelograms and Triangles
10. Circles
11. Constructions
12. Herons Formula
13. Surface Areas and Volumes
14. Statistics
15. Probability