Maharashtra Board Solutions Class 9-Maths (Part 1): Chapter 3.5- Polynomials

Class 9: Maths Chapter 3.5 solutions. Complete Class 9 Maths Chapter 3.5 Notes.

Maharashtra Board Solutions Class 9-Maths (Part 1): Chapter 3.5- Polynomials

Maharashtra Board 9th Maths Chapter 3.5, Class 9 Maths Chapter 3.5 solutions

Question 1.
Find the value of the polynomial 2x – 2x³ + 7 using given values for x.
i. x = 3
ii. x = -1
iii. x = 0
Solution:
i. p(x) = 2x – 2x³ + 7
Put x = 3 in the given polynomial.
∴ p(3) = 2(3) – 2(3)³ + 7
= 6 – 2 x 27 + 7
= 6 – 54 + 7
∴ P(3) = – 41

ii. p(x) = 2x – 2x³ + 7
Put x = -1 in the given polynomial.
∴ p(- 1) = 2(- 1) – 2(-1)³ + 7
= – 2 – 2(-1) + 7
= -2 + 2 + 7
∴ p(-1) = 7

iii. p(x) = 2x – 2x³ + 7
Put x = 0 in the given polynomial.
∴ p(0) = 2(0) – 2(0)³ + 7
= 0 – 0 + 7
∴ P(0) = 7

Question 2.
For each of the following polynomial, find p(1), p(0) and p(- 2).
i. p(x) = x³
ii. p(y) = y² – 2y + 5
ii. p(y) = x⁴ – 2x² + x
Solution:
i. p(x) = x³
∴ p(1) = 1³ = 1
p(x) = x³
∴ p(0) = 0³ = 0
p(x) = x³
∴ p(-2) = (-2)³ = -8

ii. p(y) = y² – 2y + 5
∴ p(1) = 1² – 2(1) + 5
= 1 – 2 + 5
∴ P(1) = 4
p(y) = y² – 2y + 5
∴ p(0) = 0² – 2(0) + 5
= 0 – 0 + 5
∴ p(0) = 5
p(y) = y² – 2y + 5
∴ p(- 2) = (- 2)² – 2(- 2) + 5
= 4 + 4 + 5
∴ p(-2) = 13

iii. p(x) = x⁴ – 2x² – x
∴ p(1) = (1)⁴ – 2(1)² – 1
= 1 – 2 – 1
∴ p(1) = -2
∴ p(x) = x⁴ – 2x² – x
∴ p(0) = (0)⁴ – 2(0)² – 0
= 0 – 0 – 0
∴ p(0) = 0
p(x) = x⁴ – 2x² – x
∴ p(-2) = (-2)⁴ – 2(-2)² – (-2)
= 16 – 2(4) + 2
= 16 – 8 + 2
∴ p(-2) = 10

Question 3.
If the value of the polynomial m³ + 2m + a is 12 for m = 2, then find the value of a.
Solution:
p(m) = m³ + 2m + a
∴ p(2) = (2)³ + 2(2) + a
∴ 12 = 8 + 4 + a … [∵ p(2)= 12]
∴ 12 = 12 + a
∴ a = 12 – 12
∴ a = 0

Question 4.
For the polynomial mx² – 2x + 3 if p(-1) = 7, then find m.
Solution:
p(x) = mx² – 2x + 3
∴ p(- 1) = m (- 1)² – 2(- 1) + 3
∴ 7 = m(1) + 2 + 3 …[∵ p(-1) = 7]
∴ 7 = m + 5
∴ m = 7 – 5
∴ m = 2

Question 5.
Divide the first polynomial by the second polynomial and find the remainder using remainder theorem.
i. (x² – 1x + 9); (x + 1)
ii. (2x³ – 2x² + ax – a); (x – a)
iii. (54m³ + 18m² – 27m + 5); (m – 3)
Solution:
i. p(x) = x² – 7x + 9
Divisor = x + 1
∴ take x = – 1
∴ By remainder theorem,
∴ Remainder =p(-1)
p(x) = x² – 7x + 9
∴ p(-1) = (- 1)² – 7(- 1) + 9
= 1 + 7 + 9
∴ Remainder =17

ii. p(x) = 2x³ – 2x² + ax – a
Divisor = x – a
∴ take x = a
By remainder theorem,
Remainder = p(a)
p(x) = 2x³ – 2x² + ax – a
∴ p(a) = 2a³ – 2a² + a(a) – a
= 2a³– 2a² + a² – a
∴ Remainder = 2a³ – a² – a

iii. p(m) = 54m³ + 18m² – 27m + 5
Divisor = m – 3
∴ take m = 3
∴ By remainder theorem,
Remainder = p(3)
p(m) = 54m³ + 18m² – 27m + 5
∴ p(3) = 54(3)³ +18(3)² – 27(3) + 5
= 54(27) + 18(9) – 81 + 5
= 1458 + 162 – 81 + 5
∴ Remainder = 1544

Question 6.
If the polynomial y³ – 5y² + 7y + m is divided by y + 2 and the remainder is 50, then find the value of m.
Solution:
p(y) = y³ – 5y² + 7y + m
Divisor = y + 2
∴ take y = – 2
∴ By remainder theorem,
Remainder = p(- 2) = 50
P(y) = y³ – 5y² + 7y + m
∴ P(-2) = (- 2)³ – 5(- 2)² + 7(- 2) + m
∴ 50 = -8 – 5(4) – 14 + m
∴ 50 = -8 – 20 – 14 + m
∴ 50 = – 42 + m
∴ m = 50 + 42
∴ m = 92

Question 7.
Use factor theorem to determine whether x + 3 is a factor of x² + 2x – 3 or not.
Solution:
p(x) = x² + 2x – 3
Divisor = x + 3
∴ take x = – 3
∴ Remainder = p(-3)
p(x) = x² + 2x – 3
∴ p(-3) = (-3)² + 2(- 3) – 3
= 9 – 6 – 3
∴ p(-3) = 0
∴ By factor theorem, x + 3 is a factor of x² + 2x – 3.

Question 8.
If (x – 2) is a factor of x³ – mx² + 10x – 20, then find the value of m.
Solution:
p(x) = x³ – mx² + 10x – 20 x – 2 is a factor of x3 – mx2 + lOx – 20.
∴By factor theorem,
Remainder = p(2) = 0
p(x) = x³ – mx² + 10x – 20
∴ p(2) = (2)³ – m(2)² + 10(2) – 20
∴ 0 = 8 – 4m + 20 – 20
∴ 0 = 8 – 4m
∴ 4m = 8
∴ m = 2

Question 9.
By using factor theorem in the following examples, determine whether q(x) is a factor of p(x) or not.
i. p(x) = x³ – x² – x -1 ; q(x) = x – 1
ii. p(x) = 2x³ – x² – 45 ; q(x) = x – 3
Solution:
i. p(x) = x³ – x² – x – 1
Divisor = q(x) = x – 1
∴ take x = 1
Remainder = p(1)
p(x) = x³ – x² – x – 1
∴ P(1) = (1)³ – (1)² – 1 – 1
= 1 – 1 – 1 – 1
= -2 ≠ 0
∴ By factor theorem, x – 1 is not a factor of x³ – x² – x – 1.

ii. p(x) = 2x³ – x – 45
Divisor = q(x) = x – 3
take x = 3
Remainder = p(3)
p(x) = 2x³ – x² – 45
P(3) = 2(3)³ – (3)² – 45
= 2(27) – 9 – 45
= 54 – 9 – 45
= 0
∴ By factor theorem, x – 3 is a factor of 2x³ – x² – 45.

Question 10.
If (x³¹ + 31) is divided by (x + 1), then find the remainder.
Solution:
p(x) = x³¹ + 31
Divisor = x + 1
∴ take x = – 1
∴ By remainder theorem,
Remainder = p(-1)
p(x) =x³¹ + 31 …
∴ p(-1) = (-1)³¹ + 31
= -1 + 31 = 30
∴ Remainder = 30

Question 11.
Show that m – 1 is a factor of m²¹ – 1 and m²² – 1. [3 Marks]
Solution:
i. p(m) = m²¹ – 1
Divisor = m – 1
∴ take m = 1
Remainder = p(1)
p(m) = m²¹ – 1
∴ P(1) = 1²¹ – 1 = 1 – 1 = 0
∴ By factor theorem, m -1 is a factor of m²¹ -1.

ii. p(m) = m²² – 1
Divisor = m – 1
∴ take m = 1
Remainder = p(1)
p(m) = m²² – 1
∴ P(1) = 1²² – 1 = 1 – 1 = 0
∴ By factor theorem, m -1 is a factor of m²² – 1.

Question 12.
If x – 2 and x – 12 both are the factors of the polynomial nx² – 5x + m, then show that m = n = 2.
Solution:
p(x) = nx² – 5x + m
(x – 2) is a factor of nx² – 5x + m.
∴ By factor theorem,
P(2) = 0
∴ p(x) = nx² – 5x + m
∴ p(2) = n(2)² – 5(2) + m
∴ 0 = n(4) – 10 + m
∴ 4n – 10 + m = 0 …(i)
Also, ( x = 12 ) is a factor of nx² – 5x + m.
∴ By factor theorem,
p(12) = 0
p(x) = nx² – 5x + m
∴ p(12) = n(12)² – 512 + m
0 = n4 – 52 + m
∴ 0 = n- 10 +4m … [Multiplying both sides by 4]
∴ n = 10 – 4m ……(ii)
Substituting n = 10 – 4m in equation (i),
4(10 – 4m) – 10 + m = 0
∴ 40 – 16m – 10 + m = 0
∴ -15m+ 30 = 0
∴ -15m = -30
∴ m = 2
Substituting m = 2 in equation (ii),
n = 10 – 4(2)
= 10 – 8
∴ n = 2
∴ m = n = 2

Question 13.
i. If p(x) = 2 + 5x, then find the value of p(2) + p(- 2) – p(1).
Solution:
p(x) = 2 + 5x
∴ P(2) = 2 + 5(2)
= 2 + 10
= 12
p(x) = 2 + 5x
P(- 2) = 2 + 5(- 2)
= 2 – 10 = – 8
p(x) = 2 + 5x
P(1) = 2 + 5(1)
= 2 + 5 = 7
∴ P(2) + P(- 2) – p(1) = 12 + (- 8) – 7
∴ P(2) + p(- 2) – p(1) = – 3

ii. If p(x) = 2x² – 5√3 x + 5, then find the value of p(5√3 ).
Solution:
p(x) = 2x² – 5√3 x + 5
∴ p(5√3) = 2(5√3)² – 5√3 (5√3 ) + 5
= 2 (25 x 3) – 25 x 3 + 5
= 150-75 + 5
∴ p( 5√3 ) = 80

Question 1.
1. Divide p(x) = 3x² + x + 7 by x + 2. Find the remainder.
2. Find the value of p(x) = 3x² + x + 7 when x = – 2.
3. See whether remainder obtained by division is same as the value of p(-2). Take one more example and verify. (Textbook pg. no. 50)
Solution:

∴ Remainder = 17

2. p(x) = 3x² + x + 7
Substituting x = – 2, we get
p(-2) = 3(2)² + (-2) + 7
= 12 – 2 + 7
∴ p(-2) = 17

3. Yes, remainder = p(-2)

Another Example:
If the polynomial t³ – 3t² + kt + 50 is divided by (t – 3), the remainder is 62. Find the value of k.
Solution:
When given polynomial is divided by (t – 3) the remainder is 62. It means the value of the polynomial when t = 3 is 62.
p(t) = t³ – 3t³ + kt + 50
By remainder theorem,
Remainder = p(3) = 33 – 3² + k x 3 + 50
= 27 – 3 x 9 + 3k + 50
= 27 – 27 + 3k + 50
= 3k + 50
But remainder is 62.
∴ 3k + 50 = 62
∴ 3k = 62 – 50
∴ 3k = 12
∴ k = 4

Question 2.
Verify that (x – 1) is a factor of the polynomial x³ + 4x – 5. (Textbook pg. no. 51)
Solution:
Here, p(x) = x³ + 4x – 5
Substituting x = 1 in p(x), we get
p(1) = (1)³ + 4(1) – 5
= 1 + 4 – 5
P(1) = 0
∴ By remainder theorem,
Remainder = 0
∴ (x -1) is the factor of x³ + 4x – 5.

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Maharashtra Board Solutions Class 9-Maths (Part 1): Chapter 3.5- Polynomials

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Chapterwise Maharashtra Board Solutions Class 9 Maths :

Part 1

• Chapter 1.1- Sets
• Chapter 1.2- Sets
• Chapter 1.3- Sets
• Chapter 1.4- Sets
• Chapter 2.1- Real Numbers
• Chapter 2.2- Real Numbers
• Chapter 2.3- Real Numbers
• Chapter 2.4- Real Numbers
• Chapter 2.5- Real Numbers
• Chapter 3.1- Polynomials
• Chapter 3.2- Polynomials
• Chapter 3.3- Polynomials
• Chapter 3.4- Polynomials
• Chapter 3.5- Polynomials
• Chapter 3.6- Polynomials
• Chapter 4.1- Ratio and Proportion
• Chapter 4.2- Ratio and Proportion
• Chapter 4.3- Ratio and Proportion
• Chapter 4.4- Ratio and Proportion
• Chapter 4.5- Ratio and Proportion
• Chapter 5.1- Linear Equations in Two Variables
• Chapter 5.2- Linear Equations in Two Variables
• Chapter 6.1- Financial Planning
• Chapter 6.2- Financial Planning
• Chapter 7.1- Statistics
• Chapter 7.2- Statistics
• Chapter 7.3- Statistics
• Chapter 7.4- Statistics
• Chapter 7.5- Statistics

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The Maharashtra State Board of Secondary and Higher Secondary Education or MSBSHSE (Marathi: महाराष्ट्र राज्य माध्यमिक आणि उच्च माध्यमिक शिक्षण मंडळ), is an autonomous and statutory body established in 1965. The board was amended in the year 1977 under the provisions of the Maharashtra Act No. 41 of 1965.

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