Problem Set 3 Algebra 9th Std Maths Part 1 Answers Chapter 3 Polynomials
i. Which of the following is a polynomial?
(D) √2x² +
ii. What is the degree of the polynomial √7 ?
(A)
(B) 5
(C) 2
(D) 0
Answer:
(D) 0
iii. What is the degree of the polynomial ?
(A) 0
(B) 1
(C) undefined
(D) any real number
Answer:
(C) undefined
iv. What is the degree of the polynomial 2x2+ 5xsup>3 + 7?
(A) 3
(B) 2
(C) 5
(D) 7
Answer:
(A) 3
v. What is the coefficient form of x3– 1 ?
(A) (1, -1)
(B) (3, -1)
(C) (1, 0, 0, -1)
(D) (1, 3, -1)
Answer:
(C) (1, 0, 0, -1)
vi. p(x) = x2– x + 3, then p (7√7) = ?
(A) 3
(B) 7√7
(C) 42√7+3
(D) 49√7
Answer:
(D) 49√7
vii. When x = – 1, what is the value of the polynomial 2x3+ 2x ?
(A) 4
(B) 2
(C) -2
(D) -4
Answer:
(A) 4
viii. If x – 1 is a factor of the polynomial 3x2+ mx, then find the value of m.
(A) 2
(B) -2
(C) -3
(D) 3
Answer:
(C) -3
ix. Multiply (x2– 3) (2x – 7x3+ 4) and write the degree of the product.
(A) 5
(B) 3
(C) 2
(D) 0
Answer:
(A) 5
x. Which is the following is a linear polynomials?
(A) x + 5
(B) x2+ 5
(C) x3+ 5
(D) x4+ 5
Answer:
(A) x + 5
Hints:
v. x3– 1 = x3+ 0x2+ 0x – 1
vi. p(7√ 7) = (7√ 7)2(7√ 7) (7√ 7) + 3
= 3
vii. p(-1) = 2(-1)3+ 2(-1)
= -2 – 2 = -4
vii. p(1) = 0
∴ 3(1)2+ m(1) = 0
∴ 3 + m =0
∴ m = -3
ix. Here, degree of first polynomial = 2 and
degree of second polynomial 3
∴ Degree of polynomial obtained by multiplication = 2 + 3 = 5
Question 2
Maharashtra Board Solution
Write the degree of the polynomial for each of the following. i. 5 + 3x4 ii. 7 iii. ax7 + bx9 (a, b are constants)
Solution & Step-by-Step Answer:
i. 5 + 3x4 Here, the highest power of x is 4. ∴Degree of the polynomial = 4
ii. 7 = 7x° iii. ax7+ bx9
Question 3
Maharashtra Board Solution
Write the following polynomials in standard form. [1 Mark each] i. 4x2 + 7x4 – x3 – x + 9 ii. p + 2p3 + 10p2 + 5p4 – 8
Solution & Step-by-Step Answer:
i. 7x4 – x3 + 4x2 – x + 9 ii. 5p4 + 2p3 + 10p2 + p – 8
Question 4
Maharashtra Board Solution
Write the following polynomial in coefficient form. i. x4 + 16 ii. m5 + 2m2 + 3m+15
Solution & Step-by-Step Answer:
i. x4 + 16 Index form = x4 + 0x3 + 0x2 + 0x + 16 ∴ Coefficient form of the polynomial = (1,0,0,0,16)
ii. m5+ 2m2+ 3m + 15
Question 5
Maharashtra Board Solution
Write the index form of the polynomial using variable x from its coefficient form. i. (3, -2, 0, 7, 18) ii. (6, 1, 0, 7) iii. (4, 5, -3, 0)
Solution & Step-by-Step Answer:
i. Number of coefficients = 5 ∴ Degree = 5 – 1 = 4 ∴Index form = 3x4 – 2x3 + 0x2 + 7x + 18
ii. Number of coefficients = 4 iii. Number of coefficients = 4
Question 6
Maharashtra Board Solution
Add the following polynomials. i. 7x4 – 2x3 + x + 10; 3x4 + 15x3 + 9x2 – 8x + 2 ii. 3p3q + 2p2q + 7; 2p2q + 4pq – 2p3q
Solution & Step-by-Step Answer:
i. (7x4 – 2x3 + x + 10) + (3x4 + 15x3 + 9x2 – 8x + 2) = 7x4 – 2x3 + x + 10 + 3x4 + 15x3 + 9x2 – 8x + 2 = 7x4 + 3x4 – 2x3 + 1 5x3 + 9x2 + x – 8x + 10 + 2 = 10x4 + 13x3 + 9x2 – 7x + 12
ii. (3p3q + 2p2q + 7) + (2p2q + 4pq – 2p3q)
Question 7
Maharashtra Board Solution
Subtract the second polynomial from the first. i. 5x2 – 2y + 9 ; 3x2 + 5y – 7 ii. 2x2 + 3x + 5 ; x2 – 2x + 3
Solution & Step-by-Step Answer:
i. (5x2 – 2y + 9) – (3x2 + 5y – 7) = 5x2 – 2y+ 9 – 3x2 – 5y + 1 = 5x2 – 3x2 – 2y – 5y + 9 + 7 = 2x2 – 1y + 16
ii. (2x2+ 3x + 5) – (x2– 2x + 3)
Question 8
Maharashtra Board Solution
Multiply the following polynomials. i. (m3 – 2m + 3) (m4 – 2m2 + 3m + 2) ii. (5m3 – 2) (m2 – m + 3)
Solution & Step-by-Step Answer:
i. (m3 – 2m + 3) (m4 – 2m2 + 3m + 2) = m3(m4 – 2m2 + 3m + 2) – 2m(m4 – 2m2 + 3m + 2) + 3(m4 – 2m2 + 3m + 2) = m7 – 2m5 + 3m4 + 2m3 – 2m5 + 4m3 – 6m2 – 4m + 3m4 – 6m2 + 9m + 6 = m7 – 2m5 – 2m5 + 3m4 + 3m4 + 2m3 + 4m3 – 6m2 – 6m2 – 4m + 9m + 6 = m7 – 4m5 + 6m4 + 6m3 – 12m2 + 5m + 6
ii. (5m3– 2) (m2– m + 3)
Question 9
Maharashtra Board Solution
Divide polynomial 3x3 – 8x2 + x + 7 by x – 3 using synthetic method and write the quotient and remainder.
Solution & Step-by-Step Answer:
Dividend = 3x3 – 8x2 + x + 7 ∴ Coefficient form of dividend = (3, – 8, 1,7) Divisor = x – 3 ∴ Opposite of – 3 is 3 Coefficient form of quotient = (3, 1,4) ∴ Quotient = 3x2 + x + 4 and Remainder =19
Question 10
Maharashtra Board Solution
For which value of m, x + 3 is the factor of the polynomial x3 – 2mx + 21?
Solution & Step-by-Step Answer:
Here, p(x) = x3 – 2mx + 21 (x + 3) is a factor of x3 – 2mx + 21. ∴ By factor theorem, Remainder = 0 ∴ P(- 3) = 0 p(x) = x3 – 2mx + 21 ∴ p(-3) = (-3)3 – 2(m)(-3) + 21 ∴ 0 = – 27 + 6m + 21 ∴ 6 + 6m = 0 ∴ 6m = 6 ∴ m = 1 ∴ x + 3 is the factor of x3 – 2mx + 21 for m = 1.
Question 11
Maharashtra Board Solution
At the end of the year 2016, the population of villages Kovad, Varud, Chikhali is 5x2 – 3y2, 7y2 + 2xy and 9x2 + 4xy respectively. At the beginning of the year 2017, x2 + xy – y2, 5xy and 3x2 + xy persons from each of the three villages respectively went to another village for education, then what is the remaining total population of these three villages ?
Solution & Step-by-Step Answer:
Total population of villages at the end of 2016 = (5x2 – 3y2) + (7y2 + 2xy) + (9x2 + 4xy) = 5x2 + 9x2 – 3y2 + 7y2 + 2xy + 4xy = 14x2 + 4y2 + 6xy …….(i) Total number of persons who went to other village at the beginning of 2017 = (x2 + xy – y2) + (5xy) + (3x2 + xy) = x2 + 3x2 – y2 + xy + 5xy + xy = 4x2 – y2 + 7xy … (ii) Remaining total population of villages = Total population at the end of 2016 – total number of persons who went to other village at the beginning of 2017 = 14x2 + 4y2 + 6xy – (4x2 – y2 + 7xy) … [From (i) and (ii)] = 14x2 + 4y2 + 6xy – 4x2 + y2 – 7xy = 14x2 – 4x2 + 4y2 + y2 + 6xy – 7xy = 1 = 10x2 + 5y2 – xy ∴ The remaining total population of the three villages is 10x2 + 5y2 – xy.
Question 12
Maharashtra Board Solution
Polynomials bx2 + x + 5 and bx3 – 2x + 5 are divided by polynomial x – 3 and the remainders are m and n respectively. If m – n = 0, then find the value of b.
Solution & Step-by-Step Answer:
When polynomial bx2 + x + 5 is divided by (x – 3), the remainder is m. ∴ By remainder theorem, Remainder = p(3) = m p(x) = bx2 + x + 5 ∴ p(3) = b(3)2 + 3 + 5 ∴m = b(9) + 8 m = 9b + 8 …(i) When polynomial bx3 – 2x + 5 is divided by x – 3 the remainder is n ∴ remainder = p(3) = n p(x) = bx3 – 2x + 5 ∴ P(3)= b(3)3 – 2(3) + 5 ∴ n = b(27) – 6 + 5 ∴ n = 27b – 1 …(ii) Now, m – n = 0 …[Given] ∴ m = n ∴ 9b + 8 = 27b – 1 …[From (i) and (ii)] ∴ 8 + 1 = 27b – 9b ∴ 9 = 18b ∴ b =
Question 13
Maharashtra Board Solution
Simplify. (8m2 + 3m – 6) – (9m – 7) + (3m2 – 2m + 4)
Solution & Step-by-Step Answer:
(8m2 + 3m – 6) – (9m – 7) + (3m2 – 2m + 4) = 8m2 + 3m – 6 – 9m + 7 + 3m2 – 2m + 4 = 8m2 + 3m2 + 3m – 9m – 2m – 6 + 7 + 4 = 11m2 – 8m + 5
Question 14
Maharashtra Board Solution
Which polynomial is to be subtracted from x2 + 13x + 7 to get the polynomial 3x2 + 5x – 4?
Solution & Step-by-Step Answer:
Let the required polynomial be A. ∴ (x2 + 13x + 7) – A = 3x2 + 5x – 4 ∴ A = (x2 + 13x + 7) – (3x2 + 5x – 4) = x2 + 13x + 7 – 3x2 – 5x + 4 = x2 – 3x2 + 13x – 5x + 7+4 = -2x2 + 8x + 11 ∴ – 2x2 + 8x + 11 must be subtracted from x2 + 13x + 7 to get 3x2 + 5x – 4.
Question 15
Maharashtra Board Solution
Which polynomial is to be added to 4m + 2n + 3 to get the polynomial 6m + 3n + 10?
Solution & Step-by-Step Answer:
Let the required polynomial be A. ∴ (4m + 2n + 3) + A = 6m + 3n + 10 ∴ A = 6m + 3n + 10 – (4m + 2n + 3) = 6m + 3n + 10 – 4m – 2n – 3 = 6m – 4m + 3n – 2n + 10 – 3 = 2m + n + 7 ∴ 2m + n + 7 must be added to 4m + 2n + 3 to get 6m + 3n + 10.
Question 1
Maharashtra Board Solution
Read the following passage, write the appropriate amount in the boxes and discuss. Govind, who is a dry land farmer from Shiralas has a 5 acre field. His family includes his wife, two children and his old mother. He borrowed one lakh twenty five thousand rupees from the bank for one year as agricultural loan at 10 p.c.p.a. He cultivated soyabean in x acres and cotton and tur in y acres. The expenditure he incurred was as follows : He spent ₹10,000 on seeds. The expenses for fertilizers and pesticides for the soyabean crop was ₹ 2000x and ₹ 4000x2 were spent on wages and cultivation of land. He spent ₹ 8000y on fertilizers and pesticides and ₹ 9000y2 for wages and cultivation of land for the cotton and tur crops. |
Let us write the total expenditure on all the crops by using variables x and y.
₹ 10000 + 2000x + 4000×2 + 8000y + 9000y2
He harvested 5x2quintals soyabean and sold it at ₹ 2800 per quintal. The cotton crop yield was y2quintals which fetched ₹ 5000 per quintal.
The tur crop yield was 4y quintals and was sold at ₹ 4000 per quintal. Write the total income in rupees that was obtained by selling the entire farm produce, with the help of an expression using variables x and y. (Textbook pg. no. 44)
Total income = income on soyabean crop + income on cotton crop + income on tur crop
= ₹ (5x2x 2800) + ₹( y2x 5000) + ₹ (4y x 4000)
= ₹ ( 14000x2+ y2+ 16000y)
Question 2
Maharashtra Board Solution
We have seen the example of expenditure and income (in terms of polynomials) of Govind who is a dry land farmer. He has borrowed rupees one lakh twenty-five thousand from the bank as an agriculture loan and repaid the said loan at 10 p.c.p.a. He had spent ₹ 10,000 on seeds. The expenses on soyabean crop was ₹ 2000x for fertilizers and pesticides and ₹ 4000x2 was spent on wages and cultivation. He spent ₹ 8000y on fertilizers and pesticides and ₹9000y2 on cultivation and wages for cotton and tur crop. His total income was ₹ (14000x2 + y2 + 16000y) By taking x = 2, y = 3 write the income expenditure account of Govind’s farming. (Textbook pg. no. 52)
Solution & Step-by-Step Answer:
– Credit (Income) ₹ 1,25,000 Bank loan ₹ 56000 Income from soyabean ₹ 75000 Income from cotton ₹ 48000 Income from tur ₹ 304000 Total income
– Debit (Expenses) |