Practice Set 3.2 Algebra 9th Standard Maths Part 1 Chapter 3 Polynomials Solutions Maharashtra Board

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 3.2 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 3 Polynomials.

9th Standard Maths 1 Practice Set 3.2 Chapter 3 Polynomials Textbook Answers Maharashtra Board

Class 9 Maths Part 1 Practice Set 3.2 Chapter 3 Polynomials Questions With Answers Maharashtra Board

Question 1.
Use the given letters to write the answer.
i. There are β€˜a’ trees in the village Lat. If the number of trees increases every year by ’bβ€˜. then how many trees will there be after β€˜x’ years?
ii. For the parade there are y students in each row and x such row are formed. Then, how many students are there for the parade in all ?
iii. The tens and units place of a two digit number is m and n respectively. Write the polynomial which represents the two digit number.
Solution:
i. Number of trees in the village Lat = a
Number of trees increasing each year = b
∴ Number of trees after x years = a + bx
∴ There will be a + bx trees in the village Lat after x years.

ii. Total rows = x
Number of students in each row = y
∴ Total students = Total rows Γ— Number of students in each row
= x Γ— y
= xy
∴ There are in all xy students for the parade.

iii. Digit in units place = n
Digit in tens place = m
∴ The two digit number = 10 x digit in tens place + digit in units place
= 10m + n
∴ The polynomial representing the two digit number is 10m + n.

Question 2.
Add the given polynomials.
i. x 3 – 2x 2 – 9; 5x 3 + 2x + 9
ii. -7m 4 + 5m 3 + √2 ; 5m 4 – 3m 3 + 2m 2 + 3m – 6
iii. 2y 2 + 7y + 5; 3y + 9; 3y 2 – 4y – 3
Solution:
i. (x 3 – 2x 2 – 9) + (5x 3 + 2x + 9)
= x 3 – 2x 2 – 9 + 5x 3 + 2x + 9
= x 3 + 5x 3 – 2x 2 + 2x – 9 + 9
= 6x 3 – 2x 2 + 2x

ii. (-7m 4 + 5m 3 + √2 ) + (5m 4 – 3m 3 + 2m 2 + 3m – 6)
= -7m 4 + 5m 3 + √2 + 5m 4 – 3m 3 + 2m 2 + 3m – 6
= -7m 4 + 5m 4 + 5m 3 – 3m 3 + 2m 2 + 3m + √2 – 6
= -2m 4 + 2m 3 + 2m 2 + 3m + √2 – 6

iii. (2y 2 + 7y + 5) + (3y + 9) + (3y 2 – 4y – 3)
= 2y 2 + 7y + 5 + 3y + 9 + 3y 2 – 4y – 3
= 2y 2 + 3y 2 + 7y + 3y – 4y + 5 + 9 – 3
= 5y 2 + 6y + 11

Question 3.
Subtract the second polynomial from the first.
i. x 2 – 9x + √3 ; – 19x + √3 + 7x 2
ii. 2ab 2 + 3a 2 b – 4ab; 3ab – 8ab 2 + 2a 2 b
Solution:
i. x 2 – 9x + √3 -(- 19x + √3 + 7x 2 )
= x 2 – 9x + √3 + 19x – √ 3 – 7x 2
= x 2 – 7x 2 – 9x + 19x + √3 – √3
= – 6x 2 + 10x

ii. (2ab 2 + 3a 2 b – 4ab) – (3ab – 8ab 2 + 2a 2 b)
= 2ab 2 + 3a 2 b – 4ab – 3ab + 8ab 2 – 2a 2 b
= 2ab 2 + 8ab 2 + 3a 2 b – 2a 2 b – 4ab – 3ab
= 10ab 2 + a 2 b – 7ab

Question 4.
Multiply the given polynomials.
i. 2x; x 2 – 2x – 1
ii. x 5 – 1; x 3 + 2x 2 + 2
iii. 2y +1; y 2 – 2y + 3y
Solution:
i. (2x) x (x 2 – 2x – 1) = 2x 3 – 4x 2 – 2x

ii. (x 5 – 1) Γ— (x 3 + 2x 2 + 2)
= x 5 (x 3 + 2x 2 + 2) -1(x 3 + 2x 2 + 2)
= x 8 + 2x 7 + 2x 5 – x 3 – 2x 2 – 2

iii. (2y + 1) Γ— (y 2 – 2y 3 + 3y)
= 2y(y 2 – 2y 3 + 3y) + 1(y 2 – 2y 3 + 3y)
= 2y 3 – 4y 4 + 6y 2 + y 2 – 2y 3 + 3y
= -4y 4 + 2y 3 – 2y 3 + 6y 2 + y 2 + 3y
= -4y 4 + 7y 2 + 3y

Question 5.
Divide first polynomial by second polynomial and write the answer in the form β€˜Dividend = Divisor x Quotient + Remainder’.
i. x 3 – 64; x – 4
ii. 5x 5 + 4x 4 – 3x 3 + 2x 2 + 2 ; x 2 – x
Solution:
i. x 3 – 64 = x3 + 0x 2 + 0x – 64
Maharashtra-Board-Class-9-Maths-Solutions-Chapter-3-Polynomials-Practice-Set-3.2-1
∴ Quotient = x 2 + 4x + 16, Remainder = 0
Now, Dividend = Divisor x Quotient + Remainder
∴ x 3 – 64 = (x – 4)(x 2 + 4x + 16) + 0

ii. 5x 5 + 4x 4 – 3x 3 + 2x 2 + 2 = 5x 5 + 4x 4 – 3x 3 + 2x + 0x + 2
Maharashtra-Board-Class-9-Maths-Solutions-Chapter-3-Polynomials-Practice-Set-3.2-2
∴ Quotient = 5x 3 + 9x 2 + 6x + 8,
Remainder = 8x + 2
Now, Dividend = Divisor x Quotient + Remainder
∴ 5x 5 + 4x 4 – 3x 3 + 2x 2 + 2 = (x 2 – x)(5x 3 + 9x 2 + 6x + 8) + (8x + 2)

Question 6.
Write down the information in the form of algebraic expression and simplify.
There is a rectangular farm with length (2a 2 + 3b 2 ) metre and breadth (a 2 + b 2 ) metre. The farmer used a square shaped plot of the farm to build a house. The side of the plot was (a2 – b2) metre. What is the area of the remaining part of the farm? [4 Marks]
Solution:
Length of the rectangular farm = (2a 2 + 3b 2 ) m
Breadth of the rectangular farm = (a 2 + b 2 ) m
Area of the farm = length x breadth = (2a 2 + 3b 2 ) x (a 2 + b 2 )
= 2a 2 (a 2 + b 2 ) + 3b 2 (a 2 + b 2 )
= 2a 2 + 2a 2 b 2 + 3a 2 b 2 + 3b 4
= (2a 4 + 5a 2 b 2 + 3b 4 ) sq. m … (i)
The farmer used a square shaped plot of the farm to build a house.
Side of the square shaped plot = (a 2 – b 2 ) m
∴ Area of the plot = (side) 2
= (a 2 – b 2 ) 2
= (a 4 – 2a 2 b 2 + b 4 ) sq m… .(ii)

∴ Area of the remaining farm = Area of the farm – Area of the plot
= (2a 4 + 5a 2 b 2 + 3b 4 ) – (a 4 – 2a 2 b 2 + b 4 ) … [From (i) and (ii)]
= 2a 4 + 5a 2 b 2 + 3b 4 – a 4 + 2a 2 b 2 – b 4
= 2a 4 – a 4 + 5a 2 b 2 + 2a 2 b 2 + 3b 4 – b 4
= a 4 + 7a 2 b 2 + 2b 4
∴ The area of the remaining farm is (a 4 + 7a 2 b 2 + 2b 4 ) sq. m.