Practice Set 3.2 Algebra 9th Std Maths Part 1 Answers Chapter 3 Polynomials
Solution & Step-by-Step Answer:
i. There are ‘a’ trees in the village Lat. If the number of trees increases every year by ’b‘. then how many trees will there be after ‘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 × Number of students in each row
= x × 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.
Solution & Step-by-Step Answer:
i. (x3 – 2x2 – 9) + (5x3 + 2x + 9) = x3 – 2x2 – 9 + 5x3 + 2x + 9 = x3 + 5x3 – 2x2 + 2x – 9 + 9 = 6x3 – 2x2 + 2x
ii. (-7m4+ 5m3+ √2 ) + (5m4– 3m3+ 2m2+ 3m – 6)
= -7m4+ 5m3+ √2 + 5m4– 3m3+ 2m2+ 3m – 6
=-7m4+ 5m4+5m3– 3m3+ 2m2+ 3m +√2 – 6
= -2m4+ 2m3+ 2m2+ 3m + √2 – 6
iii. (2y2+ 7y + 5) + (3y + 9) + (3y2– 4y – 3)
= 2y2+ 7y + 5 + 3y + 9 + 3y2– 4y – 3
=2y2+ 3y2+7y + 3y – 4y+5 + 9 – 3
= 5y2+ 6y + 11
Solution & Step-by-Step Answer:
i. x2 – 9x + √3 -(- 19x + √3 + 7x2) = x2 – 9x + √3 + 19x – √ 3 – 7x2 = x2 – 7x2 – 9x + 19x + √3 – √3 = – 6x2 + 10x
ii. (2ab2+ 3a2b – 4ab) – (3ab – 8ab2+ 2a2b)
= 2ab2+ 3a2b – 4ab – 3ab + 8ab2– 2a2b
=2ab2+ 8ab2+3a2b – 2a2b–4ab – 3ab
= 10ab2+ a2b – 7ab
Solution & Step-by-Step Answer:
i. (2x) x (x2 – 2x – 1) = 2x3 – 4x2 – 2x
ii. (x5– 1) × (x3+ 2x2+ 2)
= x5(x3+ 2x2+ 2) -1(x3+ 2x2+ 2)
= x8+ 2x7+ 2x5– x3– 2x2– 2
iii. (2y + 1) × (y2– 2y3+ 3y)
= 2y(y2– 2y3+ 3y) + 1(y2– 2y3+ 3y)
= 2y3– 4y4+ 6y2+ y2– 2y3+ 3y
= -4y4+2y3– 2y3+6y2+ y2+ 3y
= -4y4+ 7y2+ 3y
Solution & Step-by-Step Answer:
i. x3 – 64 = x3 + 0x2 + 0x – 64 ∴ Quotient = x2 + 4x + 16, Remainder = 0 Now, Dividend = Divisor x Quotient + Remainder ∴ x3 – 64 = (x – 4)(x2 + 4x + 16) + 0
ii. 5x5+ 4x4– 3x3+ 2x2+ 2 = 5x5+ 4x4– 3x3+ 2x + 0x + 2
∴ Quotient = 5x3+ 9x2+ 6x + 8,
Remainder = 8x + 2
Now, Dividend = Divisor x Quotient + Remainder
∴ 5x5+ 4x4– 3x3+ 2x2+ 2 = (x2– x)(5x3+ 9x2+ 6x + 8) + (8x + 2)
Solution & Step-by-Step Answer:
Length of the rectangular farm = (2a2 + 3b2) m Breadth of the rectangular farm = (a2 + b2) m Area of the farm = length x breadth = (2a2 + 3b2) x (a2 + b2) = 2a2(a2 + b2) + 3b2(a2 + b2) = 2a2 + 2a2b2 + 3a2b2 + 3b4 = (2a4 + 5a2b2 + 3b4) sq. m … (i) The farmer used a square shaped plot of the farm to build a house. Side of the square shaped plot = (a2 – b2) m ∴ Area of the plot = (side)2 = (a2 – b2)2 = (a4 – 2a2b2 + b4) sq m….(ii)
∴ Area of the remaining farm = Area of the farm – Area of the plot
= (2a4+ 5a2b2+ 3b4) – (a4– 2a2b2+ b4) … [From (i) and (ii)]
= 2a4+ 5a2b2+ 3b4– a4+ 2a2b2– b4=2a4– a4+5a2b2+ 2a2b2+3b4– b4
= a4+ 7a2b2+ 2b4
∴ The area of the remaining farm is (a4+ 7a2b2+ 2b4) sq. m.