Practice Set 3.2 Polynomials Class 9th Mathematics Part I MHB Solution

Polynomials

Class 9th Mathematics Part I MHB Solution

Practice Set 3.2
  1. Use the given letters to write the answer. i. There are ‘a’ trees in the village Lat.…
  2. Add the given polynomials. i. x^3 - 2x^2 - 9; 5x^3 + 2x +9 ii. -7m^4 + 5m^3 + root 2…
  3. Subtract the second polynomial from the first. i. x^2 - 9x + root 3-19x + root 3+7x^2…
  4. Multiply the given polynomials. i. 2x; x^2 - 2x ii. x^5 - 1; x^3 + 2x^3 + 2 iii. 2y +…
  5. Divide first polynomial by second polynomial and write the answer in the form ‘Dividend…
  6. Write down the information in the form of algebraic expression and simplify. There is a…

Practice Set 3.2

Question 1.

Use the given letters to write the 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 is 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.


Answer:

i. Given:

Total number of trees in the village = a


Increase in the number of trees by = b


To find: Number of trees after x years


Number of trees in x years = (a + b) x


ii. Given:


Number of students in each row = y


Number of rows of Students = x


So total number of students = x × y


= xy


iii. For a number in tens place should be multiplied with ten and for the number in units place should be multiplied by one. So the number that can be formed as follows:


Since m is in tens place and n is in units place,


10 × m + 1 × n


10m + n


So the polynomial representing the two digit number is = 10m + n



Question 2.

Add the given polynomials.

i. x3 – 2x2 – 9; 5x3 + 2x +9

ii. 

iii. 2y2 + 5; 3y +9; 3y2 – 4y – 3


Answer:

i.


The result = 6x3 – 2x2 + 2x


ii.



The result = 2m4 + 2m3 + 2m2 + 3m – 6 +√2


iii.



The result = 5y2 + 6y + 11



Question 3.

Subtract the second polynomial from the first.

i. 

ii. 2ab2 + 3a2b – 4ab; 3ab – 8ab2 +2a2b


Answer:

In the subtraction process, the sign of the subtrahend that is the second polynomial is inverted and then the operation is carried out.


The result = 6x2 + 10x


ii. In the subtraction process, the sign of the subtrahend that is the second polynomial is inverted and then the operation is carried out.



The result = 10ab2 – a2b – 7ab



Question 4.

Multiply the given polynomials.

i. 2x; x2 – 2x

ii. x5 – 1; x3 + 2x3 + 2

iii. 2y + 1; y2 – 2y3 + 3y


Answer:

The multiplication is as follows:

⇒ 2x × (x2 – 2x – 1)


⇒ 2x.x2 – 2x.2x – 2x.1


⇒ 2x2+1 – 4x1+1 – 2x


⇒ 2x3 – 4x2 – 2x


Therefore, the product = 2x3 – 4x2 – 2x


“.” represents multiplication


ii. The multiplication is as follows:


⇒ (x5 – 1) × (x3 + 2x3 + 2)


⇒ x5.x3 + 2x3.x5 + 2.x5 – 1.x3 – 1.2x3 – 1.2


⇒ x5+3 + 2x3 + 5 +2x5 – x3 – 2x3 – 2


⇒ x8 + 2x8 + 5 + 2x5 – x3 – 2x3 – 2


⇒ 3x8 + 2x5 + 3x3 + 3


The product is = 3x8 + 2x5 + 3x3 + 3


“.” represents multiplication


iii. The multiplication is as follows:


⇒ (2y + 1) × (y2 – 2y3 + 3y)


⇒ 2y.y2 – 2y3.2y + 3y.2y + 1.y2 – 1.2y3 + 1.3y


⇒ 2y2+1 – 4y3+1 + 6y1+1 + y2 – 2y3 + 3y


⇒ 2y3 – 4y4 + 6y2 + y2 – 2y3 + 3y


⇒ -4y4 + 7y2 + 3y


The product is = -4y4 + 7y2 + 3y


“.” represents multiplication



Question 5.

Divide first polynomial by second polynomial and write the answer in the form ‘Dividend = Divisor × Quotient + Remainder’.

i. x3 – 64; x – 4

ii. 5x5 + 4x4 – 3x3 + 2x2 + 2; x2 – x


Answer:

i. The division is as follows:



x3-64 = [(x2 + 4x + 16) × (x-4)] + 0


x3-64 = (x2 + 4x + 16) × (x-4)


ii. The division is as follows:



5x5 + 4x4 – 3x3 + 2x2 + 2 = [(5x3 + 9x2 + 6x + 8) × (x2-x)] + 8x +2


Quotient = 5x3 + 9x2 + 6x + 8


Remainder = 8x + 2



Question 6.

Write down the information in the form of algebraic expression and simplify.

There is a rectangular farm with length (2a2 + 3b2) meter and breadth (a2 + b2) meter. The farmer used a square shaped plot of the farm to build a house. The side of the plot was (a2 - b2) meter. What is the area of the remaining part of the farm?


Answer:

Given:

Length of the farm, l = (2a2 + 3b2) m


Breadth of the farm, b = (a2 + b2) m


Side of the square plot, s = (a2 - b2) m


To find: Area of remaining farm


Explain:


Area of Remaining Farm = Area of Rectangle – Area of Square


Area of Rectangle = l × b


= (2a2 + 3b2) × (a2 + b2)


= 2a2.a2 + 2a2.b2 + 3b2.a2 + 3b2.b2


= 2a4 + 5a2b2 + 3b4


Area of Square = s × s


= (a2 - b2) × (a2 - b2)


= a2.a2 – a2.b2 – a2.b2 + b2.b2


= a4 – 2a2b2 + b4


Area of Remaining Part = (2a4 + 5a2b2 + 3b4) – (a4 – 2a2b2 + b4)


= 2a4 + 5a2b2 + 3b4 - a4 + 2a2b2 - b4


= a4 + 7a2b2 + 2b4


Therefore, the area of remaining portion is = a4 + 7a2b2 + 2b4

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