Maharashtra Board Class 9 Math Chapter 3 Polynomials Solution

Maharashtra Board Class 9 Math Solution Chapter 3 – Polynomials

Balbharati Maharashtra Board Class 9 Math Solution Chapter 3: Polynomials. Marathi or English Medium Students of Class 9 get here Polynomials full Exercise Solution.

Std

Maharashtra Class 9
Subject

Math Solution

Chapter

Polynomials


Page – 36

Write any five algebraic expressions which are not polynomials. Explain why these expressions are not polynomials? Justify your answer.

Is every algebraic expression a polynomial?

Is every polynomial an algebraic expression?

A:  Any five algebraic expressions which are not polynomials are:

(i) y √2 + 3

(ii) y 3/2 + y + 2

(iii) 1/y2 + 1/y – 1

(iv) 4 √x + 3 √x + 2 √x + √x + 1

(v) x3 – 4/5x2 + √x – 3/2

Each of the above expression has powers which are not whole numbers so, they are not polynomials.

Every algebraic expression is not a polynomial but, every polynomial is an algebraic expression.

Page 38:

Activity I: Write an example of a monomial, a binomial and a trinomial having variable x and degree 5.

Monomial: 5x5

Binomial: 2x5 – 3x

Trinomial: 2x5 – 3x2 + 5

 

Activity II: Give example of a binomial in two variables having degree 5.

A: X2 y3 + x3 y2

Practice set 3.1

(1) State whether the given algebraic expressions are polynomials? Justify.

(i) y + 1 y

=A: No, because power of 1/y = y-1 is -1 which is not a whole number

(ii) 2 – 5 x

= A: No because power of √x = x1/2 is 1/2 which is not a whole number.

(iii) x2 + 7x + 9

= A: Yes because power x is 1 and 2 which is a whole number.

(iv) 2m-2 + 7m – 5

= A:  No because powers of m are -2 and -2 is not a whole number.

(v) 10

= A:  Yes because 10 = 10 × 1 = 10 × x0 and power of x is 0 which is whole number

(2). Write the coefficient of m3 in each of the given polynomial.

(i) m3

= 1

(ii) -3 /2 + m – √3 m3

= √3

(iii) -2 /3 m3 – 5m2 + 7m – 1

= -2/3

(3) Write the polynomial in x using the given information.

(i) Monomial with degree 7

A: 4x7

(ii) Binomial with degree 35

A: 9x35 – x

(iii) Trinomial with degree 8

A: x8 – 10x4 + 8

(4) Write the degree of the given polynomials.

(i) √5

= A: 0

(ii) x°

= A: 0

(iii) x2

=A: 2

(iv) √2 m10 – 7

= A: 10

(v) 2p – √7

= A: 1

(vi) 7y – y3 + y5

=A:  5

(vii) xyz + x y – z

= A: 3

(viii) m3 n7 – 3m5 n + m n

= A: 10

(5) Classify the following polynomials as linear, quadratic and cubic polynomial.

(i) 2x2 + 3x + 1

= A: Quadratic.

(ii) 5p

= A: Linear.

(iii) √2 y – 1/ 2

=A:  Linear.

(iv) m3 + 7m2 + 5 /2 m – √7

= A: Cubic.

(v) a2

= A: Quadratic.

(vi) 3r3

= A: Cubic.

(6)  Write the following polynomials in standard form.

(i) m3 + 3 + 5m

= A: m3 + 5m + 3

(ii) – 7y + y5 + 3y3 – 1 /2 + 2y4 – y2

A: y5 + 2y4 + 3y3 – y2 – 7y – 1/2

(7) Write the following polynomials in coefficient form.

(i) x3 – 2

= A: x3 – 2 = x3 + 0x2 + 0x – 2

Coefficient form = (1, 0, 0, -2)

(ii) 5y

= A: 5y = 5y + 0

Coefficient form = (5,0)

(iii) 2m4 – 3m2 + 7

= A: 2m4 – 3m2 + 7 = 2m4 + 0m3 -3m2 + 0m + 7

Coefficient form = (2, 0, -3, 0, 7)

(iv) – 2 /3

= A: -2/3 = -2/3 × x0

Therefore, Coefficient form = (-2/3)

(8) Write the polynomials in standard form.

(i) (1, 2, 3)

= A: x2 + 2x + 3

(ii) (5, 0, 0, 0, – 1)

= A:  5x4 + 0x3 + 0x2 + 0x – 1

= 5x4 – 1

(iii) (- 2, 2, – 2, 2)

= -2x3 + 2x2 – 2x + 2

(9)  Write the appropriate polynomials in the boxes.

A: Quadratic polynomial: x2, 2x2 + 5x + 10, 3x2 + 5x

Cubic polynomial: x3 + x2 + x + 5, x3 + 9,

Linear polynomial: x + 7

Binomial: x + 7, x3 + 9, 3x2 + 5x

Trinomial: 2x2 + 5x + 10,

Monomial: x2

Page 42: Let’s recall:

Degree of one polynomial is 3 and the degree of other polynomials is 5. Then what is the degree of their product?

= A: 3+5 = 8

What is the relation between degree of multiplicand and degree of a multiplier with degree of their product?

= A: (Degree of multiplicand) + (Degree of multiplier)

= (Degree of their product)

Practice set 3.2

(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?

= A: Number of trees increasing every year = b

Numbers of thus increased after x years = bx

Therefore, Total trees after x years = a + bx

(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?

= A: Number of rows = x

Number of students in a row = y

Therefore, Total number of students = xy

(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.

= A: Tens place digits = m and units place digit = n

Therefore, the number is 10m + n

(2) Add the given polynomials.

(i) x3 – 2x2 – 9 ; 5x3 + 2x + 9

= A: (x3 – 2x2 -9) + (5x3 + 2x + 9)

= (x3 + 5x3) – 2x2 + 2x – 9 + 9

= 6x3 – 2x2 + 2x

(ii) – 7m4 + 5m3 + √2 ; 5m4 – 3m3 + 2m2 + 3m – 6

= A: (-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

= A: (2y2 + 7y + 5) + (3y + 9) + (3y2 – 4y – 3)

= (2y2 + 3y2) + (7y + 3y – 4y) + (5 + 9-3)

= 5y2 + 6y + 11

(3) Subtract the second polynomial from the first.

(i) x2 – 9x + √3 ; – 19x + √3 + 7x2

= A: (x2 – 9x + √3) – (-19x + √3 + 7x2)

= (x2 – 7x2) + (-9x + 19x) + (√3 – √3)

= -6x2 + 10x

(ii) 2ab2 + 3a2 b – 4ab; 3ab – 8ab2 + 2a2 b

= (2ab2 + 3a2b – 4ab) – (3ab – 8ab2 + 2a2 b)

= (2ab2 + 8ab2) + (3a2b – 2a2b) + (-4ab – 3ab)

= 10 ab2 + a2b – 7ab

(4) Multiply the given polynomials.

 (i) 2x ; x2 – 2x -1

= A: 2x × (x2 – 2x – 1) = 2x × x2 – 2x × 2x – 1 × 2x

= 2x3 – 4x2 – 2x

(ii) x5 -1 ; x3 + 2x2 +2

= A: (x5-1) × (x3 + 2x2 + 2)

= x3(x5-1) + 2x2 (x5-1) + 2(x5-1)

= x8 + 2x7 + 2x5 – x3 – 2x2 – 2

(iii) 2y +1; y2 – 2y3 + 3y

= A: (2y + 1) × (y2 – 2y3 + 3y + 3y)

= 2y (y2 – 2y3 + 3y) + 1(y2 – 2y3 + 3y)

= 2y3 – 4y4 + 6y2 + y2 – 2y3 + 3y

= 3y + 7y2 – 4y4

(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?

A: Given, Length of rectangular form = (2a2 + 3b2) m

Breadth of rectangular form = (a2 + b2) m

Therefore, Area of rectangular form = length × breadth

= (2a2 + 3b2) × (a2 + b2) m2

= [2a2 (a2 + b2) + 3b2 (a2+b2)] m2

= [2a4 + 2a2b2 + 3a2b2 + 3b4] m2

= [2a4 + 5a2b2+3b4] m2

Sides of square plot = (a2 – b2) m

Therefore, Area of square plot = (a2 – b2)2 m2

= (a4 + b4 – 2a2 b2) m2

Therefore, Area of remaining part = area of rectangular form – area of square plot

= {2a4 + 5a2b2 + 3b4] m2 – [a4+b4-2a2b2] m2

= {(2a4 – a4) + (3b4 – b4) + (5a2b2 + 2a2b2)} m2

= [a4 + 2b4 + 7a2b2] m2

Page 44

Activity:

A: (i) 10,000 + 2000x + 4000 x2 + 8000 y + 9000 y2 rupees

(2) 5x2 × 2800 + 5/3 y2 × 5000 + 4y × 4000 rupees

= 14000 x2 + 25000/3 y2 + 16000 y rupees

Practice Set 3.3

(1) For x = 0 find the value of the polynomial x2 – 5x + 5.

= A:  Let, p(x) = x2 – 5x + 5

Therefore, p(0)  = 02 – 5 × 0 + 5

= 5

(2) If p(y) = y2 – 3 √2 y + 1 then find p (3 √2).

= A: p(y) = y2 – 3√2 y + 1

= P (3√2) = (3√2)2 – 3√2 × 3√2 + 1

= 18-18 + 1

= 1

(3) If p (m) = m3 + 2m2 – m + 10 then p(a) + p(- a) = ?

= A: p (m) = m3 + 2m2 – m + 10

Therefore, p (a) = a3 + 2a2 – a +10

And, p (-a) = -a3 + 2(-a)2 – (-a) + 10

= -a3 + 2a2 + a + 10

Therefore, p (a) + p (-a) = a3 + 2a2 – a + 10 – a3 + 2a2 + a + 10

= 4a2 + 20

(4) If p(y) = 2y3 – 6y2 – 5y + 7 then find p (2).

= p (y) = 23y3 – 6y2 – 5y + 7

= p (2) = 2 × (2)3 – 6(2)2 – 5(2) + 7

= 16-24-10+7

= 23-34

= -11

Practice set 3.5

(1) Find the value of the polynomial 2x – 2x3 + 7 using given values for x.

(i) x = 3

= A: p (x) = 2x – 2x3 + 7

P (3) = 2 × 3 – 2(3)3 + 7

(ii) x = – 1

= A: p (x) = 2x – 2x3 + 7

= p (-1) = 2(-1) -2(-1)3 + 7

= -2 + 2 +7

= 7

(iii) x = 0

= A: p (x) = 2x – 2x3 + 7

P (0) = 2 (0) -2 (0)3 + 7

= 7

(2) For each of the following polynomial, find p(1), p(0) and p(- 2).

(i) p(x) = x3

= a: p(1) = 13 = 1

P(0) = 03 = 0

P(-2) = (-2)3 = -8

(ii) p(y) = y2 – 2y + 5

= A: P(1) = 12 – 2(1) + 5

= 4

P(0) = 02– 2(0) + 5

= 5

P(-2) = (-2)2 – 2(-2) + 5

= 4+4+5 = 13

P(0) = 02 – 2(0) + 5

= 5

(iii) p(x) = x4 – 2x2 – x

P(1) = 14 – 2(1)2 – 1

= 1-2-1

= -2

P(0) = 04 – 2(0)2 – 0 = 0

P(-2) = (-2)4 – 2(-2)2 – (-2)

=16 – 8 + 2

= 10

(3) If the value of the polynomial m3 + 2m + a is 12 for m = 2, then find the value of a.

= A: Let, p(x) = m3 + 2m + a

Given, p(2) = 23 + 2(2) + a

= 8 + 4 + a

= 12+a

Therefore, 12+a = 12

=> a = 12-12

= 0

(4) For the polynomial mx2 – 2x + 3 if p(- 1) = 7 then find m.

A: Given, P(-1) = 7

Also, P(-1) = m (-1)2 – 2(-1) + 3

=> 7 = m+2+3

= m+5

=> m = 7-5

= 2

(6) If the polynomial y3 – 5y2 + 7y + m is divided by y + 2 and the remainder is 50 then find the value of m

= A: By remainder theorem

P(-2) = 50

Where p(x) = y3 – 5y2 + 7y + m

Therefore, p(-2) = (-2)3 – 5(-2)2 + 7(-2) + m

= -8 -20 -14 + m

= m – 42

Therefore, m – 42 = 50

=> m = 50 + 42

= 92

(7) Use factor theorem to determine whether x + 3 is factor of x2 + 2x – 3 or not.

= Let, p(x) = x2 + 2x – 3

= Therefore, P(-3) = (-3)2 + 2 (-3) – 3

= 9-6-3

= 0

As per remainder theorem,

Remainder = 0

Therefore, x + 3 is a factor of x2 + 2x – 3

(8)

A: Because, (x-2) is a factor of x3 – mx2 + 10x – 20 =p(x)

Therefore, p(2) = 0

=> 23 – m × 22 + 10 x 2 – 20 = 0

=> 8 – 4m+ 20-20 = 0

=> 4m = 8

=> m = 2

Updated: July 24, 2021 — 2:07 am

Leave a Reply

Your email address will not be published. Required fields are marked *

19 − sixteen =