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[PDF] NCERT Solution For Class 9 Maths Chapter 2- Polynomials

NCERT Solution For Class 9 Maths Chapter 2- Polynomials 2 Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following




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We shall also study the Remainder Theorem and Factor Theorem and their use in the factorisation of polynomials In addition to the above, we shall study some 

[PDF] NCERT Solutions for cbse class 9 Maths Chapter 2- Polynomials

NCERT Solution For Class 9 Maths Chapter 2- Polynomials Exercise 2 4 Page: 44 2 Use the Factor Theorem to determine whether g(x) is a factor of p(x) in 

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NCERT Solutions for Class 9 Maths Chapter 2- Polynomials Polynomials Therefore, according to factor theorem, x+1 is a factor of the given polynomial

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Factor Theorem: Let ( ) P x be a polynomial or dividend and a linear polynomial or divisor or factor of ( ) P x if and only if k is a zero of polynomial ( )




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Solution 1 When setting up the synthetic division tableau, we need to enter 0 for the coefficient of x in the dividend Doing so 

[PDF] CBSE NCERT Solutions for Class 9 Mathematics Chapter 2

3 ? x2 ? 2 + ?2 x + ?2 2 Use the factor theorem to determine whether g(x) is a factor of p(x) in each of 

[PDF] NCERT Solutions for Class 9 Maths Chapter 2 Exercise 24

NCERT Solutions Class 9 Maths Chapter 2 Exercise 2 4 Page 3 Q2 Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each

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9th Class Maths Polynomials-Remainder Theorem We are now going to discuss an important topic for class 9 Remainder NCERT Maths solutions class 9




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2 sept 2021 · Explain Factor theorem with model problems Solution: P(x)= + 14 + 48 CLASS:9 POLYNOMIALS ACTIVITY SHEET-07 FACTOR THEOREM

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RS Aggarwal Solutions for Class 9 Maths Chapter 2 - Polynomials Exercise 2(D) PAGE: 90 Using factor theorem, show that g(x) is a factor of p(x), when

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Taking 5 as a common factors ( )2 Exercise 5 1 Q1 Factorize 3 2 3 2 2 9 15 3 st st st + - Solution: Given 3 Example 26: using factor theorem, prove that

[PDF] CBSE NCERT Solutions for Class 9 Mathematics - cloudfrontnet

CBSE NCERT Solutions for Class 9 Mathematics Chapter 2 Back of Chapter Use the factor theorem to determine whether g(x) is a factor of p(x) in each of the

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quadratics whose solutions are integers or rational numbers remainder and factor theorems to factorise and to solve polynomials that are of degree 9 × × x −1 3x2 − x + 2 3x + 2 −(3x2 − 3x) 2x + 2 −(2x − 2) 4 Quotient Dividend

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Math Class X 1 Solution Let f(x) = 3x 3 + 5x 2 – 11x – 4 Since, 3x + 1 = 3 1 1 + = 3 – – 3 3 ⌈ ⌉ ⌈ By remainder theorem, the required remainder =

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CBSE NCERT Solutions for Class

9 Mathematics Chapter 2 Back of Chapter Questions

E xercise: 2.1

1.Which of the following expressions are polynomials in one variable and which are

not? State reasons for your answer. (i)4ݔ ଶ െ3x+ 7 (ii)y ଶ +ξ2(iii)3ξt+ tξ2 (iv)y + ଶ ୷ (v)ݔ ଵ଴ + y ଷ + t ହ଴ So lution: (i)Given expression is a polynomial

It is of the form a

୬ ݔ ௡ + a୬ିଵ ݔ ୬ିଵ ڮ+ ଵ

ݔ+ a

଴ where a ୬ ,a ୬ିଵ ,...a ଴ a re constants. Hence given expression 4x ଶ െ3x+ 7 is a polynomial. (ii)Given expression is a polynomial

It is of the form a୬

x ୬ + a ୬ିଵ x ୬ିଵ ڮ+ ଵ x +a ଴ where a ୬ ,a ୬ିଵ ,...a ଴ ar e constants. Hence given expression y ଶ +ξ2 is a polynomial. (iii)Given expression is not a polynomial. It is not in the form of a୬ x ୬ + a ୬ିଵ 2 ୬ିଵ ڮ+ ଵ x +a ଴ w here a ୬ ,a ୬ିଵ ,...a ଴ all constants.

Hence given expression

3ξt+ tξ2 is not a polynomial.(iv)Given expression is not a polynomial

y + 2 y = y+ 2. y ିଵ

It is not of form

a ୬ x ୬ + a ୬ିଵ x ୬ିଵ ڮ+ ଴ , where a ୬ ,a ୬ିଵ ,...a ଴ are constants. Hence given expression y + ଶ ୷ is not a polynomial. (v)Given expression is a polynomial in three variables. It has three variables x,y,t.

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Hence the given expression x

ଵ଴ + y ଷ + t ହ଴ is not a polynomial in one variable . 2. Write the coefficients of x ଶ in each of the following: (i) 2 +x ଶ + x (ii) 2െx ଶ + x ଷ (iii) ஠ 6 x ଶ + x (iv) ξ2xെ1

Solution:

(i) The constant multiplied with the term x ଶ is called the coefficient of the x ଶ .

Given polynomial is 2 +x

ଶ + x.

Hence, the coefficient of x

ଶ in given polynomial is equal to 1. (ii) The constant multiplied with the term x ଶ is called the coefficient of the x ଶ .

Given polynomial is 2െx

ଶ + x ଷ .

Hence, the coefficient of x

ଶ in given polynomial is equal to െ1. (iii) The constant multiplied with the term x ଶ is called the coefficient of the x ଶ .

Given polynomial is

஠ ଶ x ଶ + x.

Hence, the

coefficient of x ଶ in given polynomial is equal to ஠ ଶ . (iv) The constant multiplied with the term x ଶ is called the coefficient of the x ଶ .

Given polynomial is ξ2xെ1.

In the given polynomial, there is no x

ଶ term.

Hence, the coefficient of x

ଶ in given polynomial is equal to 0. 3. Give one example each of a binomial of degree 35 and of a monomial of degree 100
୭ .

Solution:

Degree of polynomial is highest power of variable in the polynomial. And n umber of terms in monomial and binomial respectively equals to one and two.

A binomial of

degree 35 can be x ଷହ + 7

A monomial of degree 100 can be 2x

ଵ଴଴ + 9 4. Write the degree of each of the following polynomials

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(i) 5x ଷ + 4x ଶ +7x (ii) 4െy ଶ (iii) 5ݐെξ 7 (iv) 3

Solution:

(i) Degree of polynomial is highest power of variable in the polynomial.

Given polynomial is 5x

ଷ + 4x ଶ +7x Hence, the degree of given polynomial is equal to 3. (ii) Degree of polynomial is highest power of variable in the polynomial.

Given polynomial is 4െy

Hence, the degree of given polynomial is 2.

(iii) Degree of polynomial is highest power of variable in the polynomial

Given polynomial is 5tെξ7

Hence, the degree of given polynomial is 1.

(iv) Degree of polynomial 1, highest power of variable in the polynomial.

Given polynomial is 3.

Hence, the degree of given polynomial is 0.

5. Classify the following as linear, quadratic and cubic polynomials. (i) x ଶ + x (ii) xെx ଷ (iii) y +y ଶ + 4 (iv) 1 +x (v) 3t (vi) r ଶ (vii) 7x ଷ

Solution:

(i) Linear, quadratic, cubic polynomials have degrees

1,2,3 respectively.

Given polynomial is x

ଶ + x

It is a

quadratic polynomial as its degree is 2.

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(ii) Linear, quadratic, cubic polynomials have its degree 1,2,3 respectively.

Given polynomial is xെx

ଷ .

It is a cubic polynomial as its degree is 3.

(iii) Linear, quadratic, cubic polynomial has its degree 1,2,3 respectively.

Given polynomial is y +y

ଶ + 4. It is a quadratic polynomial as its degree is 2. (v) Linear, quadratic, cubic polynomial has its degree 1,2,3 respectively.

Given polynomial is 1 +x .

It is a linear polynomial as its degree is 1. (v) Linear, quadratic, cubic polynomial has its degree 1,2,3 respectively.

Given polynomial is 3t

It is

a linear polynomial as its degree is 1. (vi) Linear, quadratic, cubic polynomial has its degree 1,2,3 respectively.

Given polynomial is r

ଶ . It is a quadratic polynomial as its degree is 2. (vii) Linear, quadratic, cubic polynomial has its degree 1,2,3 respectively.

Given polynomial is 7x

ଷ . It is a cubic polynomial as its degree is 3.

Exercise

: 2.2 1. Find the value of the polynomial 5xെ4x ଶ + 3 at (i) x =0 (ii) x =െ1 (iii) x =2

Solution:

(i) Given polynomial is 5xെ4x ଶ + 3

Value of polynomial at x =0 is 5(0)െ4(0)

ଶ + 3 = 0െ0 +3 = 3

Therefore, value of polynomial 5xെ4x

ଶ + 3 at x =0 is equal to 3.

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(ii) Given polynomial is 5xെ4x ଶ + 3

Value of given polynomial at

x =െ1 is 5(െ1)െ4(െ1) ଶ + 3 =െ5െ4 +3 =െ6

Therefore, value of polynomial 5xെ4x

ଶ + 3 at x =െ1 is equal to െ6. (iii) Given polynomial is 5xെ4x ଶ + 3

Value of given polynomial at

x =2 is 5(2)െ4(2) ଶ + 3 =10െ16+ 3 =െ3

Therefore, value of polynomial 5xെ4x

ଶ + 3 at x =2 is equal to െ3 2. Find P (0),P(1) and P(2) for each of the following polynomials. (i) P(y)= y ଶ െy +1 (ii) P(t)= 2+ t+ 2t ଶ െt ଷ (iii) P(x)= x ଷ (iv) P(x)=(xെ1)(x +1 )

Solution:

(i) Given polynomial is P(y)= y ଶ െy +1 P (0)=(0) ଶ െ0 +1 = 1 P (1)=(1) ଶ െ1 +1 = 1 P (2)=(2) ଶ െ2 +1 = 4െ2 +1 = 3 (ii) Given polynomial is P(t)= 2+ t+ 2t ଶ െt ଷ P (0)= 2+ 0+ 2. (0) ଶ െ(0) ଷ = 2 P (1)= 2+ 1+ 2(1) ଶ െ(1) ଷ = 4

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P(2)= 2+ 2+ 2. (2)

ଶ െ(2) ଷ = 4 (iii) Given polynomial is P(x)= x ଷ P (0)=(0) ଷ = 0 P (1)=(1) ଷ = 1 P (2)=(2) ଷ = 8 (iv) Given polynomial is p(x)=(xെ1)(x +1 ) P (0)=(0െ1)(0 +1 ) = (െ1)(1) =െ1 P (1)=(1െ1)(1 +1 ) = (0)(2) = 0 P (2)=(2െ1)(2 +1 ) = 3 3. Verify whether the following are zeroes of the polynomial, indicated against them. (i) P(x) =3x+ 1,x=െ ଵ ଷ (ii) P(x)=5xെɎ,x= ସ ହ (iii) P(x)= x ଶ െ1,x=1 ,െ1 (iv) P(x)=(x +1 )(xെ2),x= െ1,2 (v) P(x)= x ଶ ,x= 0 (vi) P(x) =lx+ m,x=െ ୫ ୪ (vii) P(x)= 3x ଶ െ1,x= ିଵ ξ 7 , ଶ 

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