NCERT Exemplar Class 9 Maths Solutions for Chapter 2 - Polynomials

NCERT Exemplar Solutions Class 9 Maths Chapter 2 – Free PDF Download

NCERT Exemplar Solutions Class 9 Maths Chapter 2 Polynomials are provided here for students to prepare well for exams. Students can use these materials as a reference tool for studying as it contains solved questions relevant to the exercise problems present in the NCERT Exemplar textbook. The exemplar problems are designed by subject experts in accordance with the CBSE Syllabus for Class 9, which covers the following topics of Chapter 2, Polynomials:

• Finding the degree of polynomials
• Polynomials in one variable
• Coefficients and zeroes of polynomials
• Remainder theorem for polynomials
• Division of one polynomial from another
• The factorisation of polynomials using the Factor theorem
• Algebraic identities to factorise polynomials

Before solving the questions, let us learn what a polynomial is. A polynomial is an expression of algebraic terms (more than two), especially the sum of several terms that contain different powers of the same variable. To facilitate easy learning and help students understand the concepts of polynomials, free NCERT Exemplars are provided here, which can be downloaded in the form of a PDF. They can practise questions related to polynomials by solving the NCERT Exemplar for Class 9 Maths Chapter 2 Polynomials. To download the PDF of these solutions, click on the below link.

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Exercise 2.1 Page No: 14

Write the correct answer in each of the following:

1. Which one of the following is a polynomial?

Solution:

(C)

Explanation:

(A)

The equation contains the terms x² and -2x^-2.

Here, the exponent of x in the second term = – 2, which is not a whole number.

Hence, the given algebraic expression is not a polynomial.

(B)

The equation contains the term √2x^½ .

Here, the exponent of x in the first term = ½ , which is not a whole number.

Hence, the given algebraic expression is not a polynomial.

(C)

The equation contains the term x² and 3x .

Here, the exponent of x in first term and second term= 2 and 1, respectively, which is a whole number.

Hence, the given algebraic expression is a polynomial.

(D)

The equation is a rational function.

Here, the given equation is not in the standard form of a polynomial.

Hence, the given algebraic expression is not a polynomial.

Hence, option C is the correct answer

2. √2 is a polynomial of degree

(A) 2

(B) 0

(C) 1

(D) ½

Solution:

(B) 0

Explanation:

√2 can be written as √2x⁰

i.e., √2 = √2x⁰

Therefore, the degree of the polynomial = 0

Hence, option B is the correct answer

3. Degree of the polynomial 4×4 + 0x3 + 0x5 + 5x + 7 is

(A) 4

(B) 5

(C) 3

(D) 7

Solution:

(A) 4

Explanation:

Degree of a polynomial = Highest power of the variable in a polynomial.

The highest power of variable x in the polynomial 4×4 + 0x3 + 0x5 + 5x + 7 is 4.

Therefore, degree of the polynomial of 4×4 + 0x3 + 0x5 + 5x + 7 = 4

Hence, option A is the correct answer

4. Degree of the zero polynomial is

(A) 0

(B) 1

(C) Any natural number

(D) Not defined

Solution:

(D) Not defined

Explanation:

The degree of a zero polynomial is not defined.

Hence, option D is the correct answer

5. If p(x)= x² – 2√2x + 1, then p(2√2) is equal to

(A) 0

(B) 1

(C) 4√2

(D) 8√2 +1

Solution:

(B) 1

Explanation:

According to the question,

p(x) = x² – 2√2x + 1

To get p(2√2),

We substitute x = 2√2,

p(2√2) = (2√2)² – (2√2 × (2√2)) + 1

= (4 × 2) – (4 × 2) + 1

= 8 – 8 + 1

= 1

Hence, option B is the correct answer

6. The value of the polynomial 5x – 4x² + 3, when x = – 1 is

(A) – 6

(B) 6

(C) 2

(D) – 2

Solution:

(A) – 6

Explanation:

According to the question,

p(x) = 5x – 4x² + 3

To get p(– 1),

We substitute x = – 1,

p(– 1) = 5(– 1) – 4(– 1)² + 3

= 5(– 1) – 4(1) + 3

= – 5 – 4 + 3

= – 9 + 3

= – 6

Hence, option A is the correct answer

7. If p(x) = x + 3, then p(x) + p(–x) is equal to

(A) 3

(B) 2x

(C) 0

(D) 6

Solution:

(D) 6

Explanation:

p(x) = x + 3

p(– x) = – x + 3

Therefore,

p(x) + p(–x) = (x + 3) + (– x + 3)

= x + 3 – x + 3

= 6

Hence, option D is the correct answer

8. Zero of the zero polynomial is

(A) 0

(B) 1

(C) Any real number

(D) Not defined

Solution:

(C) Any real number

Explanation:

A zero polynomial is a constant polynomial whose coefficients are all equal to 0.

Zero of a polynomial is the value of the variable that makes the polynomial equal to zero.

Therefore, zero of the zero polynomial is any real number.

Hence, option C is the correct answer

9. Zero of the polynomial p(x) = 2x + 5 is

(A) – 2/5

(B) – 5/2

(C) 2/5

(D) 5/2

Solution:

(B) – 5/2

Explanation:

Zero of the polynomial ⇒ p(x) = 0

p(x) = 0

2x + 5 = 0

2x = – 5

x = – 5/2

Hence, option B is the correct answer

10. One of the zeroes of the polynomial 2x² + 7x –4 is

(A) 2

(B) ½

(C) – ½

(D) –2

Solution:

(B) ½

Explanation:

Zero of the polynomial ⇒ p(x) = 0

p(x) = 0

2x² + 7x – 4 = 0

2x² – 1x + 8x – 4 = 0

x(2x – 1) + 4(2x – 1) = 0

(x + 4)(2x – 1) = 0

Consider, x + 4

x + 4 = 0

x = – 4

Consider, 2x – 1

2x – 1 = 0

2x = 1

x = ½

Hence, option B is the correct answer

Exercise 2.2 Page No: 16

1. Which of the following expressions are polynomials? Justify your answer:

Solution:

(i) 8

8 can be written as 8x⁰.

i.e., 8 = 8x⁰,

Here, the power of x = 0, which is a whole number.

Hence, 8 is a polynomial.

(ii) √3x² – 2x

√3x² – 2x

Here, the power of x are 2 and 1, respectively

2 and 1 both are whole numbers.

Hence, √3x² – 2x is a polynomial.

(iii) 1 – √(5x)

1 – √5√x = 1 – √5 x^½

Here, the power of x = ½, which is not a whole number.

Hence, 1 – √5x is not a polynomial

(iv)

1/5x^–2 + 5x + 7 = 5x² + 5x + 7

Here, the power of x are 2 and 1 respectively

2 and 1 both are whole numbers.

Hence, 1/5x^-2 + 5x + 7 is a polynomial.

(v)

((x – 2)(x – 4))/x = (x² – 4x – 2x + 8)/x

= (x² – 6x + 8)/x

= x – 6 + (8/x)

= x – 6 + 8x^–1

Here, the power of x = – 1, which is not a whole number, but a negative number.

Hence, ((x – 2)(x – 4))/x is not a polynomial

(vi)

1/(x+1) = (x+1)^{– 1}

Here, the power of x is not a whole number.

Hence, 1/(x+1) is not a polynomial

(vii)

(1/7)a³ – (2/√3)a² + 4a – 7

Here, the power of a are 3, 2 and 1, respectively

3, 2 and 1 are all whole numbers.

Hence, (1/7)a³ – (2/√3)a² + 4a – 7 is a polynomial.

(viii)

1/2x = (x^–1/2)

Here, the power of x = – 1, which is not a whole number, but a negative number.

Hence, 1/2x is not a polynomial

Exercise 2.3 Page No: 18

1. Classify the following polynomials as polynomials in one variable, two variables etc.

(i) x² + x + 1
(ii) y³ – 5y
(iii) xy + yz + zx
(iv) x² – 2xy + y² + 1

Solution:

(i) x² + x + 1

Here, the polynomial contains only one variable, i.e., x.

Hence, the given polynomial is a polynomial in one variable.

(ii) y³ – 5y

Here, the polynomial contains only one variable, i.e., y.

Hence, the given polynomial is a polynomial in one variable.

(iii) xy + yz + zx

Here, the polynomial contains three variables, i.e., x, y and z.

Hence, the given polynomial is a polynomial in three variables.

(iv) x² – 2xy + y² + 1

Here, the polynomial contains two variables, i.e., x and y.

Hence, the given polynomial is a polynomial in two variables.

2. Determine the degree of each of the following polynomials:

(i) 2x – 1
(ii) –10
(iii) x³ – 9x + 3x⁵
(iv) y³ (1 – y⁴)

Solution:

Degree of a polynomial in one variable = highest power of the variable in an algebraic expression

(i) 2x – 1

Power of x = 1

The highest power of the variable x in the given expression = 1

Hence, the degree of the polynomial 2x – 1 = 1

(ii) –10

There is no variable in the given term.

Let us assume that the variable in the given expression is x.

– 10 = –10x⁰

Power of x = 0

The highest power of the variable x in the given expression = 0

Hence, the degree of the polynomial – 10 = 0

(iii) x³ – 9x + 3x⁵

Powers of x = 3, 1 and 5, respectively.

The highest power of the variable x in the given expression = 5

Hence, the degree of the polynomial x³ – 9x + 3x⁵= 5

(iv) y³ (1 – y⁴)

The equation can be written as,

y³ (1 – y⁴) = y³ – y⁷

Powers of y = 3 and 7, respectively.

The highest power of the variable y in the given expression = 7

Hence, degree of the polynomial y³ (1 – y⁴) = 7

3. For the polynomial 

, write

(i) the degree of the polynomial
(ii) the coefficient of x³
(iii) the coefficient of x⁶
(iv) the constant term

Solution:

The given polynomial is

(i)Powers of x = 3, 1, 2 and 6, respectively.

The highest power of the variable x in the given expression = 6

Hence, the degree of the polynomial = 6

(ii) The given equation can be written as,

Hence, the coefficient of x³ in the given polynomial is 1/5.

(iii) The coefficient of x⁶ in the given polynomial is – 1

(iv) Since the given equation can be written as,

The constant term in the given polynomial is 1/5 as it has no variable x associated with it.

4. Write the coefficient of x² in each of the following:

(i) (π/6)x + x² – 1
(ii) 3x – 5
(iii) (x –1) (3x – 4)
(iv) (2x – 5) (2x² – 3x + 1)

Solution:

(i) (π/6) x + x²−1

(π/6) x + x²−1 = (π/6) x + (1) x²−1

The coefficient of x² in the polynomial (π/6) x + x²−1 = 1.

(ii) 3x – 5

3x – 5 = 0x² + 3x – 5

The coefficient of x² in the polynomial 3x – 5 = 0, zero.

(iii) (x – 1) (3x – 4)

(x – 1)(3x – 4) = 3x² – 4x – 3x + 4

= 3x² – 7x + 4

The coefficient of x² in the polynomial 3x² – 7x + 4 = 3.

(iv) (2x – 5) (2x² – 3x + 1)

(2x – 5) (2x² – 3x + 1)

= 4x³ – 6x² + 2x – 10x² + 15x– 5

= 4x³ – 16x² + 17x – 5

The coefficient of x² in the polynomial (2x – 5) (2x² – 3x + 1) = – 16

5. Classify the following as a constant, linear, quadratic and cubic polynomial:

(i) 2 – x² + x³ 

(ii) 3x³
(iii) 5t – √7

(iv) 4 – 5y²
(v) 3

(vi) 2 + x
(vii) y³ – y

(viii) 1 + x + x²
(ix) t² 

(x) √2x – 1

Solution:

Constant polynomials: The polynomial of the degree zero.

Linear polynomials: The polynomial of degree one.

Quadratic polynomials: The polynomial of degree two.

Cubic polynomials: The polynomial of degree three.

(i) 2 – x² + x³

Powers of x = 2 and 3, respectively.

The highest power of the variable x in the given expression = 3

Hence, the degree of the polynomial = 3

Since it is a polynomial of degree 3, it is a cubic polynomial.

(ii) 3x³

Power of x = 3.

The highest power of the variable x in the given expression = 3

Hence, the degree of the polynomial = 3

Since it is a polynomial of degree 3, it is a cubic polynomial.

(iii) 5t – √7

Power of t = 1.

The highest power of the variable t in the given expression = 1

Hence, the degree of the polynomial = 1

Since it is a polynomial of degree 1, it is a linear polynomial.

(iv) 4 – 5y²

Power of y = 2.

The highest power of the variable y in the given expression = 2

Hence, the degree of the polynomial = 2

Since it is a polynomial of degree 2, it is a quadratic polynomial.

(v) 3

There is no variable in the given expression.

Let us assume that x is the variable in the given expression.

3 can be written as 3x⁰.

i.e., 3 = x⁰

Power of x = 0.

The highest power of the variable x in the given expression = 0

Hence, the degree of the polynomial = 0

Since it is a polynomial of the degree 0, it is a constant polynomial.

(vi) 2 + x

Power of x = 1.

The highest power of the variable x in the given expression = 1

Hence, the degree of the polynomial = 1

Since it is a polynomial of degree 1, it is a linear polynomial.

(vii) y³ – y

Powers of y = 3 and 1, respectively.

The highest power of the variable x in the given expression = 3

Hence, the degree of the polynomial = 3

Since it is a polynomial of degree 3, it is a cubic polynomial.

(viii) 1 + x + x²

Powers of x = 1 and 2, respectively.

The highest power of the variable x in the given expression = 2

Hence, the degree of the polynomial = 2

Since it is a polynomial of degree 2, it is a quadratic polynomial.

(ix) t²

Power of t = 2.

The highest power of the variable t in the given expression = 2

Hence, the degree of the polynomial = 2

Since it is a polynomial of degree 2, it is a quadratic polynomial.

(x) √2x – 1

Power of x = 1.

The highest power of the variable x in the given expression = 1

Hence, the degree of the polynomial = 1

Since it is a polynomial of degree 1, it is a linear polynomial.

6. Give an example of a polynomial, which is:

(i) monomial of degree 1
(ii) binomial of degree 20
(iii) trinomial of degree 2

Solution:

(i) Monomial = an algebraic expression that contains one term

An example of a polynomial, which is a monomial of degree 1 = 2t

(ii) Binomial = an algebraic expression that contains two terms

An example of a polynomial, which is a binomial of degree 20 = x²⁰ + 5

(iii) Trinomial = an algebraic expression that contains three terms

An example of a polynomial, which is a trinomial of degree 2 = y² + 3y + 11

7. Find the value of the polynomial 3𝑥³ – 4𝑥² + 7𝑥 – 5, when x = 3 and also when x = –3.

Solution:

Given that,

p(x) = 3𝑥³ – 4𝑥² + 7𝑥 – 5

According to the question,

When x = 3,

p(x) = p(3)

p(x) = 3𝑥³ – 4𝑥² + 7𝑥 – 5

Substituting x = 3,

p(3)= 3(3)³ – 4(3)² + 7(3) – 5

p(3) = 3(3)³ – 4(3)² + 7(3) – 5

= 3(27) – 4(9) + 21 – 5

= 81 – 36 + 21 – 5

= 102 – 41

= 61

When x = – 3,

p(x) = p(– 3)

p(x) = 3𝑥³ – 4𝑥² + 7𝑥 – 5

Substituting x = – 3,

p(– 3)= 3(– 3)³ – 4(– 3)² + 7(– 3) – 5

p(– 3) = 3(–3)³ – 4(–3)² + 7(–3) – 5

= 3(–27) – 4(9) – 21 – 5

= –81 – 36 – 21 – 5

= –143

8. If p(𝑥) =𝑥² – 4𝑥 + 3, evaluate: 𝑝(2)− 𝑝(−1) + 𝑝(½).

Given that,

p(𝑥) =𝑥² – 4𝑥 + 3

According to the question,

When x = 2,

p(x) = p(2)

p(𝑥) =𝑥² – 4𝑥 + 3

Substituting x = 2,

p(2) = (2)² – 4(2) + 3

= 4 – 8 + 3

= – 4 + 3

= – 1

When x = – 1,

p(x) = p(– 1)

p(𝑥) =𝑥² – 4𝑥 + 3

Substituting x = – 1,

p(– 1) = (– 1)² – 4(– 1) + 3

= 1 + 4 + 3

= 8

When x = ½ ,

p(x) = p(½)

p(𝑥) =𝑥² – 4𝑥 + 3

Substituting x = ½,

p(½) = (½)² – 4(½) + 3

= ¼ – 2 + 3

= ¼ + 1

= 5/4

Now,

p(2)− p(−1) + p(½) = – 1 – 8 + (5/4)

= – 9 + (5/4)

= ( – 36 + 5)/4

= – 31/4

9. Find p(0), p(1),𝑝(−2) for the following polynomials:

(i) (𝑥)=10𝑥−4𝑥² –3

(ii) (𝑦)=(y + 2) (y – 2)

Solution:

(i) According to the question,

p(x) = 10𝑥−4𝑥² –3

When x = 0,

p(x) = p(0)

Substituting x = 0,

p(0) = 10(0)−4(0)² –3

= 0 – 0 – 3

= – 3

When x = 1,

p(x) = p(1)

Substituting x = 1,

p(1) = 10(1)−4(1)² –3

= 10 – 4 – 3

= 6 – 3

= 3

When x = – 2,

p(x) = p(– 2)

Substituting x = – 2,

p(– 2) = 10(– 2)−4(– 2)² –3

= – 20 – 16 – 3

= – 36 – 3

= – 39

(ii) According to the question,

p(𝑦)=(y + 2) (y – 2)

When y = 0,

p(y) = p(0)

Substituting y = 0,

p(0) =(0 + 2) (0 – 2)

= (2)(– 2)

= – 4

When y = 1,

p(y) = p(1)

Substituting y = 1,

p(1) =(1 + 2) (1 – 2)

=(3) (– 1)

= – 3

When y = – 2,

p(y) = p(– 2)

Substituting y = – 2,

p(– 2) =(– 2 + 2) (– 2 – 2)

= (0) (– 4)

= 0

10. Verify whether the following are true or false:

(i) –3 is a zero of x – 3
(ii) – 1/3 is a zero of 3x + 1
(iii) – 4/5 is a zero of 4 –5y
(iv) 0 and 2 are the zeroes of t² – 2t
(v) –3 is a zero of y² + y – 6

Solution:

(i) –3 is a zero of x – 3

False

Zero of x – 3 is given by,

x – 3 = 0

⇒ x=3

(ii) – 1/3 is a zero of 3x + 1

True

Zero of 3x + 1 is given by,

3x + 1 = 0

⇒ 3x = – 1

⇒ x = – 1/3

(iii) – 4/5 is a zero of 4 –5y

False

Zero of 4 – 5y is given by,

4 – 5y =0

⇒ – 5y = – 4

⇒ y = 4/5

(iv) 0 and 2 are the zeroes of t² – 2t

True

Zeros of t² – 2t is given by,

t² – 2t = t(t – 2) = 0

⇒ t = 0 or 2

(v) –3 is a zero of y² + y – 6

True

Zero of y² + y – 6 is given by,

y² + y – 6 = 0

⇒ y² + 3x – 2x – 6 = 0

⇒ y (y + 3) – 2(x + 3) = 0

⇒ (y – 2) (y + 3) =0

⇒ y = 2 or – 3

11. Find the zeroes of the polynomial in each of the following:

(i) p(x) = x – 4
(ii) g(x) = 3 – 6x
(iii) q(x) = 2x –7
(iv) h(y) = 2y

Solution:

(i) p(x) = x – 4

Zero of the polynomial p(x) ⇒ p(x) = 0

P(x) = 0

⇒ x – 4= 0

⇒ x = 4

Therefore, the zero of the polynomial is 4.

(ii) g(x) = 3 – 6x

Zero of the polynomial g(x) ⇒ g(x) = 0

g(x) = 0

⇒3 – 6x = 0

⇒ x = 3/6 = ½

Therefore, the zero of the polynomial is ½

(iii) q(x) = 2x –7

Zero of the polynomial q(x) ⇒ q(x) = 0

q(x) = 0

⇒2x – 7 = 0

⇒ x = 7/2

Therefore, the zero of the polynomial is 7/2

(iv) h(y) = 2y

Zero of the polynomial h(y) ⇒ h(y) = 0

h(y) = 0

⇒2y =0

⇒ y = 0

Therefore, the zero of the polynomial is 0

12. Find the zeroes of the polynomial:

p(𝑥)= (𝑥 –2)²−(𝑥 + 2)²

Solution:

p(x) = (𝑥 –2)²−(𝑥 + 2)²

We know that,

Zero of the polynomial p(x) = 0

Hence, we get,

⇒ (x–2)²−(x + 2)² = 0

Expanding using the identity, a² – b² = (a – b) (a + b)

⇒ (x – 2 + x + 2) (x – 2 –x – 2) = 0

⇒ 2x ( – 4) = 0

⇒ – 8 x= 0

Therefore, the zero of the polynomial = 0

13. By actual division, find the quotient and the remainder when the first polynomial is divided by the second polynomial: x⁴ + 1; x –1

Solution:

Performing the long division method, we get,

Hence, from the above long division method, we get,

Quotient = x³ + x² + x + 1

Remainder = 2.

14. By Remainder Theorem find the remainder, when p(x) is divided by g(x), where

(i) p(𝑥) = 𝑥³ – 2𝑥² – 4𝑥 – 1, g(𝑥) = 𝑥 + 1
(ii) p(𝑥) = 𝑥³ – 3𝑥² + 4𝑥 + 50, g(𝑥) = 𝑥 – 3
(iii) p(𝑥) = 4𝑥³ – 12𝑥² + 14𝑥 – 3, g(𝑥) = 2𝑥 – 1
(iv) p(𝑥) = 𝑥³ – 6𝑥² + 2𝑥 – 4, g(𝑥) = 1 – 3/2 𝑥

Solution:

(i) Given p(x) = 𝑥³ – 2𝑥² – 4𝑥 – 1 and g(x) = x + 1

Here zero of g(x) = – 1

By using the remainder theorem

P(x) divided by g(x) = p( – 1)

P ( – 1) = ( – 1)³ – 2 ( – 1)² – 4 ( – 1) – 1 = 0

Therefore, the remainder = 0

(ii) given p(𝑥) = 𝑥³ – 3𝑥² + 4𝑥 + 50, g(𝑥) = 𝑥 – 3

Here zero of g(x) = 3

By using the remainder theorem p(x) divided by g(x) = p(3)

p(3) = 3³ – 3 × (3)² + 4 × 3 + 50 = 62

Therefore, the remainder = 62

(iii) p(x) = 4x³ – 12x² + 14x – 3, g(x) = 2x – 1

Here zero of g(x) = ½

By using the remainder theorem p(x) divided by g(x) = p (½)

P( ½ ) = 4( ½ )³ – 12( ½ )² + 14 ( ½ ) – 3

= 4/8 – 12/4 + 14/2 – 3

= ½ + 1

= 3/2

Therefore, the remainder = 3/2

(iv) p(𝑥) = 𝑥³ – 6𝑥² + 2𝑥 – 4, g(𝑥) = 1 – 3/2 𝑥

Here zero of g(x) = 2/3

By using the remainder theorem p(x) divided by g(x) = p(2/3)

p(2/3) = (2/3)³ – 6(2/3)² + 2(2/3) – 4

= – 136/27

Therefore, the remainder = – 136/27

15. Check whether p(𝑥) is a multiple of g(𝑥) or not:

(i) p(𝑥) = 𝑥³ – 5𝑥² + 4𝑥 – 3, g(𝑥) = 𝑥 – 2
(ii) p(𝑥)= 2𝑥³ – 11𝑥²− 4𝑥 + 5, 𝑔(𝑥)= 2𝑥 + 1

Solution:

(i)

According to the question,

g(x)=x – 2,

Then, zero of g(x),

g(x) = 0

x – 2 = 0

x = 2

Therefore, zero of g(x) = 2

So, substituting the value of x in p(x), we get,

p(2) =(2)³ – 5(2)² + 4(2) – 3

= 8 – 20 + 8 – 3

= – 7 ≠ 0

Hence, p(x) is not the multiple of g(x), the remainder ≠ 0.

(ii)

According to the question,

g(x)= 2𝑥 + 1

Then, zero of g(x),

g(x) = 0

2x + 1 = 0

2x = – 1

x = – ½

Therefore, zero of g(x) = – ½

So, substituting the value of x in p(x), we get,

p(–½) = 2 × ( – ½ )³ – 11 × ( – ½ )² – 4 × ( – 1/2) + 5

= – ¼ – 11/4 + 7

= 16/4

= 4 ≠ 0

Hence, p(x) is not the multiple of g(x),  the remainder ≠ 0.

16. Show that:

(i) 𝑥 + 3 is a factor of 69 + 11𝑥−𝑥² + 𝑥³.
(ii) 2𝑥−3 is a factor of 𝑥 + 2𝑥³ – 9𝑥² + 12

Solution:

(i)According to the question,

Let p(x) = 69 + 11x − x² + x³ and g(x) = x + 3

g(x) = x + 3

zero of g(x) ⇒ g(x) = 0

x + 3 = 0

x = – 3

Therefore, zero of g(x) = – 3

So, substituting the value of x in p(x), we get,

p( – 3) = 69 + 11( – 3) –( – 3)² + ( – 3)³

= 69 – 69

= 0

Since, the remainder = zero,

We can say that,

g(x) = x + 3 is factor of p(x) = 69 + 11x − x² + x³

(ii) According to the question,

Let p(x) = x + 2x³ – 9x² + 12 and g(x) =2x−3

g(x) = 2x – 3

zero of g(x) ⇒ g(x) = 0

2x – 3 = 0

x = 3/2

Therefore, zero of g(x) = 3/2

So, substituting the value of x in p(x), we get,

P(3/2) = 3/2 + 2(3/2)³ – 9(3/2)² + 12

= (81 – 81) / 4

= 0

Since, the remainder = zero,

We can say that,

g(x) = 2x – 3 is factor of p(x) = x + 2x³ – 9x² + 12

17. Determine which of the following polynomials has x – 2 a factor:

(i) 3𝑥² + 6𝑥−24.
(ii) 4𝑥² + 𝑥−2.

Solution:

(i) According to the question,

Let p(x) =3𝑥² + 6𝑥−24 and g(x) = x – 2

g(x) = x – 2

zero of g(x) ⇒ g(x) = 0

x – 2 = 0

x = 2

Therefore, zero of g(x) = 2

So, substituting the value of x in p(x), we get,

p(2) = 3(2)² + 6 (2) – 24

= 12 + 12 – 24

= 0

Since, the remainder = zero,

We can say that,

g(x) = x – 2 is factor of p(x) = 3𝑥² + 6𝑥−24

(ii) According to the question,

Let p(x) = 4𝑥² + 𝑥−2 and g(x) = x – 2

g(x) = x – 2

zero of g(x) ⇒ g(x) = 0

x – 2 = 0

x = 2

Therefore, zero of g(x) = 2

So, substituting the value of x in p(x), we get,

p(2) = 4(2)² + 2−2

= 16 ≠ 0

Since the remainder ≠ zero,

We can say that,

g(x) = x – 2 is not a factor of p(x) = 4𝑥² + 𝑥−2

18. Show that p – 1 is a factor of p¹⁰ – 1 and also of p¹¹ – 1.

Solution:

According to the question,

Let h(p) = 𝑝 ¹⁰ − 1,and g(p) = 𝑝 – 1

zero of g(p) ⇒ g(p) = 0

p – 1 = 0

p = 1

Therefore, zero of g(x) = 1

We know that,

According to factor theorem if g(p) is a factor of h(p) , then h(1) should be zero

So,

h(1) = (1)¹⁰ − 1 = 1 − 1 = 0

⟹ g (p) is a factor of h(p).

Now, we have h(p) = 𝑝 ¹¹ − 1, g (p) = 𝑝 – 1

Putting g (p) = 0 ⟹ 𝑝 − 1 = 0 ⟹ 𝑝 = 1

According to the factor theorem, if g (p) is a factor of h(p),

Then h(1) = 0

⟹ (1)¹¹ – 1 = 0

Therefore, g(p) = 𝑝 – 1 is the factor of h(p) = 𝑝 ¹⁰ – 1

19. For what value of m is 𝑥³ – 2𝑚𝑥² + 16 divisible by x + 2?

Solution:

According to the question,

Let p(x) = x ³ – 2mx² + 16, and g(x) = x + 2

g(x) = 0

⟹ x + 2 = 0

⟹ x = – 2

Therefore, zero of g(x) = – 2

We know that,

According to the factor theorem,

if p(x) is divisible by g(x), then the remainder p(−2) should be zero.

So, substituting the value of x in p(x), we get,

p( – 2) = 0

⟹ ( – 2)³ – 2m( – 2)² + 16 = 0

⟹ 0 – 8 – 8m + 16 = 0

⟹ 8m = 8

⟹ m = 1

20. If 𝑥 + 2𝑎 is a factor of 𝑥⁵ – 4𝑎²𝑥³ + 2𝑥 + 2𝑎 + 3, find a.

Solution:

According to the question,

Let p(x) = x⁵ – 4a²x³ + 2x + 2a + 3 and g(x) = x + 2a

g(x) = 0

⟹ x + 2a = 0

⟹ x = – 2a

Therefore, zero of g(x) = – 2a

We know that,

According to the factor theorem,

If g(x) is a factor of p(x), then p( – 2a) = 0

So, substituting the value of x in p(x), we get,

p ( – 2a) = ( – 2a)⁵ – 4a²( – 2a)³ + 2( – 2a) + 2a + 3 = 0

⟹ – 32a⁵ + 32a⁵ – 2a + 3 = 0

⟹ – 2a = – 3

⟹ a = 3/2

Exercise 2.4 Page No: 23

1. If the polynomials az³ + 4z² + 3z – 4 and z³ – 4z + a leave the same remainder when divided by z – 3, find the value of a.

Solution:

Zero of the polynomial,

g₁(z) = 0

z-3 = 0

z = 3

Therefore, zero of g(z) = – 2a

Let p(z) = az³+4z²+3z-4

So, substituting the value of z = 3 in p(z), we get,

p(3) = a(3)³+4(3)²+3(3)-4

⇒p(3) = 27a+36+9-4

⇒p(3) = 27a+41

Let h(z) = z³-4z+a

So, substituting the value of z = 3 in h(z), we get,

h(3) = (3)³-4(3)+a

⇒h(3) = 27-12+a

⇒h(3) = 15+a

According to the question,

We know that,

The two polynomials, p(z) and h(z), leaves same remainder when divided by z-3

So, h(3)=p(3)

⇒15+a = 27a+41

⇒15-41 = 27a – a

⇒-26 = 26a

⇒a = -1

2. The polynomial p(x) = x⁴ – 2x³ + 3x² – ax + 3a – 7 when divided by x + 1 leaves the remainder 19. Find the values of a. Also, find the remainder when p(x) is divided by x + 2.

Solution:

p(x) = x⁴ – 2x³ + 3x² – ax + 3a – 7.

Divisor = x + 1

x + 1 = 0

x = -1

So, substituting the value of x = – 1 in p(x), we get,

p(-1) = (-1)⁴ – 2(-1)³ + 3(-1)² – a(-1) + 3a – 7.

19 = 1 + 2 + 3 + a + 3a – 7

19 = 6 – 7 + 4a

4a – 1 = 19

4a = 20

a = 5

Since a = 5.

We get the polynomial,

p(x) = x⁴ – 2x³ + 3x² – (5)x + 3(5) – 7

p(x) = x⁴ – 2x³ + 3x² – 5x + 15 – 7

p(x) = x⁴ – 2x³ + 3x² – 5x + 8

As per the question,

When the polynomial obtained is divided by (x + 2),

We get,

x + 2 = 0

x = – 2

So, substituting the value of x = – 2 in p(x), we get,

p(-2) = (-2)⁴ – 2(-2)³ + 3(-2)² – 5(-2) + 8

⇒ p(-2) = 16 + 16  + 12 + 10 + 8

⇒ p(-2) = 62

Therefore, the remainder = 62.

3. If both x – 2 and x – ½ are factors of px² + 5x + r, show that p = r.

Solution:

Given, f(x) = px²+5x+r and factors are x-2, x – ½

g₁(x) = 0,

x – 2 = 0

x = 2

Substituting x = 2 in place of equation, we get

f(x) = px²+5x+r

f(2) = p(2)²+5(2)+r=0

=  4p + 10 + r = 0   … eq.(i)

x – ½ = 0

x = ½

Substituting x = ½ in place of the equation, we get,

f(x) = px²+5x+r

f( ½ ) = p( ½ )² + 5( ½ ) + r =0

= p/4 + 5/2 + r = 0

= p + 10 + 4r = 0  … eq(ii)

On solving eq(i) and eq(ii),

We get,

4p + r = – 10 and p + 4r = – 10

Since the RHS of both equations are the same,

We get,

4p + r = p + 4r

3p=3r

p = r.

Hence Proved.

4. Without actual division, prove that 2x⁴ – 5x³ + 2x² – x + 2 is divisible by x² – 3x + 2.

[Hint: Factorise x² – 3x + 2]

Solution:

x²-3x+2

x²-2x-1x+2

x(x-2)-1(x-2)

(x-2)(x-1)

Therefore,(x-2)(x-1)are the factors.

Considering (x-2),

x-2=0

x=2

Then, p(x) becomes,

p(x)=2

p(x)=2x⁴-5x³+2x²-x+2

p(2)=2(2)⁴-5(2)³+2(2)²-2+2

=32-40+8

= -40+40=0

Therefore, (x-2) is a factor.

Considering (x-1),

x-1=0

x=1

Then, p(x) becomes,

p(x)=1

p(x)=2x⁴-5x³+2x²-x+2

p(1)=2(1)⁴-5(1)³+2(1)²-1+2

=2-5+2-1+2

=6-6

=0

Therefore, (x-1) is a factor.

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