SEBA Class 9 Maths Chapter 2 Polynomials

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SEBA Class 9 Maths Chapter 2 Polynomials

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Polynomials

Chapter: 2

Exercise 2.1

Q.1. Which of the following expressions are polynomials in one vari-able and which are not? State reasons for your answer.

(i) 4x²-3x+7

Ans: Yes, 4 x²-3 x+7 is a polynomial of one variable. As x has degree 2 in it and only x is one variable.

(ii) y²+√2

Ans: Yes, y is only one variable.

(iii) 3√t+t√2

Ans: No as 3√t+r√2 can be written as 3,t¹/² +√2 Here, the exponent of t in 3 t ¹/² which is not a whole number.

(iv) y+2/y

Ans: No, as y+2/y can be written as y+2 y¹ where the exponent of y in 2/y is -1, which is not a whole number.

(v) x¹⁰+y³+t⁵⁰

Ans: Yes, It is a polynomial in three variables, x, y and t.

Q.2. Write the coefficients of x² in each of the following:

(i) 2 + x² + x

Ans: We have given that the equation 2 + x²+ x We can also write 1x²+ 1x + 2

Therefore, the coefficient of x² in this equation is 1.

(ii) 2 – x² + x³

Ans: We have given that the equation: 2 – x² + x³ We can also write x³- x² + 2

Therefore, the coefficient of x² in this equation is -1.

(iii) π/2 x² + x

Ans: We have given that the equation π/2x² + x

Therefore, the coefficient of x² in this equation is 11/.7

(iv) √2x- 1

Ans: We have given that the equation √2x -1

We can also write ox²+ √2x) – 1

Therefore, the coefficient of x² in this equation is 0.

(v) (2x – 3)(x² – 3x + 1)

Ans: (2x – 3)(x² – 3x + 1)

= 2x(x²- 3x + 1) – 3(x² – 3x + 1)

= 2x³- 6x² + 2x – 3x² + 9x – 3

= 2x³ – 9x² + 11x – 3

∴ Coefficient x ^ 2 = – 9

Q. 3. Give one example each of a binomial of degree 35, and a mono- mial of degree 100.

Ans: We know that polynomials having only two terms are called binomials.

Therefore, the example of a binomial of degree 35 is ax³⁵ + b where a and b are any real number.

Again, the example of a monomial of degree 100 is ax¹⁰⁰ where a is any real number.

Q.4. Write the degree of each of the following polynomials:

(i) 5x³ + 4x²+ 7x

Ans: We know that the highest power of variable in a polyno- mial is called degree of the polynomial, In polynomial 5x³ + 4x² + 7x

The highest power of variable x is 3.

Therefore, the degree of the polynomial 5x³ + 4x² + 7x is 3.

(ii) 4 – y²

Ans: In polynomial 4 – y² the highest power of the variable y is 2.

Therefore, the degree of the polynomial 4 – y² is 2.

(iii) 5t -√7

Ans: In polynomial 5t – √5 the highest power of the variable t is 1.

Therefore, the degree of polynomial 5t -√5 is 1.

(iv) 3hh.

The only term here is 3 which can be written as 3x⁰ .So the highest power of the variable x is 0.

Therefore, the degree of the polynomial 3 is 0.

Q.5. Classify the following as linear, quadratic and cubic polynomials.

(i) x²+ x

Ans: In polynomial x² + x the highest power of the variable x is 2.

So the degree of the polynomial is 2. We know that the polynomial of degree 2 is called a quadratic polynomial.

Therefore, the polynomial x² + x is a quadratic polynomial.

(ii) x – x³

Ans: In polynomial x – x³, the highest power of the variable x is 3. So the degree of the polynomial is 3.

We know that the polynomial of degree 3 is called a cubic polynomial.

Therefore, the polynomial x – x³ is a cubic polynomial.

(iii) y + y²+ 4

Ans: In polynomial y + y² + 4 the highest power of the variable y is 2. So the degree of the polynomial is 2. We know that the polynomial of degree 2 is called quadratic polynomial.

Therefore, the polynomial y + y² + 4 is a quadratic polynomial.

(iv) 1 + x

Ans: In polynomial l + x the highest power of the variable x is 1. So the degree of the polynomial is 1.

We know that the polynomial of degree 1 is called linear polynomial.

Therefore, polynomial 1 + x is a linear polynomial.

(v) 3t

Ans: In polynomial 3t, the highest power of the variable t is 1. So, the degree of the polynomial is 1.

We know that the polynomial of degree 1 is called a linear polynomial.

Therefore, polynomial 3t is a linear polynomial

(vi) r²

Ans: In polynomial r² the highest power of the variable r is 2. So, the degree of the polynomial is 2. We know that the polynomial of degree 2 is called quadratic polynomial.

Therefore, polynomial r ^ 2 is a quadratic polynomial.

(vii) 7x²

Ans: In polynomial 7x³ the highest power of the variable x is 3. So the degree of the polynomial is 3. We know that the polynomial of degree 3 is called a cubic polynomial.

Therefore, polynomial 7x³ is a cubic polynomial.

Q. 6. Which one is a polynomial?

(a) x- + 3x – 1

(b) y + 1/y

(c) 3y/2 + x

(d) 2√x+ 1

Ans: (c) 3y/2 + x

6. Two statements are given below.

Statement-1 : The degree of the zero polynomial is not defined.

Statement-2 : The degree of the non zero constant polynomial is zero.

Choose the correct option

(a) Statement 1 and statement 2 both are true

(b) Statement 1 and statement 2 both are false

(d) Statement 1 is false but statement 2 is true.

Ans. (a) Statement 1 and statement 2 both are true

Explanation :

The degree of zero polynomial is not defined because in zero polynomial, the coefficient of any variable is zero, i.e.

0x², 0x³ or 0x⁷, ……….etc.

Hence we can not exactly determine the degree of variable

Statement I → True

A non-zero constant polynomial like 2 can be written as 2x⁰, Hence 2 is a polynomial of degree 0, because exponent of x is 0.

Statement II → True

7. Which of the following are constant polynomials?

(i) 5

(ii) x

(iii) 0

(iv) 5x

Choose the correct option

(a) (i) and (ii)

(b) (iii) and (iv)

(d) (i) and (iv)

Ans. (c) (i) and (iii)

Explanation:

We know that a polynomial of degree zero is called a constant polynomial.

5 = 5x⁰ (degree is 0)

0 = 0.x⁰ (degree is 0)

8. Match column-I with column-II

Column-I	Column-II
(A) Zero of the zero polynomial is	(P) Zero
(B) √2 is a polynomial of degree	(Q) Not defined
(C) Degree of the zero polynomial is	(R) Any real number

Choose the correct alternative.

(a) A → P, B → Q, C → R

(b) A → R, B → P, C → Q

(d) A → P, B → R, C → Q

Ans: (A) zero of the zero polynomial is any real number.

Let the zero polynomial be p(x) = 0. Let k be any real number. Since the equation p(x) = 0 is satisfied for every real value of k, therefore every real number is a zero of the zero polynomial.

Column I (A) → Column II (R)

(B) √2 = √2 x⁰,

Hence √2 is a polynomial of degree 0, because exponent of x is 0.

∴ Column I (B) → Column II (P)

i.e. 0.x² or 0.x³ ……etc.

We can not exactly determine the degree of variable…….

∴ Column I (C) → Column II (Q)

Hence the correct option (b)

9. If (3x − 1)⁷ = a₇x⁷ + a₆x⁶ + a₅x⁵ + a₄x⁴ + a₃x³ + a₂x² + a₁x + a₀ then

a₇ + a₆ + a₅ + a₄ + a₃ + a₂ + a₁ + a₀ is.

(a) 0

(b) 128

(d) 64

Ans: Here,

(3x – 1)⁷= a₇ x⁷ + a₆x⁶ + a₅x⁵ + a₄x⁴ + a₃x³ + a₂x² + a₁x + a₀

Putting x = 1, we get

(3×1 – 1)⁷ = a₇×1⁷ + a₆×1⁶ + a₅×1⁵ + a₄×1⁴ + a₃×1³ + a₂×1² + a₁×1 + a₀

⇒ 2⁷ = a₇ + a₆ + ………. + a₀

∴ a₇ + a₆ + ………. + a₁ + a₀ = 128

The correct option is (b).

Exercise 2.2

Q.1. Find the value of the polynomial 5x-4x²+3 at

(i) x = 0

Ans: Let P(x) = 5x-4 x² – 3

Therefore, P(0) = 5 (0) – 4 (0) + 3 = 0 – 0 + 3 = 3

So, the value of P(x) at x = 0 is 3.

(ii) x = – 1

Ans: Let P(x)=5x-4x²+3

Therefore, P(-1)=5(-1)-4(-1)²+3=6

So the value of P(x) at x=-1 is -6

(iii) x = 2

Ans: Let P(x)=5x-4x²+3

Therefore, P(2)=5×2-4(2)²+3=10-16+3=-3

So, the value of P(x) at x=2 is-3

Q.2. Find p(¹⁰) and p(²) for each of the following polynomials:

(i) p(y )= y² – y + 1

Ans: We have given P(y) = y²-y+1

Therefore, the value of polynomial p(y) at y = 0 is

P(0)=0²-0+1=1

Again, the value of polynomial p(y) at y = 1 is P(1)=1²-1+1=1

p(2) = 2²-2+1=3

(ii) p(t) = 2+1+2r²-t³

Ans: We have given, p(t) = 2+t+2r²-t³

Therefore, the value of polynomial p(t) at 1 = 0 is

P(0)=2+0+2.(0)² – (0)³ = 2

Again, the value of polynomial p(t) at t = 1 is

p(1)=2+1+2(1)² – (1)³=4

Again, the value of polynomial p(t) at t = 2 is

p(2)=2+2+2 (2)² – (2)³ = 4

(iii) p(x) = x³

Ans: We have given, p(x) = x³

Therefore, the value of polynomial p(x) at x=0 is

P(0)=(0)³ = 0

Again the value of polynomial p(x) at x = 1 is

p(1) = (1)³ = 1

Again, the value of polynomial p(x) at x = 2 is is

P(2)=(2)³=8

(iv) p(x)=(x-1)(x+1)

Ans: We have given p(x)=(x-1)(x+1)

Therefore, the value of polynomial p(x) at x=0 is

P(0)=(0-1)(0+1)=(-1)x(+1)=-1

Again, the value of polynomial p(x) at x = 1 is

p(1) = (1-1)x(1+1)=0x2=0

Again, the value of polynomial p(x) at x = 2 is

p(2)=(2-1) (2+1)=1×3=3

Q.3. Verify whether the following are zeros of the polynomial indi-cated against them:

(i) P(x)=3x+1,x=-1/3

Ans: We have given that, p(x) = 3x + 1

Therefore, the value of polynomial p(x) at x = – 1/3 is

Yes x = – ⅓ is the zero of polynomial p(x)

(ii) P(x) = 5x − π, x = -4/5

Ans: x = – We have given that, p(x) = 5x – π

Therefore, the value of polynomial p(x) at x = 4/5 is

No, x = 4/5 is not the zero of p(x) = 5x – π.

(iii) p(x) = x²-1, x = 1,-1

Ans: We have given that, p(x) = x² – 1

Therefore, the value of polynomial p(x) at x = 1 is

p(1) = 1²- 1 = 0

Again, the value of polynomial p(x) at x = – 1 is

p(- 1) = (- 1)²- 1 = 0

Yes, x = 1, -1 are the zero of polynomial p(x) = x²- 1

(iv) p(x) = (x+1)(x-2), x = – 1,2

Ans: We have given that, p(x) = (x + 1)(x – 2)

Therefore, the value of polynomial p(x) at x = – 1 is

p(- 1) = (- 1 + 1)(- 1 – 2) = 0(- 3) = 0

Again, the value of polynomial p(x) at x = 2 is

p(2) = (2 + 1)(2 – 2) = 3 x 0 = 0

Yes x = – 1, 2 are the zero of polynomial

p(x) = x + 1(x – 2)

Yes x = – 1/3 is the zero of polynomial p(x)

(v) p(x) = x², x=0

Ans: We have given that, p(x) = x²

Therefore, the value of polynomial p(x) at x = 0 is

p(0) = 0² = 0

Yes x = 0 is the zero of polynomial p(x) = x²

(vi) P(x)=lx+m, x=-m/1

Ans: We have given that, p(x) = lx + m

Therefore, the value of polynomial p(x) at x = – m is

Yes, x = – m/l is the zero of polynomial p(x) = lx + m

(vii) p(x)=3x²-3x²-1,x=1/√3,2-√3

Ans: We have given that, p(x) = 3x² – 1

Therefore, the value of polynomial p(x) at x = – 1/√3 is

Again, the value of polynomial p(x) at x = 2/√3 is

Therefore, x = – 1/√3 is the zero of polynomial and p(x) and x = 2/√3

is not the zero of polynomial p(x).

(viii) p(x) =2x+1,x=1/2

Ans: We have given that, p(x) = 2x + 1

Therefore, the value of polynomial p(x) at x = 1/2 is

Therefore, x = 1/2 is not the zero of polynomial p(x).

Q.4. Find the zero of the polynomial in each of the following cases:

(i) p(x) = x+5

Ans: We have given that, p(x) = x+5 ..(1)

To find the zero of polynomial p(x), we can take p(x) = 0…(2)

From equations (1) and (2)

x+5=0

∴ x=-5

Therefore, x = -5 is the zero of the polynomial p(x)=x+5.

(ii) p(x) = x-5

Ans: We have given that, p(x)=x-5 …(1)

To find the zero of polynomial p(x) we can take p(x) = 0 …(2)

From equations (1) and (2)

x-5=0

∴ x=5

Therefore x = 5 is the zero of polynomial p(x)=x-5.

(iii) p(x) = 2x+5

Ans: To find the zero of polynomial p(x), we can take p(x) = 0 …(2)

From equations (1) and (2)

Therefore, x = – 5/2 is the zero of polynomial p(x) = 2x+5

(iv) p(x) = 3x – 2

Ans: We have given that, p(x)=3x-2 …(1)

To find the zero of polynomial p(x) we can take p(x)=0 … (2)

Therefore, x=-5/3 is the zero of polynomial p(x)=3x-2

(v) p(x) = 3x

Ans: We have given that, p(x)= ax, a ≠ 0 …(1)

To find the zero of polynomial p(x) we can take p(x) = 0 …(2)

From equations (1) and (2)

ax = 0

∴ x = 0/a = 0

Therefore, x = 0 is the zero of polynomial p(x) = ax, a ≠ 0.

(vi) p(x) = ax, a ≠ 0

Ans: We have given that, p(x)=3x …(1)

To find the zero of polynomial p(x) we can take p(x)=0 …(2)

From equations (1) and (2), 3x=0 ∴ x=0

Therefore, x = 0 is the zero of polynomial p(x) = 3x

(vii) p(x)=cx+d, c ≠ 0, c; d are real numbers.

Ans: ) We have given that, p(x) = cx + d …(1)

c ≠ 0, c, d are real numbers.

To find the zero of polynomial p(x) we can take

p(x) = 0 …(2)

From equations (1) and (2)

cx + d = 0

x = – d/c x where c ≠ 0x and are real numbers

Therefore, x=-d/c, where c, d and c,d

are real numbers is the solution of polynomial

p(x) = cx + d.

5. Which of the following are the zeroes of the polynomial x²-x-6?

(i) 3

(ii) -3

(iii) 2

(iv) -2

Choose the correct alternative-

(a) (i) and (ii)

(b) (ii) and (iii)

(d) (iii) and (iv)

Ans: x²-x-6= x²-(3-2)x-6

= x² – 3x + 2x-6 = x(x-3)+2(x – 3)

= (x – 3) (x + 2)

Either, x – 3 = 0 or, ⇒ x = 3 ⇒x=-2 x + 2 = 0

Zeroes of the polynomial are 3 & -2

The correct option is (c)

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SEBA Class 9 Maths Chapter 2 Polynomials