Class 9 Maths - Polynomials NCERT Solutions

Chapter 2: Polynomials (NCERT Solutions)

Exercise 2.1

Q1. Which of the following expressions are polynomials in one variable and which are not? State reasons.
(i) 4x² - 3x + 7 (ii) y² + √2 (iii) 3√t + t√2 (iv) y + 2/y (v) x¹⁰ + y³ + t⁵⁰

(i) 4x² - 3x + 7: Yes. Polynomial in one variable x. All exponents are non-negative integers.

(ii) y² + √2: Yes. Polynomial in one variable y. √2 is a constant term.

(iii) 3√t + t√2: No. 3√t = 3t^1/2. Exponent 1/2 is not a whole number.

(iv) y + 2/y: No. 2/y = 2y^-1. Exponent -1 is not a non-negative integer.

(v) x¹⁰ + y³ + t⁵⁰: Yes, but it is a polynomial in three variables, not one variable.

Q2. Write the coefficients of x² in each of the following:
(i) 2 + x² + x (ii) 2 - x² + x³ (iii) (π/2)x² + x (iv) √2x - 1

(i) 2 + x² + x: Coefficient of x² = 1.

(ii) 2 - x² + x³: Coefficient of x² = -1.

(iii) (π/2)x² + x: Coefficient of x² = π/2.

(iv) √2x - 1: There is no x² term. Coefficient of x² = 0.

Q3. Give one example each of a binomial of degree 35, and of a monomial of degree 100.

Binomial of degree 35: x³⁵ + 1 (has two terms, highest exponent is 35).

Monomial of degree 100: 5x¹⁰⁰ (has one term, exponent is 100).

Q4. Write the degree of each of the following polynomials:
(i) 5x³ + 4x² + 7x (ii) 4 - y² (iii) 5t - √7 (iv) 3

(i) 5x³ + 4x² + 7x: Degree = 3 (highest exponent).

(ii) 4 - y²: Degree = 2.

(iii) 5t - √7: Degree = 1.

(iv) 3: 3 is a non-zero constant. Degree = 0.

Q5. Classify the following as linear, quadratic, and cubic polynomials:
(i) x² + x (ii) x - x³ (iii) y + y² + 4 (iv) 1 + x (v) 3t (vi) r² (vii) 7x³

Linear (degree 1): (iv) 1 + x, (v) 3t.

Quadratic (degree 2): (i) x² + x, (iii) y + y² + 4, (vi) r².

Cubic (degree 3): (ii) x - x³, (vii) 7x³.

Exercise 2.2

Q1. Find the value of the polynomial 5x - 4x² + 3 at (i) x = 0 (ii) x = -1 (iii) x = 2.

Let p(x) = 5x - 4x² + 3.

(i) x = 0: p(0) = 5(0) - 4(0)² + 3 = 0 - 0 + 3 = 3.

(ii) x = -1: p(-1) = 5(-1) - 4(-1)² + 3 = -5 - 4 + 3 = -6.

(iii) x = 2: p(2) = 5(2) - 4(2)² + 3 = 10 - 16 + 3 = -3.

Q2. Find p(0), p(1) and p(2) for each of the following polynomials:
(i) p(y) = y² - y + 1 (ii) p(t) = 2 + t + 2t² - t³ (iii) p(x) = x³ (iv) p(x) = (x-1)(x+1)

(i) p(y) = y² - y + 1:
p(0) = 0 - 0 + 1 = 1. p(1) = 1 - 1 + 1 = 1. p(2) = 4 - 2 + 1 = 3.

(ii) p(t) = 2 + t + 2t² - t³:
p(0) = 2. p(1) = 2 + 1 + 2 - 1 = 4. p(2) = 2 + 2 + 8 - 8 = 4.

(iii) p(x) = x³:
p(0) = 0. p(1) = 1. p(2) = 8.

(iv) p(x) = (x-1)(x+1) = x² - 1:
p(0) = -1. p(1) = 0. p(2) = 3.

Q3. Verify whether the following are zeroes of the polynomial, indicated against them:
(i) p(x) = 3x + 1, x = -1/3 (ii) p(x) = 5x - π, x = 4/5 (iii) p(x) = x² - 1, x = 1 and -1
(iv) p(x) = (x+1)(x-2), x = -1 and 2 (v) p(x) = x², x = 0 (vi) p(x) = lx + m, x = -m/l
(vii) p(x) = 3x² - 1, x = -1/√3 and 2/√3 (viii) p(x) = 2x + 1, x = 1/2

(i) p(-1/3): 3(-1/3) + 1 = -1 + 1 = 0. Yes, -1/3 is a zero.

(ii) p(4/5): 5(4/5) - π = 4 - π ≠ 0. No, 4/5 is not a zero.

(iii) p(1) = 1-1 = 0. p(-1) = 1-1 = 0. Yes, both 1 and -1 are zeroes.

(iv) p(-1) = (-1+1)(-1-2) = 0. p(2) = (2+1)(2-2) = 0. Yes, both are zeroes.

(v) p(0) = 0² = 0. Yes, 0 is a zero.

(vi) p(-m/l) = l(-m/l) + m = -m + m = 0. Yes, -m/l is a zero.

(vii) p(-1/√3) = 3(1/3) - 1 = 0. Yes. p(2/√3) = 3(4/3) - 1 = 4 - 1 = 3 ≠ 0. No, 2/√3 is NOT a zero.

(viii) p(1/2) = 2(1/2) + 1 = 1 + 1 = 2 ≠ 0. No, 1/2 is not a zero.

Q4. Find the zero of the polynomial in each of the following cases:
(i) p(x) = x + 5 (ii) p(x) = x - 5 (iii) p(x) = 2x + 5 (iv) p(x) = 3x - 2
(v) p(x) = 3x (vi) p(x) = ax, a ≠ 0 (vii) p(x) = cx + d, c ≠ 0

Set p(x) = 0 and solve for x:

(i) x + 5 = 0 ⇒ x = -5.

(ii) x - 5 = 0 ⇒ x = 5.

(iii) 2x + 5 = 0 ⇒ x = -5/2.

(iv) 3x - 2 = 0 ⇒ x = 2/3.

(v) 3x = 0 ⇒ x = 0.

(vi) ax = 0 ⇒ x = 0.

(vii) cx + d = 0 ⇒ x = -d/c.

Exercise 2.3

Q1. Find the remainder when x³ + 3x² + 3x + 1 is divided by:
(i) x + 1 (ii) x - 1/2 (iii) x (iv) x + π (v) 5 + 2x

Let p(x) = x³ + 3x² + 3x + 1. By Remainder Theorem, remainder = p(value that makes divisor zero).

(i) x + 1 = 0 ⇒ x = -1: p(-1) = (-1)³ + 3(-1)² + 3(-1) + 1 = -1 + 3 - 3 + 1 = 0.

(ii) x = 1/2: p(1/2) = (1/2)³ + 3(1/2)² + 3(1/2) + 1 = 1/8 + 3/4 + 3/2 + 1 = 1/8 + 6/8 + 12/8 + 8/8 = 27/8.

(iii) x = 0: p(0) = 0 + 0 + 0 + 1 = 1.

(iv) x + π = 0 ⇒ x = -π: p(-π) = (-π)³ + 3(-π)² + 3(-π) + 1 = -π³ + 3π² - 3π + 1 = 1 - 3π + 3π² - π³.

(v) 5 + 2x = 0 ⇒ x = -5/2: p(-5/2) = (-5/2)³ + 3(-5/2)² + 3(-5/2) + 1 = -125/8 + 75/4 - 15/2 + 1 = -125/8 + 150/8 - 60/8 + 8/8 = -27/8.

Q2. Find the remainder when x³ - ax² + 6x - a is divided by (x - a).

By Remainder Theorem, put x = a in p(x) = x³ - ax² + 6x - a:
p(a) = a³ - a(a)² + 6(a) - a = a³ - a³ + 6a - a = 5a.

Q3. Check whether 7 + 3x is a factor of 3x³ + 7x.

7 + 3x = 0 ⇒ x = -7/3.
p(-7/3) = 3(-7/3)³ + 7(-7/3)
= 3(-343/27) + (-49/3)
= -343/9 - 49/3
= -343/9 - 147/9
= -490/9 ≠ 0.
Since the remainder is not 0, 7 + 3x is NOT a factor of 3x³ + 7x.

Exercise 2.4

Q1. Determine which of the following polynomials has (x + 1) as a factor:
(i) x³ + x² + x + 1 (ii) x⁴ + x³ + x² + x + 1 (iii) x⁴ + 3x³ + 3x² + x + 1 (iv) x³ - x² - (2+√2)x + √2

(x + 1) is a factor if p(-1) = 0.

(i) p(-1) = -1 + 1 - 1 + 1 = 0. Yes, (x+1) is a factor.

(ii) p(-1) = 1 - 1 + 1 - 1 + 1 = 1 ≠ 0. Not a factor.

(iii) p(-1) = 1 - 3 + 3 - 1 + 1 = 1 ≠ 0. Not a factor.

(iv) p(-1) = -1 - 1 + (2+√2) + √2 = 0 + 2√2 = 2√2 ≠ 0. Not a factor.

Q2. Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each case:
(i) p(x) = 2x³ + x² - 2x - 1, g(x) = x + 1
(ii) p(x) = x³ + 3x² + 3x + 1, g(x) = x + 2
(iii) p(x) = x³ - 4x² + x + 6, g(x) = x - 3

(i) g(x) = x + 1 ⇒ x = -1: p(-1) = 2(-1)³ + (-1)² - 2(-1) - 1 = -2 + 1 + 2 - 1 = 0. Yes, g(x) is a factor.

(ii) g(x) = x + 2 ⇒ x = -2: p(-2) = -8 + 12 - 6 + 1 = -1 ≠ 0. No, g(x) is not a factor.

(iii) g(x) = x - 3 ⇒ x = 3: p(3) = 27 - 36 + 3 + 6 = 0. Yes, g(x) is a factor.

Q3. Find the value of k, if x - 1 is a factor of p(x) in each case:
(i) p(x) = x² + x + k (ii) p(x) = 2x² + kx + √2 (iii) p(x) = kx² - √2x + 1 (iv) p(x) = kx² - 3x + k

(x - 1) is a factor when p(1) = 0.

(i) p(1) = 1 + 1 + k = 0 ⇒ k = -2.

(ii) p(1) = 2 + k + √2 = 0 ⇒ k = -2 - √2.

(iii) p(1) = k - √2 + 1 = 0 ⇒ k = √2 - 1.

(iv) p(1) = k - 3 + k = 0 ⇒ 2k = 3 ⇒ k = 3/2.

Q4. Factorise: (i) 12x² - 7x + 1 (ii) 2x² + 7x + 3 (iii) 6x² + 5x - 6 (iv) 3x² - x - 4

(i) 12x² - 7x + 1: Find two numbers with product 12 and sum -7: -3 and -4.
= 12x² - 3x - 4x + 1 = 3x(4x - 1) - 1(4x - 1) = (3x - 1)(4x - 1).

(ii) 2x² + 7x + 3: Product = 6, sum = 7: 1 and 6.
= 2x² + x + 6x + 3 = x(2x + 1) + 3(2x + 1) = (x + 3)(2x + 1).

(iii) 6x² + 5x - 6: Product = -36, sum = 5: 9 and -4.
= 6x² + 9x - 4x - 6 = 3x(2x + 3) - 2(2x + 3) = (3x - 2)(2x + 3).

(iv) 3x² - x - 4: Product = -12, sum = -1: -4 and 3.
= 3x² - 4x + 3x - 4 = x(3x - 4) + 1(3x - 4) = (x + 1)(3x - 4).

Q5. Factorise: (i) x³ - 2x² - x + 2 (ii) x³ - 3x² - 9x - 5 (iii) x³ + 13x² + 32x + 20 (iv) 2y³ + y² - 2y - 1

(i) x³ - 2x² - x + 2:
Try x = 1: 1 - 2 - 1 + 2 = 0. So (x - 1) is a factor.
Divide: x³ - 2x² - x + 2 = (x - 1)(x² - x - 2) = (x - 1)(x - 2)(x + 1) = (x - 1)(x + 1)(x - 2).

(ii) x³ - 3x² - 9x - 5:
Try x = 5: 125 - 75 - 45 - 5 = 0. So (x - 5) is a factor.
Divide: = (x - 5)(x² + 2x + 1) = (x - 5)(x + 1)².

(iii) x³ + 13x² + 32x + 20:
Try x = -1: -1 + 13 - 32 + 20 = 0. So (x + 1) is a factor.
Divide: = (x + 1)(x² + 12x + 20) = (x + 1)(x + 2)(x + 10) = (x + 1)(x + 2)(x + 10).

(iv) 2y³ + y² - 2y - 1:
Try y = 1: 2 + 1 - 2 - 1 = 0. So (y - 1) is a factor.
Divide: = (y - 1)(2y² + 3y + 1) = (y - 1)(2y + 1)(y + 1) = (y - 1)(y + 1)(2y + 1).

Exercise 2.5 (Selected Key Questions)

Q1. Use suitable identities to find the following products:
(i) (x + 4)(x + 10) (ii) (x + 8)(x - 10) (iii) (3x + 4)(3x - 5) (iv) (y² + 3/2)(y² - 3/2) (v) (3 - 2x)(3 + 2x)

Using (x + a)(x + b) = x² + (a+b)x + ab and (a+b)(a-b) = a² - b²:

(i) (x+4)(x+10): x² + 14x + 40.

(ii) (x+8)(x-10): x² - 2x - 80.

(iii) (3x+4)(3x-5): 9x² - 3x - 20.

(iv) (y² + 3/2)(y² - 3/2): y⁴ - 9/4.

(v) (3 - 2x)(3 + 2x): 9 - 4x².

Q9. Verify that x³ + y³ + z³ - 3xyz = (1/2)(x + y + z)[(x - y)² + (y - z)² + (z - x)²].

We know the identity: x³ + y³ + z³ - 3xyz = (x + y + z)(x² + y² + z² - xy - yz - zx).
Now expand the RHS of what needs to be verified:
(1/2)(x+y+z)[(x-y)² + (y-z)² + (z-x)²]
= (1/2)(x+y+z)[x² - 2xy + y² + y² - 2yz + z² + z² - 2zx + x²]
= (1/2)(x+y+z)[2x² + 2y² + 2z² - 2xy - 2yz - 2zx]
= (x+y+z)[x² + y² + z² - xy - yz - zx]
= x³ + y³ + z³ - 3xyz. (Verified).

Q13. If x + y + z = 0, show that x³ + y³ + z³ = 3xyz.

Using the identity: x³ + y³ + z³ - 3xyz = (x + y + z)(x² + y² + z² - xy - yz - zx).
Since x + y + z = 0 (given):
x³ + y³ + z³ - 3xyz = 0 × (x² + y² + z² - xy - yz - zx) = 0.
Therefore x³ + y³ + z³ = 3xyz. (Proved)

Class 9 Maths - Polynomials Practice
Chapter 2: Polynomials (Practice Questions)
RD Sharma / Extra Practice
Q1. If p(x) = x² - 4x + 3, find p(0), p(1), p(-1), and p(3).
p(0) = 0 - 0 + 3 = 3.
p(1) = 1 - 4 + 3 = 0.
p(-1) = 1 + 4 + 3 = 8.
p(3) = 9 - 12 + 3 = 0.
(So x = 1 and x = 3 are zeroes of p(x).)
Q2. Find the remainder when p(x) = x⁴ - 2x³ + 3x² - ax + 3a - 7 is divided by (x + 1), given the remainder is 19.
p(-1) = 1 + 2 + 3 + a + 3a - 7 = 4a - 1.
Given: 4a - 1 = 19 ⇒ 4a = 20 ⇒ a = 5.
Q3. Using identities, evaluate 105².
105² = (100 + 5)² = 100² + 2 × 100 × 5 + 5² = 10000 + 1000 + 25 = 11025.
Q4. Evaluate 999² using an identity.
999² = (1000 - 1)² = 1000² - 2 × 1000 × 1 + 1² = 1000000 - 2000 + 1 = 998001.
Q5. If a + b = 10 and ab = 21, find a² + b².
Using (a + b)² = a² + 2ab + b²:
100 = a² + 2(21) + b²
a² + b² = 100 - 42 = 58.
Q6. If x - y = 7 and xy = 9, find x² + y².
(x - y)² = x² - 2xy + y²
49 = x² + y² - 18
x² + y² = 67.
Q7. Expand (2x - 3y + 4z)² using identity.
Using (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca:
Here a = 2x, b = -3y, c = 4z.
= 4x² + 9y² + 16z² + 2(2x)(-3y) + 2(-3y)(4z) + 2(4z)(2x)
= 4x² + 9y² + 16z² - 12xy - 24yz + 16zx.
Q8. Factorise: 8a³ + 27b³ + 36a²b + 54ab².
= 8a³ + 27b³ + 36a²b + 54ab²
= (2a)³ + (3b)³ + 3(2a)²(3b) + 3(2a)(3b)²
Using (x + y)³ = x³ + 3x²y + 3xy² + y³:
= (2a + 3b)³.
Q9. Factorise: 27x³ - 125y³.
Using a³ - b³ = (a - b)(a² + ab + b²):
a = 3x, b = 5y.
= (3x - 5y)(9x² + 15xy + 25y²) = (3x - 5y)(9x² + 15xy + 25y²).
Q10. Evaluate 103 × 107 using a suitable identity.
103 × 107 = (105 - 2)(105 + 2) = 105² - 4 = 11025 - 4 = 11021.
Q11. If x + 1/x = 4, find x² + 1/x².
(x + 1/x)² = x² + 2 + 1/x² = 16.
x² + 1/x² = 16 - 2 = 14.
Q12. Factorise: x³ - 23x² + 142x - 120.
Try x = 1: 1 - 23 + 142 - 120 = 0. So (x - 1) is a factor.
Divide: x³ - 23x² + 142x - 120 = (x - 1)(x² - 22x + 120).
Factorise x² - 22x + 120: find two numbers with product 120 and sum -22: -10 and -12.
= (x - 1)(x - 10)(x - 12) = (x - 1)(x - 10)(x - 12).
Q13. Without actually calculating, find the value of 48³ + (-30)³ + (-18)³.
Let a = 48, b = -30, c = -18.
a + b + c = 48 - 30 - 18 = 0.
When a + b + c = 0, we know: a³ + b³ + c³ = 3abc.
= 3 × 48 × (-30) × (-18) = 3 × 48 × 540 = 3 × 25920 = 77760.
Q14. If the polynomials 2x³ + ax² + 3x - 5 and x³ + x² - 4x + a leave the same remainder when divided by (x - 2), find the value of a.
Remainder when divided by (x - 2) is p(2) in each case.
p(2) = 2(8) + 4a + 6 - 5 = 16 + 4a + 1 = 17 + 4a.
q(2) = 8 + 4 - 8 + a = 4 + a.
Setting equal: 17 + 4a = 4 + a ⇒ 3a = -13 ⇒ a = -13/3.
Q15. If x² + y² = 49 and xy = 10, find the value of x + y and x - y (considering positives).
(x + y)² = x² + 2xy + y² = 49 + 20 = 69 ⇒ x + y = √69.
(x - y)² = x² - 2xy + y² = 49 - 20 = 29 ⇒ x - y = √29.
Class 9 Maths - Polynomials Summary
Chapter 2: Polynomials (Concepts & Formulas)
1. Key Definitions
Polynomial: An expression of the form a_nxⁿ + ... + a₁x + a₀ where all exponents are non-negative integers.
Degree: The highest power of the variable in the polynomial.
Linear: Degree 1. Quadratic: Degree 2. Cubic: Degree 3.
Zero of Polynomial: A value of x for which p(x) = 0.
Monomial: 1 term. Binomial: 2 terms. Trinomial: 3 terms.
2. Remainder Theorem
If a polynomial p(x) is divided by (x - a), the remainder = p(a).
3. Factor Theorem
(x - a) is a factor of p(x) if and only if p(a) = 0.
4. Algebraic Identities
(a + b)² = a² + 2ab + b²
(a - b)² = a² - 2ab + b²
(a + b)(a - b) = a² - b²
(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
(a + b)³ = a³ + 3a²b + 3ab² + b³
(a - b)³ = a³ - 3a²b + 3ab² - b³
a³ + b³ = (a + b)(a² - ab + b²)
a³ - b³ = (a - b)(a² + ab + b²)
a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - ab - bc - ca)
If a + b + c = 0, then a³ + b³ + c³ = 3abc
(x + a)(x + b) = x² + (a + b)x + ab

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