Chapter 2: Polynomials (NCERT Solutions)

Exercise 2.1

Q1. The graphs of y = p(x) are given in Fig. 2.10 below, for some polynomials p(x). Find the number of zeroes of p(x), in each case.

The number of zeroes is equal to the number of times the graph intersects the x-axis.

Graph does not intersect x-axis. No zeroes.
Graph intersects x-axis at 1 point. 1 zero.
Graph intersects x-axis at 3 points. 3 zeroes.
Graph intersects x-axis at 2 points. 2 zeroes.
Graph intersects x-axis at 4 points. 4 zeroes.
Graph intersects (touches) x-axis at 3 points. 3 zeroes.

Exercise 2.2

Q1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.

(i) x² - 2x - 8
x² - 4x + 2x - 8 = x(x - 4) + 2(x - 4) = (x + 2)(x - 4).
Zeroes: -2, 4.
Sum = (-2) + 4 = 2. -b/a = -(-2)/1 = 2. Verified.
Product = (-2)(4) = -8. c/a = -8/1 = -8. Verified.

(ii) 4s² - 4s + 1
(2s - 1)² = 0. Zeroes: 1/2, 1/2.
Sum = 1. -b/a = -(-4)/4 = 1. Verified.
Product = 1/4. c/a = 1/4. Verified.

(iii) 6x² - 3 - 7x
6x² - 7x - 3 = 6x² - 9x + 2x - 3 = 3x(2x - 3) + 1(2x - 3) = (3x + 1)(2x - 3).
Zeroes: -1/3, 3/2.
Sum = 7/6. -b/a = -(-7)/6 = 7/6. Verified.
Product = -1/2. c/a = -3/6 = -1/2. Verified.

(iv) 4u² + 8u
4u(u + 2) = 0. Zeroes: 0, -2.
Sum = -2. -b/a = -8/4 = -2. Verified.
Product = 0. c/a = 0/4 = 0. Verified.

(v) t² - 15
(t - √15)(t + √15) = 0. Zeroes: ±√15.
Sum = 0. -b/a = -0/1 = 0. Verified.
Product = -15. c/a = -15/1 = -15. Verified.

(vi) 3x² - x - 4
3x² - 4x + 3x - 4 = x(3x - 4) + 1(3x - 4) = (x + 1)(3x - 4).
Zeroes: -1, 4/3.
Sum = 1/3. -b/a = -(-1)/3 = 1/3. Verified.
Product = -4/3. c/a = -4/3. Verified.

Q2. Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.

Formula: x² - (Sum)x + Product

(i) 1/4, -1:
x² - (1/4)x + (-1) = x² - x/4 - 1.
Multiply by 4: 4x² - x - 4

(ii) √2, 1/3:
x² - √2x + 1/3.
Multiply by 3: 3x² - 3√2x + 1

(iii) 0, √5:
x² - 0x + √5 ⇒ x² + √5

(iv) 1, 1:
x² - x + 1

(v) -1/4, 1/4:
x² - (-1/4)x + 1/4 = x² + x/4 + 1/4.
Multiply by 4: 4x² + x + 1

(vi) 4, 1:
x² - 4x + 1

Polynomials - RD Sharma Important Questions

Q1. Find the zeroes of the polynomial 4x² - 3x - 1.

4x² - 4x + x - 1 = 4x(x - 1) + 1(x - 1) = (4x + 1)(x - 1).
Zeroes are x = -1/4, 1.

Q2. If α and β are the zeroes of the quadratic polynomial f(x) = x² - 5x + 4, find the value of 1/α + 1/β - 2αβ.

α + β = 5, αβ = 4.
1/α + 1/β = (α + β)/αβ = 5/4.
Value = 5/4 - 2(4) = 5/4 - 8 = (5 - 32)/4 = -27/4.

Q3. If the squared difference of the zeroes of the quadratic polynomial f(x) = x² + px + 45 is equal to 144, find the value of p.

(α - β)² = 144.
(α + β)² - 4αβ = 144.
(-p)² - 4(45) = 144.
p² - 180 = 144 ⇒ p² = 324 ⇒ p = ±18.

Q4. Find a quadratic polynomial whose zeroes are reciprocals of the zeroes of the polynomial f(x) = ax² + bx + c.

Original zeroes α, β. New zeroes 1/α, 1/β.
Sum = (α+β)/αβ = (-b/a) / (c/a) = -b/c.
Product = 1/αβ = 1/(c/a) = a/c.
Poly: k[x² - (-b/c)x + a/c] = k[cx² + bx + a].

Q5. If one zero of the polynomial (a² + 9)x² + 13x + 6a is reciprocal of the other, find the value of a.

Let zeroes be α and 1/α.
Product = α(1/α) = 1.
c/A = 1 ⇒ 6a / (a² + 9) = 1.
a² - 6a + 9 = 0 ⇒ (a-3)² = 0 ⇒ a = 3.

Q6. If the sum of the zeroes of the polynomial f(x) = (k² - 14)x² - 2x - 12 is 1, find k.

Sum = -(-2) / (k² - 14) = 1.
2 = k² - 14 ⇒ k² = 16 ⇒ k = ±4.

Q7. If α, β are zeroes of x² - p(x + 1) - c, show that (α + 1)(β + 1) = 1 - c.

Poly: x² - px - (p + c).
α + β = p, αβ = -(p + c).
LHS = αβ + α + β + 1.
= -(p + c) + p + 1 = -p - c + p + 1 = 1 - c. Proved.

Q8. Find a quadratic polynomial, the sum and product of whose zeroes are √2 and -3/2 respectively. Also find its zeroes.

Poly: k[x² - √2x - 3/2]. Let k=2.
2x² - 2√2x - 3.
Finding zeroes: D = (-2√2)² - 4(2)(-3) = 8 + 24 = 32.
x = (2√2 ± 4√2) / 4.
x₁ = 6√2 / 4 = 3√2 / 2.
x₂ = -2√2 / 4 = -√2 / 2.

Q9. If α and β are the zeroes of the quadratic polynomial f(x) = x² - 1, find a quadratic polynomial whose zeroes are 2α/β and 2β/α.

Zeroes of x² - 1 are 1, -1. (So α=1, β=-1).
New zeroes: 2(1)/(-1) = -2 and 2(-1)/1 = -2.
Sum = -4, Product = 4.
Poly: x² + 4x + 4.

Q10. If the sum of the squares of zeroes of the polynomial 6x² + x + k is 25/36, find the value of k.

α + β = -1/6, αβ = k/6.
α² + β² = (α + β)² - 2αβ = 25/36.
1/36 - 2k/6 = 25/36.
1 - 12k = 25 ⇒ -12k = 24 ⇒ k = -2.

Q11. If one zero of 2x² + 3x + k is 1/2, find value of k.

Put x = 1/2. 2(1/4) + 3(1/2) + k = 0.
1/2 + 3/2 + k = 0 ⇒ 2 + k = 0 ⇒ k = -2.

Q12. If α, β are zeroes of x² + x - 2, find value of 1/α - 1/β.

Zeroes are (x+2)(x-1) ⇒ -2, 1.
Case 1: 1/(-2) - 1/1 = -1/2 - 1 = -3/2.
Case 2: 1/1 - 1/(-2) = 1 + 1/2 = 3/2.

Q13. If zeroes of x² - lx + m differ by 1, then show that l² - 4m - 1 = 0.

(α - β)² = 1.
(α + β)² - 4αβ = 1.
l² - 4m = 1 ⇒ l² - 4m - 1 = 0.

Q14. Find a polynomial whose zeroes are 2α and 2β where α, β are zeroes of x² - 5x + 6.

Zeroes of x² - 5x + 6 are 2, 3.
New zeroes: 4, 6.
Poly: x² - (10)x + 24.

Q15. Find value of k such that 3x² + 2kx + x - k - 5 has sum of zeroes equal to half of their product.

Poly: 3x² + (2k+1)x - (k+5).
Sum = -(2k+1)/3. Product = -(k+5)/3.
Sum = 1/2 * Product.
-(2k+1)/3 = -(k+5)/6.
2(2k+1) = k+5 ⇒ 4k+2 = k+5 ⇒ 3k = 3 ⇒ k = 1.

Q16. If sum of squares of zeroes of kx² + 4x + 4 is 24, find k.

α + β = -4/k, αβ = 4/k.
16/k² - 8/k = 24.
2 - k = 3k² ⇒ 3k² + k - 2 = 0.
(3k - 2)(k + 1) = 0 ⇒ k = 2/3, -1.

Q17. If α, β are zeroes of x² - 6x + a, find value of 'a' if 3α + 2β = 20.

α + β = 6.
Multiply by 2: 2α + 2β = 12.
Subtract from given eq: (3α + 2β) - (2α + 2β) = 20 - 12.
α = 8.
Then β = -2.
a = αβ = -16.

Q18. Find a quadratic polynomial whose zeroes are (3 + √5)/5 and (3 - √5)/5.

Sum = 6/5.
Product = (9 - 5)/25 = 4/25.
Poly: x² - 6/5x + 4/25.
Multiply by 25: 25x² - 30x + 4.

Q19. Form a quadratic polynomial with zeroes 3 and -1/3.

(x - 3)(x + 1/3) = x² - 3x + x/3 - 1.
Multiply by 3: 3x² - 8x - 3.

Q20. If x+2 is a factor of x² + ax + 2b and a+b=4, find a and b.

P(-2) = 0 ⇒ 4 - 2a + 2b = 0 ⇒ a - b = 2.
Given a + b = 4.
Adding: 2a = 6 ⇒ a = 3.
b = 1.

Polynomials - Formulas & PYQs

Key Formulas & Concepts

1. Linear Polynomial

A polynomial of degree 1. General form: ax + b.

Zero = -b/a.

2. Quadratic Polynomial

A polynomial of degree 2. General form: ax² + bx + c.

Let α and β be the zeroes.

Sum of Zeroes (α + β): -b/a

Product of Zeroes (αβ): c/a

3. Cubic Polynomial (For References)

General form: ax³ + bx² + cx + d.

α + β + γ = -b/a

αβ + βγ + γα = c/a

αβγ = -d/a

4. Formation of Quadratic Polynomial

k[x² - (Sum of zeroes)x + (Product of zeroes)]

Previous Year Questions (CBSE/JKBOSE)

Q1. If α and β are the zeroes of the polynomial x² - 5x + 6, find the value of α² + β². (CBSE 2012)

α + β = 5, αβ = 6.
α² + β² = (α + β)² - 2αβ
= (5)² - 2(6) = 25 - 12 = 13.

Q2. Find the zeroes of the quadratic polynomial 6x² - 3 - 7x and verify the relationship between zeroes and coefficients. (CBSE 2008, 2016)

(Already solved in Notes.html Example).

Q3. If the sum of the zeroes of the polynomial (a+1)x² + (2a+3)x + (3a+4) is -1, find the product of its zeroes. (CBSE 2012)

Sum = -(2a+3)/(a+1) = -1
2a + 3 = a + 1 ⇒ a = -2.
Product = (3a+4)/(a+1) = (-6+4)/(-1) = 2.

Q4. Write a quadratic polynomial sum of whose zeroes is 2 and product is -8. (CBSE 2023)

x² - (Sum)x + (Product)
x² - 2x - 8.

Q5. If one zero of the polynomial (k² + 4)x² + 13x + 4k is reciprocal of the other, find k. (CBSE 2019)

Product = c/a = 1.
4k / (k² + 4) = 1.
k² - 4k + 4 = 0 ⇒ (k - 2)² = 0 ⇒ k = 2.

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