April 10, 2026
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Chapter 2: Polynomials
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Exercise 2.1 (Page 18)
11 mark each
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.
Answer:
(i) The number of zeroes is 1.
(ii) The number of zeroes is 2.
(iii) The number of zeroes is 3.
(iv) The number of zeroes is 1.
(v) The number of zeroes is 1.
(vi) The number of zeroes is 4.
(i) The number of zeroes is 1.
(ii) The number of zeroes is 2.
(iii) The number of zeroes is 3.
(iv) The number of zeroes is 1.
(v) The number of zeroes is 1.
(vi) The number of zeroes is 4.
Exercise 2.2 (Page 23)
Q1 (i)3 marks
Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.
(i) x² – 2x – 8
(i) x² – 2x – 8
Answer:
x² – 2x – 8 = (x – 4)(x + 2)
Zeroes are: 4 and -2.
Verification:
Sum of zeroes = 4 + (-2) = 2 = -(-2)/1 = -(coefficient of x)/coefficient of x²
Product of zeroes = 4 × (-2) = -8 = (-8)/1 = constant term/coefficient of x²
x² – 2x – 8 = (x – 4)(x + 2)
Zeroes are: 4 and -2.
Verification:
Sum of zeroes = 4 + (-2) = 2 = -(-2)/1 = -(coefficient of x)/coefficient of x²
Product of zeroes = 4 × (-2) = -8 = (-8)/1 = constant term/coefficient of x²
Q1 (ii)3 marks
(ii) 4s² – 4s + 1
Answer:
4s² – 4s + 1 = (2s – 1)²
Zeroes are: ½ and ½.
Verification:
Sum = ½ + ½ = 1 = -(-4)/4 = -(coefficient of s)/coefficient of s²
Product = ½ × ½ = ¼ = 1/4 = constant term/coefficient of s²
4s² – 4s + 1 = (2s – 1)²
Zeroes are: ½ and ½.
Verification:
Sum = ½ + ½ = 1 = -(-4)/4 = -(coefficient of s)/coefficient of s²
Product = ½ × ½ = ¼ = 1/4 = constant term/coefficient of s²
Q1 (iii)3 marks
(iii) 6x² – 3 – 7x
Answer:
Rewrite as 6x² – 7x – 3 = (3x + 1)(2x – 3)
Zeroes are: -1/3 and 3/2.
Verification:
Sum = -1/3 + 3/2 = 7/6 = -(-7)/6 = -(coefficient of x)/coefficient of x²
Product = (-1/3)(3/2) = -1/2 = -3/6 = constant term/coefficient of x²
Rewrite as 6x² – 7x – 3 = (3x + 1)(2x – 3)
Zeroes are: -1/3 and 3/2.
Verification:
Sum = -1/3 + 3/2 = 7/6 = -(-7)/6 = -(coefficient of x)/coefficient of x²
Product = (-1/3)(3/2) = -1/2 = -3/6 = constant term/coefficient of x²
Q1 (iv)2 marks
(iv) 4u² + 8u
Answer:
4u² + 8u = 4u(u + 2)
Zeroes are: 0 and -2.
Verification:
Sum = 0 + (-2) = -2 = -8/4 = -(coefficient of u)/coefficient of u²
Product = 0 × (-2) = 0 = 0/4 = constant term/coefficient of u²
4u² + 8u = 4u(u + 2)
Zeroes are: 0 and -2.
Verification:
Sum = 0 + (-2) = -2 = -8/4 = -(coefficient of u)/coefficient of u²
Product = 0 × (-2) = 0 = 0/4 = constant term/coefficient of u²
Q1 (v)2 marks
(v) t² – 15
Answer:
t² – 15 = (t – √15)(t + √15)
Zeroes are: √15 and -√15.
Verification:
Sum = √15 + (-√15) = 0 = -0/1 = -(coefficient of t)/coefficient of t²
Product = (√15)(-√15) = -15 = -15/1 = constant term/coefficient of t²
t² – 15 = (t – √15)(t + √15)
Zeroes are: √15 and -√15.
Verification:
Sum = √15 + (-√15) = 0 = -0/1 = -(coefficient of t)/coefficient of t²
Product = (√15)(-√15) = -15 = -15/1 = constant term/coefficient of t²
Q1 (vi)3 marks
(vi) 3x² – x – 4
Answer:
3x² – x – 4 = (3x – 4)(x + 1)
Zeroes are: 4/3 and -1.
Verification:
Sum = 4/3 + (-1) = 1/3 = -(-1)/3 = -(coefficient of x)/coefficient of x²
Product = (4/3)(-1) = -4/3 = -4/3 = constant term/coefficient of x²
3x² – x – 4 = (3x – 4)(x + 1)
Zeroes are: 4/3 and -1.
Verification:
Sum = 4/3 + (-1) = 1/3 = -(-1)/3 = -(coefficient of x)/coefficient of x²
Product = (4/3)(-1) = -4/3 = -4/3 = constant term/coefficient of x²
Q2 (i)2 marks
Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.
(i) 1/4, -1
(i) 1/4, -1
Answer:
Let α and β be the zeroes. Then α + β = 1/4, αβ = -1.
Quadratic polynomial = x² – (α + β)x + αβ = x² – (1/4)x + (-1)
= x² – x/4 – 1
Multiplying by 4: 4x² – x – 4.
Let α and β be the zeroes. Then α + β = 1/4, αβ = -1.
Quadratic polynomial = x² – (α + β)x + αβ = x² – (1/4)x + (-1)
= x² – x/4 – 1
Multiplying by 4: 4x² – x – 4.
Q2 (ii)2 marks
(ii) √2, 1/3
Answer:
α + β = √2, αβ = 1/3.
Polynomial = x² – √2 x + 1/3.
Multiplying by 3: 3x² – 3√2 x + 1.
α + β = √2, αβ = 1/3.
Polynomial = x² – √2 x + 1/3.
Multiplying by 3: 3x² – 3√2 x + 1.
Q2 (iii)2 marks
(iii) 0, √5
Answer:
α + β = 0, αβ = √5.
Polynomial = x² + √5.
α + β = 0, αβ = √5.
Polynomial = x² + √5.
Q2 (iv)2 marks
(iv) 1, 1
Answer:
α + β = 1, αβ = 1.
Polynomial = x² – x + 1.
α + β = 1, αβ = 1.
Polynomial = x² – x + 1.
Q2 (v)2 marks
(v) -1/4, 1/4
Answer:
α + β = -1/4, αβ = 1/4.
Polynomial = x² – (-1/4)x + 1/4 = x² + x/4 + 1/4.
Multiplying by 4: 4x² + x + 1.
α + β = -1/4, αβ = 1/4.
Polynomial = x² – (-1/4)x + 1/4 = x² + x/4 + 1/4.
Multiplying by 4: 4x² + x + 1.
Q2 (vi)2 marks
(vi) 4, 1
Answer:
α + β = 4, αβ = 1.
Polynomial = x² – 4x + 1.
α + β = 4, αβ = 1.
Polynomial = x² – 4x + 1.
Key Concepts Summary
Note 1Degree & Zeroes
What is the relationship between the degree of a polynomial and its number of zeroes?
Answer:
A polynomial of degree n can have at most n zeroes. This is because the graph of y = p(x) intersects the x-axis at most n points.
A polynomial of degree n can have at most n zeroes. This is because the graph of y = p(x) intersects the x-axis at most n points.
Note 2Quadratic Polynomial
If α and β are the zeroes of the quadratic polynomial ax² + bx + c, what are the formulas for sum and product of zeroes?
Answer:
Sum of zeroes = α + β = –b/a = –(coefficient of x)/coefficient of x²
Product of zeroes = αβ = c/a = constant term/coefficient of x²
Sum of zeroes = α + β = –b/a = –(coefficient of x)/coefficient of x²
Product of zeroes = αβ = c/a = constant term/coefficient of x²
Note 3Cubic Polynomial
If α, β, γ are the zeroes of the cubic polynomial ax³ + bx² + cx + d, what are the relationships between zeroes and coefficients?
Answer:
α + β + γ = –b/a
αβ + βγ + γα = c/a
αβγ = –d/a
α + β + γ = –b/a
αβ + βγ + γα = c/a
αβγ = –d/a
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