Chapter 2: Polynomials
Exercise & In-Text Questions & Answers

Answer & Explanation:
Based on the geometrical meaning of zeroes of a polynomial (Section 2.2, Page 7-9), the number of zeroes is equal to the number of times the graph intersects the x-axis.

(i) The graph does not intersect the x-axis → 0 zeroes
(ii) The graph intersects the x-axis at one point → 1 zero
(iii) The graph intersects the x-axis at three points → 3 zeroes
(iv) The graph intersects the x-axis at two points → 2 zeroes
(v) The graph intersects the x-axis at four points → 4 zeroes
(vi) The graph intersects the x-axis at three points → 3 zeroes

Reference: Section 2.2 "Geometrical Meaning of the Zeroes of a Polynomial", Page 7-9

Answer & Explanation:
For a quadratic polynomial ax² + bx + c, if α and β are zeroes, then:
α + β = -b/a (Sum of zeroes)
αβ = c/a (Product of zeroes)
Reference: Section 2.3, Page 10-11

(i) x² - 2x - 8
Factorizing: x² - 2x - 8 = (x - 4)(x + 2)
Zeroes: x = 4, -2
Sum: 4 + (-2) = 2 = -(-2)/1 ✓
Product: 4 × (-2) = -8 = -8/1 ✓

(ii) 4s² - 4s + 1
Factorizing: 4s² - 4s + 1 = (2s - 1)²
Zeroes: s = 1/2, 1/2
Sum: 1/2 + 1/2 = 1 = -(-4)/4 ✓
Product: (1/2) × (1/2) = 1/4 = 1/4 ✓

(iii) 6x² - 3 - 7x
Rewriting: 6x² - 7x - 3
Factorizing: 6x² - 7x - 3 = (2x - 3)(3x + 1)
Zeroes: x = 3/2, -1/3
Sum: 3/2 + (-1/3) = 7/6 = -(-7)/6 ✓
Product: (3/2) × (-1/3) = -1/2 = -3/6 ✓

(iv) 4u² + 8u
Factorizing: 4u² + 8u = 4u(u + 2)
Zeroes: u = 0, -2
Sum: 0 + (-2) = -2 = -8/4 ✓
Product: 0 × (-2) = 0 = 0/4 ✓

(v) t² - 15
Factorizing: t² - 15 = (t - √15)(t + √15)
Zeroes: t = √15, -√15
Sum: √15 + (-√15) = 0 = -0/1 ✓
Product: (√15) × (-√15) = -15 = -15/1 ✓

(vi) 3x² - x - 4
Factorizing: 3x² - x - 4 = (3x - 4)(x + 1)
Zeroes: x = 4/3, -1
Sum: 4/3 + (-1) = 1/3 = -(-1)/3 ✓
Product: (4/3) × (-1) = -4/3 = -4/3 ✓

Reference: Section 2.3 "Relationship between Zeroes and Coefficients of a Polynomial", Page 10-13

Answer & Explanation:
For a quadratic polynomial with sum of zeroes = S and product of zeroes = P, the polynomial is:
x² - Sx + P
Reference: Example 4, Page 12

(i) 1/4, -1
Polynomial: x² - (1/4)x + (-1) = x² - x/4 - 1
Multiplying by 4: 4x² - x - 4

(ii) √2, 1/3
Polynomial: x² - √2x + 1/3
Multiplying by 3: 3x² - 3√2x + 1

(iii) 0, √5
Polynomial: x² - 0x + √5 = x² + √5

(iv) 1, 1
Polynomial: x² - 1x + 1 = x² - x + 1

(v) -1/4, 1/4
Polynomial: x² - (-1/4)x + 1/4 = x² + x/4 + 1/4
Multiplying by 4: 4x² + x + 1

(vi) 4, 1
Polynomial: x² - 4x + 1 = x² - 4x + 1

Reference: Section 2.3, Page 12-13

Answer & Explanation:
p(-1) = (-1)² - 3(-1) - 4 = 1 + 3 - 4 = 0
Since p(-1) = 0, -1 is a zero of the polynomial.
Reference: Section 2.1, Page 5

Answer & Explanation:
The zero of the linear polynomial ax + b is -b/a.
This is obtained by solving ax + b = 0 → x = -b/a.
Reference: Section 2.1, Page 5

Answer & Explanation:
The three cases for the graph of y = ax² + bx + c (a ≠ 0) are:
Case (i): The graph cuts x-axis at two distinct points → Two distinct zeroes
Case (ii): The graph touches x-axis at exactly one point → One zero (repeated)
Case (iii): The graph does not intersect x-axis → No zeroes

Reference: Section 2.2, Page 8-9

Answer & Explanation:
For a cubic polynomial ax³ + bx² + cx + d with zeroes α, β, γ:
α + β + γ = -b/a
αβ + βγ + γα = c/a
αβγ = -d/a

Reference: Section 2.3, Page 14

Answer & Explanation:
Given p(x) = 3x³ - 5x² - 11x - 3
p(3) = 3(27) - 5(9) - 11(3) - 3 = 81 - 45 - 33 - 3 = 0 ✓
p(-1) = 3(-1) - 5(1) - 11(-1) - 3 = -3 - 5 + 11 - 3 = 0 ✓
p(-1/3) = 3(-1/27) - 5(1/9) - 11(-1/3) - 3 = -1/9 - 5/9 + 11/3 - 3 = (-6/9) + (33/9) - 27/9 = 0 ✓

Now verifying relationships:
Sum of zeroes: 3 + (-1) + (-1/3) = 5/3 = -(-5)/3 ✓
Sum of products: 3(-1) + (-1)(-1/3) + (-1/3)(3) = -3 + 1/3 - 1 = -11/3 = -11/3 ✓
Product of zeroes: 3(-1)(-1/3) = 1 = -(-3)/3 ✓

Reference: Example 5, Page 14-15

Answer & Explanation:
The values of y = x² - 3x - 4 for different values of x are:

x	-2	-1	0	1	2	3	4	5
y = x² - 3x - 4	6	0	-4	-6	-6	-4	0	6

Reference: Table 2.1 and Fig. 2.2, Page 7-8

Answer & Explanation:
The values of y = x³ - 4x for different values of x are:

x	-2	-1	0	1	2
y = x³ - 4x	0	3	0	-3	0

Reference: Table 2.2 and Fig. 2.6, Page 9