Finding Zeroes: Factor x² + 5x + 6 x² + 5x + 6 = x² + 2x + 3x + 6 = x(x + 2) + 3(x + 2) = (x + 2)(x + 3) Setting equal to zero: (x + 2)(x + 3) = 0 x + 2 = 0 → x = -2 x + 3 = 0 → x = -3 Zeroes: α = -2, β = -3 Verification: For polynomial ax² + bx + c, we have a = 1, b = 5, c = 6 Sum of zeroes = α + β = -2 + (-3) = -5 From formula: Sum = -b/a = -5/1 = -5 Product of zeroes = αβ = (-2)(-3) = 6 From formula: Product = c/a = 6/1 = 6 Both relationships verified.Part (a): 2x² - 5x + 7 Compare with ax² + bx + c: a = 2, b = -5, c = 7 Discriminant D = b² - 4ac = (-5)² - 4(2)(7) = 25 - 56 = -31 Since D < 0, the polynomial has no real zeroes Since a = 2 > 0, the graph opens upward. Part (b): -3x² + 4x - 1 Compare with ax² + bx + c: a = -3, b = 4, c = -1 Discriminant D = b² - 4ac = (4)² - 4(-3)(-1) = 16 - 12 = 4 Since D > 0, the polynomial has two distinct real zeroes Since a = -3 < 0, the graph opens downward Part (c): x² - 6x + 9 Compare with ax² + bx + c: a = 1, b = -6, c = 9 Discriminant D = b² - 4ac = (-6)² - 4(1)(9) = 36 - 36 = 0 Since D = 0, the polynomial has one real zero (repeated root) Since a = 1 > 0, the graph opens upward.Part (a): Parabola opening upward with vertex above x-axis When a parabola opens upward and vertex is above x-axis The graph does not touch x-axis at any point Number of zeroes = 0 Part (b): Parabola opening downward with vertex above x-axis When a parabola opens downward and vertex is above x-axis The graph must cross the x-axis at two points Number of zeroes = 2 Part (c):Cubic curve intersecting x-axis at three points Each intersection point with x-axis represents a zero Number of zeroes = 3 Part (d): A line passing through x-axis, touching x-axis at one point When a line touches the x-axis at exactly one point Number of zeroes = 1 Part (e): Polynomial curve intersecting x-axis at four points Each intersection point with x-axis represents a zero Number of zeroes = 4 Part (f):Parabola opening downwards touching x-axis at one point When a parabola touches the x-axis at exactly one point This point is the vertex (minimum point) This represents a repeated root Number of zeroes = 1 (or 2 equal zeroes)Let the two zeroes be α and 1/α (since they are reciprocals) Product of zeroes = α × (1/α) = 1 For polynomial ax² + bx + c, product of zeroes = c/a Here a = k² + 4, b = 13, c = 4k Therefore: 4k/(k² + 4) = 1 Cross multiply: 4k = k² + 4 Rearrange: k² - 4k + 4 = 0 This is a perfect square: (k - 2)² = 0 Therefore: k - 2 = 0 Answer: k = 2Given zeroes: α = -1 and β = 2 If these are zeroes, then (x + 1) and (x - 2) are factors Combined factor: (x + 1)(x - 2) = x² - 2x + x - 2 = x² - x - 2 Now divide the polynomial by this factor: 2x³ - x² - 5x - 2 ÷ (x² - x - 2) Using polynomial division: 2x³ - x² - 5x - 2 = (x² - x - 2)(2x + 1) The quotient is (2x + 1) Setting 2x + 1 = 0: x = -1/2 All zeroes: -1, 2, and -1/2For polynomial x² - 2kx + (k² - 4) Compare with ax² + bx + c: a = 1, b = -2k, c = k² - 4 Calculate discriminant: D = b² - 4ac = (-2k)² - 4(1)(k² - 4) D = 4k² - 4k² + 16 = 16 Notice: D = 16 (constant, independent of k) Part (a): Two distinct real zeroes Condition: D > 0 Since D = 16 > 0 always For all real values of k Part (b): Two equal real zeroes Condition: D = 0 Since D = 16 ≠ 0 Answer: Not possible for any value of k Part (c): No real zeroes Condition: D < 0 Since D = 16 > 0 always Answer: Not possible for any value of kIdentifying coefficients: Polynomial: 2x³ - 5x² - 14x + 8 Compare with ax³ + bx² + cx + d a = 2, b = -5, c = -14, d = 8 Part (a): Finding α + β + γ Sum of zeroes = α + β + γ = -b/a α + β + γ = -(-5)/2 = 5/2 Answer: α + β + γ = 5/2 Part (b): Finding αβ + βγ + γα Sum of products of zeroes taken two at a time = αβ + βγ + γα = c/a αβ + βγ + γα = -14/2 = -7 Answer: αβ + βγ + γα = -7 Part (c): Finding αβγ Product of all zeroes = αβγ = -d/a αβγ = -8/2 = -4 Answer: αβγ = -4Given zeroes: α = (3-√3)/5 and β = (3+√3)/5 Step 1: Find sum of zeroes α + β = (3-√3)/5 + (3+√3)/5 = (3-√3 + 3+√3)/5 = (6)/5 = 6/5 Step 2: Find product of zeroes αβ = [(3-√3)/5] × [(3+√3)/5] = [(3-√3)(3+√3)]/25 Using identity (a-b)(a+b) = a² - b²: = [3² - (√3)²]/25 = [9 - 3]/25 = 6/25 Step 3: Form the polynomial For a quadratic polynomial with zeroes α and β: Polynomial = x² - (α + β)x + αβ = x² - (6/5)x + 6/25 Step 4: Clear the fractions (multiply by 25) 25x² - 25(6/5)x + 25(6/25) = 25x² - 30x + 6 Answer: 25x² - 30x + 6 Or in general form: k(25x² - 30x + 6) where k is any non-zero constantGiven zeroes: α = -2, β = -3, γ = -1 Method 1: Using factors If α, β, γ are zeroes, then (x - α), (x - β), (x - γ) are factors Factor 1: x - (-2) = x + 2 Factor 2: x - (-3) = x + 3 Factor 3: x - (-1) = x + 1 Forming the polynomial: p(x) = k(x + 2)(x + 3)(x + 1), where k is any non-zero constant Taking k = 1: p(x) = (x + 2)(x + 3)(x + 1) Step 1: Multiply first two factors (x + 2)(x + 3) = x² + 3x + 2x + 6 = x² + 5x + 6 Step 2: Multiply result with third factor p(x) = (x² + 5x + 6)(x + 1) = x²(x + 1) + 5x(x + 1) + 6(x + 1) = x³ + x² + 5x² + 5x + 6x + 6 = x³ + 6x² + 11x + 6 Answer: K(x³ + 6x² + 11x + 6) where K is any non-zero constant (Or any constant multiple: k(x³ + 6x² + 11x + 6) where k ≠ 0)Part (a): Finding k From polynomial x² - 5x + k, we have: Sum of zeroes: α + β = -(-5)/1 = 5 ... (equation 1) Given: α - β = 1 ... (equation 2) Adding equations (1) and (2): 2α = 6 → α = 3 Subtracting equation (2) from (1): 2β = 4 → β = 2 Product of zeroes: k = αβ = 3 × 2 = 6 Answer: k = 6 Part (b): Finding both zeroes From calculations from (a) bit α = 3, β = 2 Part (c): Direction of graph The polynomial becomes x² - 5x + 6 Coefficient of x² is a = 1 Since a > 0, the graph opens upward
