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Here's a more detailed explanation for question 1, covering both parts. 1. (i) Understanding the Factor Theorem The Factor Theorem is a fundamental concept in algebra that links factors of a polynomial to its roots. It states: If (x-c) is a factor of a polynomial f(x), then f(c) = 0. This means that when you substitute c into the polynomial, the result is zero. Conversely, if f(c) = 0, then (x-c) is a factor of f(x). Step 1: Use the Factor Theorem to find a. We are given the polynomial f(x) = ax^4 + x^3 - 12x^2 - x + 2. We are told that (x-1) is a factor of f(x). According to the Factor Theorem, if (x-1) is a factor, then f(1) must be equal to 0. So, we substitute x=1 into the polynomial f(x): f(1) = a(1)^4 + (1)^3 - 12(1)^2 - (1) + 2 Now, we set this expression equal to 0: a(1) + 1 - 12(1) - 1 + 2 = 0 a + 1 - 12 - 1 + 2 = 0 Combine the constant terms: a + (1 - 12 - 1 + 2) = 0 a + (-11 - 1 + 2) = 0 a + (-12 + 2) = 0 a - 10 = 0 To solve for a, add 10 to both sides: a = 10 The value of the constant a is 10. Step 2: Verify that f(-1) = 0. Now that we know a=10, we can write the complete polynomial: f(x) = 10x^4 + x^3 - 12x^2 - x + 2 To verify that f(-1) = 0, we substitute x=-1 into this polynomial: f(-1) = 10(-1)^4 + (-1)^3 - 12(-1)^2 - (-1) + 2 Remember that: (-1)^even power = 1 (-1)^odd power = -1 So, we have: f(-1) = 10(1) + (-1) - 12(1) - (-1) + 2 f(-1) = 10 - 1 - 12 + 1 + 2 Now, combine the terms: f(-1) = (10 - 1) - 12 + 1 + 2 f(-1) = 9 - 12 + 1 + 2 f(-1) = -3 + 1 + 2 f(-1) = -2 + 2 f(-1) = 0 Since f(-1) = 0, this verifies the statement. This also means that (x-(-1)), or (x+1), is another factor of f(x). 1. (ii) Understanding Vieta's Formulas Vieta's formulas provide a relationship between the roots of a polynomial and its coefficients. For a quadratic equation of the form Ax^2 + Bx + C = 0, if the roots are r_1 and r_2, then: Sum of roots: r_1 + r_2 = -(B)/(A) Product of roots: r_1 · r_2 = (C)/(A) Step 1: Define the roots and identify coefficients. The given quadratic equation is x^2 + (k+1)x + k = 0. Comparing this to the standard form Ax^2 + Bx + C = 0: A = 1 (coefficient of x^2) B = (k+1) (coefficient of x) C = k (constant term) We are told that the roots of this equation are and 2. Step 2: Apply Vieta's formulas to set up equations. Using the sum of roots formula: + 2 = -((k+1))/(1) 3 = -(k+1) (1) Using the product of roots formula: · 2 = (k)/(1) 2^2 = k (2) Step 3: Solve the system of equations for k. We have a system of two equations with two unknowns ( and k). We can solve this by substitution. From equation (1), we can express in terms of k: 3 = -(k+1) = -(k+1)/(3) Now, substitute this expression for into equation (2): 2(-(k+1)/(3))^2 = k Square the term inside the parenthesis: 2((k+1)^2)/(3^2) = k 2((k^2 + 2k + 1))/(9) = k Multiply both sides by 9 to eliminate the denominator: 2(k^2 + 2k + 1) = 9k Distribute the 2 on the left side: 2k^2 + 4k + 2 = 9k Rearrange the equation to form a standard quadratic equation (ax^2+bx+c=0): 2k^2 + 4k - 9k + 2 = 0 2k^2 - 5k + 2 = 0 Step 4: Solve the quadratic equation for k. We can solve this quadratic equation by factoring, using the quadratic formula, or completing the square. Let's use factoring. We need two numbers that multiply to (2)(2)=4 and add to -5. These numbers are -4 and -1. Rewrite the middle term: 2k^2 - 4k - k + 2 = 0 Factor by grouping: 2k(k - 2) - 1(k - 2) = 0 (2k - 1)(k - 2) = 0 Set each factor equal to zero to find the possible values of k: 2k - 1 = 0 2k = 1 k = (1)/(2) k - 2 = 0 k = 2 The values of the constant k are (1)/(2) and 2.