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 $\boxed{10}$. 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: $(-1)^{\text{even power}} = 1$ $(-1)^{\text{odd power}} = -1$ 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: $r_1 + r_2 = -\frac{B}{A}$ Product of roots: $r_1 \cdot r_2 = \frac{C}{A}$ 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 $\alpha$ and $2\alpha$. Step 2: Apply Vieta's formulas to set up equations. Using the sum of roots formula: $$\alpha + 2\alpha = -\frac{(k+1)}{1}$$ $$3\alpha = -(k+1) \quad (1)$$ Using the product of roots formula: $$\alpha \cdot 2\alpha = \frac{k}{1}$$ $$2\alpha^2 = k \quad (2)$$ Step 3: Solve the system of equations for $k$. We have a system of two equations with two unknowns ($\alpha$ and $k$). We can solve this by substitution. From equation (1), we can express $\alpha$ in terms of $k$: $$3\alpha = -(k+1)$$ $$\alpha = -\frac{k+1}{3}$$ Now, substitute this expression for $\alpha$ into equation (2): $$2\left(-\frac{k+1}{3}\right)^2 = k$$ Square the term inside the parenthesis: $$2\frac{(k+1)^2}{3^2} = k$$ $$2\frac{(k^2 + 2k + 1)}{9} = k$$ 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 $k$: $$2k - 1 = 0 \implies 2k = 1 \implies k = \frac{1}{2}$$ $$k - 2 = 0 \implies k = 2$$ The values of the constant $k$ are $\boxed{\frac{1}{2} \text{ and } 2}$.