Please more explanation on this question under polynomial

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 $\boxed{10}$.

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: $r_1 \cdot r_2 = \frac{C}{A}$

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 $2\alpha$.

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:

$$\alpha \cdot 2\alpha = \frac{k}{1}$$ $$2\alpha^2 = k \quad (2)$$

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 ($ax^2+bx+c=0$):

$$2k^2 + 4k - 9k + 2 = 0$$ $$2k^2 - 5k + 2 = 0$$

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 $\boxed{\frac{1}{2} \text{ and } 2}$.