Let's work through the problem step by step.

Part (i):

You are given the polynomial $p(x) = ax^3 - x^2 + 4x - a$ and the information that $(2x - 1)$ is a factor of $p(x)$.

Since $(2x - 1)$ is a factor, it implies that when $x = \frac{1}{2}$, $p(x) = 0$. Let's substitute $x = \frac{1}{2}$ into the polynomial to find the value of $a$.

$p\left( \frac{1}{2} \right) = a\left( \frac{1}{2} \right)^3 - \left( \frac{1}{2} \right)^2 + 4\left( \frac{1}{2} \right) - a = 0$

Simplifying each term:

$p\left( \frac{1}{2} \right) = a\left( \frac{1}{8} \right) - \frac{1}{4} + 2 - a = 0$

Multiply out the terms involving $a$:

$\frac{a}{8} - \frac{1}{4} + 2 - a = 0$

Now, multiply through by 8 to eliminate the fractions:

$a - 2 + 16 - 8a = 0$

Simplifying:

$-7a + 14 = 0$

Solving for $a$:

$7a = 14 \implies a = 2$

Thus, the value of $a$ is $2$.

Now, substitute $a = 2$ into the original polynomial $p(x)$:

$p(x) = 2x^3 - x^2 + 4x - 2$

Factorization of $p(x)$:

Since $(2x - 1)$ is a known factor, we can perform synthetic or polynomial division of $p(x)$ by $2x - 1$.

Dividing $2x^3 - x^2 + 4x - 2$ by $2x - 1$ gives:

$p(x) = (2x - 1)(x^2 + 0x + 2)$

Thus, the factorization of $p(x)$ is:

$p(x) = (2x - 1)(x^2 + 2)$

Part (ii):

Now that $a = 2$, we need to express $\frac{8x - 13}{p(x)}$ in partial fractions. We are given that:

$\frac{8x - 13}{p(x)} = \frac{A}{2x - 1} + \frac{Bx + C}{x^2 + 2}$

Multiply both sides by $p(x) = (2x - 1)(x^2 + 2)$:

$8x - 13 = A(x^2 + 2) + (Bx + C)(2x - 1)$

Expand both sides:

$8x - 13 = A(x^2 + 2) + (Bx + C)(2x - 1)$ $8x - 13 = A(x^2 + 2) + (2Bx^2 - Bx + 2Cx - C)$ $8x - 13 = (A + 2B)x^2 + (-B + 2C)x + (2A - C)$

Now, equate the coefficients of $x^2$, $x$, and the constant terms:

• For $x^2$: $A + 2B = 0$
• For $x$: $-B + 2C = 8$
• For the constant: $2A - C = -13$

Solve this system of equations:

1. $A + 2B = 0$ implies $A = -2B$
2. Substitute $A = -2B$ into $2A - C = -13$: $2(-2B) - C = -13 \implies -4B - C = -13 \implies C = -4B + 13$
3. Substitute $C = -4B + 13$ into $-B + 2C = 8$: $-B + 2(-4B + 13) = 8 \implies -B - 8B + 26 = 8 \implies -9B = -18 \implies B = 2$
4. Substitute $B = 2$ into $A = -2B$: $A = -2(2) = -4$
5. Finally, substitute $B = 2$ into $C = -4B + 13$: $C = -4(2) + 13 = -8 + 13 = 5$

Thus, $A = -4$, $B = 2$, and $C = 5$.

The partial fraction decomposition is:

$\frac{8x - 13}{(2x - 1)(x^2 + 2)} = \frac{-4}{2x - 1} + \frac{2x + 5}{x^2 + 2}$

Would you like further clarifications on any part of the solution?

Related Questions:

1. How do you perform polynomial division by a binomial like $2x - 1$?
2. Why does $(2x - 1)$ being a factor imply $p(\frac{1}{2}) = 0$?
3. What is the method for solving a system of linear equations involving constants?
4. How can you identify the structure of partial fractions for different types of factors in the denominator?
5. What are the steps for expanding and simplifying products of binomials?

Tip:

When solving for unknown constants in partial fractions, always match coefficients of powers of $x$ on both sides of the equation to set up a system of equations.