Let's solve the problems shown in the image step by step.

Problem 12:

(a)
The polynomial given is $8x^3 + 10x^2 - px - 5$ and it is divisible by $2x + 1$, meaning the remainder is zero.

We can use synthetic division or substitution via the Remainder Theorem, which states that if a polynomial $f(x)$ is divisible by $(2x + 1)$, then substituting the value of $x$ that makes $2x + 1 = 0$ into the polynomial should result in zero. Let's solve:

1. $2x + 1 = 0 \Rightarrow x = -\frac{1}{2}$
2. Substitute $x = -\frac{1}{2}$ into $f(x) = 8x^3 + 10x^2 - px - 5$: $8\left(-\frac{1}{2}\right)^3 + 10\left(-\frac{1}{2}\right)^2 - p\left(-\frac{1}{2}\right) - 5 = 0$ Simplifying: $8\left(-\frac{1}{8}\right) + 10\left(\frac{1}{4}\right) + \frac{p}{2} - 5 = 0$ $-1 + 2.5 + \frac{p}{2} - 5 = 0$ $-3.5 + \frac{p}{2} = 0$ $\frac{p}{2} = 3.5 \Rightarrow p = 7$ So, the value of $p$ is 7.

(b)
The polynomial given is $x^6 + x^4 - 2x^2 + k$, and when it is divided by $1 + x^2$, the remainder is 5.

We can write this division as: $x^6 + x^4 - 2x^2 + k = (1 + x^2) \cdot Q(x) + 5$ Where $Q(x)$ is the quotient and 5 is the remainder.

Since we are looking for the value of $k$, the polynomial should give the remainder 5 when evaluated at $x = i$ (where $i^2 = -1$) because $1 + x^2 = 0$ when $x = i$. Let's substitute $x = i$ into the polynomial: $i^6 + i^4 - 2i^2 + k = 5$ Now simplify: $i^6 = -1, \quad i^4 = 1, \quad i^2 = -1$ Substitute these values: $-1 + 1 - 2(-1) + k = 5$ $-1 + 1 + 2 + k = 5$ $k + 2 = 5 \Rightarrow k = 3$ So, the value of $k$ is 3.

Problem 13:

The polynomial is $x^3 + px^2 - x - 2$ and it has $x - 1$ as a factor. Since $x - 1$ is a factor, we can use the Factor Theorem, which states that if $x - 1$ is a factor, then substituting $x = 1$ into the polynomial should result in zero.

Substitute $x = 1$: $1^3 + p(1^2) - 1 - 2 = 0$ Simplifying: $1 + p - 1 - 2 = 0$ $p - 2 = 0 \Rightarrow p = 2$ So, the value of $p$ is 2.

Would you like more detailed explanations or any clarifications? Here are five related questions that might help deepen your understanding:

1. How is synthetic division used to solve polynomials divisible by linear factors?
2. What is the Remainder Theorem, and how is it applied to polynomial division?
3. Can you solve other polynomial division problems using substitution, as shown in Problem 12(b)?
4. How does the Factor Theorem simplify finding roots of polynomials?
5. How can we use complex numbers like $i$ to solve polynomial remainder problems?

Tip: When using the Remainder or Factor Theorems, always ensure the divisor is in the form $ax + b$, and substitute $x = -\frac{b}{a}$ into the polynomial.