We are given the polynomial $p(x) = x^3 - 2x^2 - 7x - 4$. Let's factor this polynomial and address the questions step by step.

Factorization of the polynomial

To factor the cubic polynomial $p(x) = x^3 - 2x^2 - 7x - 4$, we can apply the Rational Root Theorem or synthetic division to find the roots. Let's find a potential rational root by testing factors of the constant term $-4$ and the leading coefficient $1$.

1. Test $x = 1$: $p(1) = 1^3 - 2(1^2) - 7(1) - 4 = 1 - 2 - 7 - 4 = -12$ So, $x = 1$ is not a root.

2. Test $x = -1$: $p(-1) = (-1)^3 - 2(-1)^2 - 7(-1) - 4 = -1 - 2 + 7 - 4 = 0$ So, $x = -1$ is a root.

Now that we know $x + 1$ is a factor, we can perform synthetic division to divide $p(x)$ by $x + 1$.

After performing synthetic division, we get: $p(x) = (x + 1)(x^2 - 3x - 4)$ Now, we can factor the quadratic part: $x^2 - 3x - 4 = (x - 4)(x + 1)$ So the complete factorization of $p(x)$ is: $p(x) = (x + 1)^2(x - 4)$

Answering the questions:

• a) Number of prime factors: The prime factors are $(x + 1)$ and $(x - 4)$, so there are 2 distinct prime factors.

• b) Factor with the highest repetition: The factor $(x + 1)$ appears twice, so it has the highest repetition.

• c) Quadratic prime factor: There are no irreducible quadratic factors, so the correct answer here would be none.

• d) Sum of prime factors: The prime factors are $x + 1$ and $x - 4$. Their sum is: $(x + 1) + (x - 4) = 2x - 3$

Do you want further details or have any questions?

Here are 5 questions based on the above explanation:

1. How can the Rational Root Theorem help in factorizing cubic polynomials?
2. What other methods can be used to factor polynomials when synthetic division is not easy?
3. Why is synthetic division useful for finding polynomial roots?
4. How can you check if a quadratic factor is irreducible?
5. Can a polynomial have both real and complex roots when factored?

Tip: When factoring polynomials, always look for common factors first before applying complex methods.