# Factoring the Polynomials ## (a) \( p(z) = z^6 + 2z^5(2 + i) + 3 + 4i \) Let's write the polynomial fully: \[ p(z) = z^6 + [2(2+i)]z^5 + (3 + 4i) \] \[ = z^6 + (4 + 2i)z^5 + (3 + 4i) \] ### Factoring Attempt This is a degree 6 polynomial with a constant term \(3 + 4i\) and a \(z^5\) term. There is no \(z^4, z^3, z^2\), or \(z\) term. #### Rational Root Theorem The Rational Root Theorem is not helpful here due to complex coefficients. Let us factor out \(z^5\) from the first two terms: \[ = z^5(z + 4 + 2i) + (3 + 4i) \] There is no common factor in all terms, and the constant term prevents factoring out \(z\) directly. Let's attempt factoring by grouping or substitution. #### Substitution Let’s check if \(z = a\) is a root (try small integers and simple complex numbers): - For \(z = \): \(p() = 3 + 4i\) (not zero) - For \(z = - (4+2i)\): \[ p(-(4+2i)) = [-(4+2i)]^6 + (4 + 2i)[-(4+2i)]^5 + (3+4i) \] This is complex and not promising without further context. #### Conclusion **This polynomial does not factor nicely over the rationals or with simple complex roots. Unless instructed otherwise or unless there is a typo, it cannot be factored further using elementary methods.** **Final Factored Form:** \[ \boxed{p(z) = z^6 + (4 + 2i)z^5 + (3 + 4i)} \] If you have a correction or more context, please provide it. --- ## (b) \( q(z) = z^6 - 64 \) This is a difference of two sixth powers. ### Step 1: Recognize as a Difference of Squares \[ z^6 - 64 = (z^3)^2 - 8^2 = (z^3 - 8)(z^3 + 8) \] ### Step 2: Factor Each Cubic (Sum and Difference of Cubes) - \(z^3 - 8 = (z - 2)(z^2 + 2z + 4)\) - \(z^3 + 8 = (z + 2)(z^2 - 2z + 4)\) ### Step 3: Write the Complete Factorization \[ q(z) = (z^3 - 8)(z^3 + 8) \] \[ = (z - 2)(z^2 + 2z + 4)(z + 2)(z^2 - 2z + 4) \] ### Step 4: Optional - Factor Quadratics Further Quadratics can be factored over the complex numbers. - \(z^2 + 2z + 4 = (z + 1 - i\sqrt{3})(z + 1 + i\sqrt{3})\) - \(z^2 - 2z + 4 = (z - 1 - i\sqrt{3})(z - 1 + i\sqrt{3})\) So, the full factorization over complex numbers: \[ \boxed{ q(z) = (z - 2)(z + 2)(z + 1 - i\sqrt{3})(z + 1 + i\sqrt{3})(z - 1 - i\sqrt{3})(z - 1 + i\sqrt{3}) } \] --- ## Summary Table | Polynomial | Complete Factorization | |------------|-----------------------| | \(p(z)\) | Not easily factorable over rationals/complex (see above) | | \(q(z)\) | \((z - 2)(z + 2)(z + 1 - i\sqrt{3})(z + 1 + i\sqrt{3})(z - 1 - i\sqrt{3})(z - 1 + i\sqrt{3})\) |