n this question, we will use techniques involving complex numbers to study the sexticpolynomial defined byP (z) = z6 − 4z3 + 8.(a) (1 mark) Use the quadratic formula to find the set of possible values of z3 such thatP (z) = 0. (Note that we are not trying to find the values of z yet - only the values ofz3.)(b) (2 marks) In the WebWork part of this assignment, you found the three solutions ofz3 = 2 − 2i in exponential polar form.Use a similar method to find the three solutions of z3 = 2 + 2i. Express your answerin exponential polar form.2(c) (1 mark) Use the results of the previous parts to find the six values of z such thatP (z) = 0. Express your answers in exponential polar form.(d) (3 marks) Use your answers from part (c) to factor the polynomial P (z) into threequadratics with real coefficients. (You may use the facts that cos( π12 ) = √6+√24 andsin( π12 ) = √6−√24 ).Before you submit this work, make sure to check your answer. If you multiply outyour quadratic factors, you should recover P (z).

This is a complex problem, but let's break it down step by step:

(a) We can rewrite the equation P(z) = z^6 - 4z^3 + 8 = 0 as a quadratic equation in terms of z^3. Let's set u = z^3. Then we have u^2 - 4u + 8 = 0. Using the quadratic formula, we find the roots of this equation to be u = 2 ± 2i.

(b) To find the roots of the equation z^3 = 2 + 2i, we first convert the right side to polar form. The magnitude is sqrt(2^2 + 2^2) = 2sqrt(2), and the argument is arctan(2/2) = π/4. So, 2 + 2i = 2sqrt(2)(cos(π/4) + isin(π/4)). By De Moivre's theorem, the three roots of this equation are z = sqrt[2]{2sqrt(2)}(cos((π/4 + 2πk)/3) + isin((π/4 + 2πk)/3)), where k = 0, 1, 2.

(c) Similarly, the roots of the equation z^3 = 2 - 2i are z = sqrt[2]{2sqrt(2)}(cos((7π/4 + 2πk)/3) + isin((7π/4 + 2πk)/3)), where k = 0, 1, 2. So, the six roots of the original equation are these six values of z.

(d) The polynomial P(z) can be factored as P(z) = (z - z1)(z - z2)(z - z3)(z - z4)(z - z5)(z - z6), where z1, z2, z3, z4, z5, z6 are the six roots found in part (c). To express this as a product of three quadratics, we can pair up the roots in conjugate pairs. Since the coefficients of the polynomial are real, if a complex number is a root, then its conjugate must also be a root. So, we can write P(z) = (z - z1)(z - z1*)(z - z2)(z - z2*)(z - z3)(z - z3*), where z1*, z2*, z3* are the conjugates of z1, z2, z3. This is a product of three quadratics with real coefficients.