Hi! Just wondering if any one can help me out with this roots question?

Question

(i). Factorise $z^2 - 5z + 6$ and hence, solve the equation $ z^2 - 5z + 6 = 0$

(ii). Show that $z^2 - 5z + 6$ is a factor of $z^3 + (-4 + i)z^2 + (1 - 5i)z + 6(1 + i)$.

(iii). Find the three roots of the equation $z^3 + (-4 + i)z^2 + (1 - 5i)z + 6(1 + i)$

I'm ok with the first two parts of the question i'm just a bit burned out on the third any help be much appreciated

Answer 1

Hint. One may use $$ z^3 + (-4 + i)z^2 + (1 - 5i)z + 6(1 + i)=(z^2-5z+6)(z+i+1) $$ found in $(\text{ii})$ then one may use $$ z^2-5z+6=(z-3)(z-2). $$

Answer 2

The sum of the roots of the cubic is $4-i$. Alternately, the product of the roots of the cubic is $-6(1+i)$. You know two of the roots, so you know the third.

Answer 3

(iii) :- $$z^3 + (-4 + i)z^2 + (1 - 5i)z + 6(1 + i)$$ $$= z^3 + -4z^2 + iz^2 + 1z - 5iz + 6 + 6i$$ $$= z^3 -4z^2 + z + 6 + i(z^2 - 5z + 6 )$$ By trial and error: $ z^3 -4z^2 + z + 6 = (z -2)(z^2 -2z - 3)$ $$(z -2)(z^2 -2z - 3) + i(z-3)(z -2)$$ $$= (z -2)(z^2 -2z - 3) + i(z-3)(z -2)$$ $$= (z -2)(z + 1)(z - 3) + i(z-3)(z -2)$$ $$= (z -2)(z -3)(z + 1 + i)$$