Quick questions on Polynomials and roots explained: H2 Further Mathematics

Many quantities reduce to the symmetric functions through standard identities. The most used:

If a polynomial has real coefficients and $a + bi$ (with $b \neq 0$) is a root, then its complex conjugate $a - bi$ is also a root. Complex roots of a real polynomial therefore come in conjugate pairs, so a real polynomial of odd degree has at least one real root.

To find the equation whose roots are a transformation of the originals (for example $\alpha + 1$, or $\dfrac{1}{\alpha}$), substitute. If $y = \alpha + 1$ then $\alpha = y - 1$, so substituting $x = y - 1$ into the original polynomial gives the new equation in $y$. Alternatively, recompute the symmetric functions of the new roots.

If $y = \alpha + 1$ then $\alpha = y - 1$; substitute $x = y - 1$, not $x = y + 1$.

For $x^2 - 5x + 6 = 0$ with roots $\alpha, \beta$, state $\alpha + \beta$ and $\alpha\beta$. [1 mark]

A real cubic has $1 - 2i$ as a root. State another root. [1 mark]

If $\alpha + \beta = 4$ and $\alpha\beta = 3$, find $\alpha^2 + \beta^2$. [2 marks]