Complex roots of cubic and quartic equationsEdexcel International A Level Further Maths: Revision notes

If a polynomial $\text{f}(z)$ has real coefficients and $z_1$ is a root, then so is its conjugate $z_1^*$. Complex roots therefore come in conjugate pairs. A root $p+q\mathrm{i}$ and its conjugate give the real quadratic factor $(z-p-q\mathrm{i})(z-p+q\mathrm{i})=(z-p)^2+q^2=z^2-2pz+p^2+q^2$. For the roots $1\pm2\mathrm{i}$ this is $z^2-2z+5$. Because non-real roots come in pairs, a cubic with real coefficients always has at least one real root.

To divide a cubic or quartic by a quadratic, use long division or compare coefficients. For $\text{g}(x)=x^4-6x^3+18x^2-30x+25$ with factor $x^2-2x+5$, write $\text{g}(x)=(x^2-2x+5)(x^2+px+q)$. Constant term: $5q=25$, so $q=5$. Coefficient of $x^3$: $p-2=-6$, so $p=-4$. The quotient is $x^2-4x+5$. Check the $x^2$ and $x$ terms: $5-2p+q=5+8+5=18$ and $5p-2q=-20-10=-30$. A cubic divides by a linear factor to leave a quadratic in the same way: $z^3-5z^2+17z-13=(z-1)(z^2-4z+13)$.

If one root of a real cubic is known, find the real quadratic factor and solve it. Example: $z^3+z^2-z+15=0$ has the root $1-2\mathrm{i}$. Its conjugate $1+2\mathrm{i}$ is also a root, so $z^2-2z+5$ is a factor. Writing $z^3+z^2-z+15=(z^2-2z+5)(z+c)$, the constant gives $5c=15$, so $c=3$ and the real root is $z=-3$. If instead a real root is given, such as $z=1$ for $z^3-5z^2+17z-13=0$, divide by $(z-1)$ and solve the quadratic $z^2-4z+13=0$ to get $z=2\pm3\mathrm{i}$.

A real quartic has either four real roots, two real and a conjugate pair, or two conjugate pairs. One complex root given: use its conjugate to form a real quadratic factor and divide. For $\text{g}(x)$ above with root $1+2\mathrm{i}$, the other factor $x^2-4x+5$ gives $x=2\pm\mathrm{i}$, so the roots are $1\pm2\mathrm{i}$ and $2\pm\mathrm{i}$. Two real roots given: if $\text{h}(3)=0$ and $\text{h}(-1)=0$ for $\text{h}(x)=x^4-4x^3+6x^2-4x-15$, then $(x-3)(x+1)=x^2-2x-3$ is a factor. Divide to get $x^2-2x+5$, whose roots are $1\pm2\mathrm{i}$.

The roots of $\text{h}(x)=0$ above are $3$, $-1$, $1+2\mathrm{i}$ and $1-2\mathrm{i}$. On an Argand diagram they are the points $(3,0)$, $(-1,0)$, $(1,2)$ and $(1,-2)$. Both diagonals have length $4$ and midpoint $(1,0)$, and they are perpendicular, so the points form a square of area $\frac12\times4\times4=8$. Conjugate pairs are always symmetric about the real axis, which makes such a symmetry easy to spot.