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Complex Roots of Polynomials

Complex Roots of Polynomials

The Fundamental Theorem of Algebra states that a polynomial of degree $n$ with complex coefficients has $n$ complex roots, some of which may be repeated, and hence such polynomials can always be factorised into $n$ linear factors given by these complex roots.

It follows from this that polynomials with real coefficients can be factorised into linear and irreducible quadratic factors with real coefficients.

An irreducible quadratic with real coefficients has two complex roots, one of the form $a+bi$ and the other its complex conjugate $a-bi$. This is explained in the section on the quadratic formula.

Hence a polynomial with real coefficients has an even number of complex roots (including the case when there are none) and the rest are real roots.

Worked Examples

Example 1

Find the roots of the polynomial $x^2-4x+29$.


\begin{align} x &= \frac{ 4\pm \sqrt{(-4)^2-(4\times1\times29)} }{2 \times 1}\\ \\ &=\frac{ 4\pm \sqrt{16-116} }{2}\\ \\ &=\frac{ 4\pm \sqrt{-100} }{2}\\ \\ &=\frac{ 4\pm 10i }{2}\\ \\ &=2\pm 5i \end{align}

Therefore the roots of the polynomial are $x_1=2+5i$ and $x_2=2-5i$.

Example 2

Find the roots of the polynomial $x^3-8x^2+22x-20$.


As with solving any cubic, we can use trial and error to find the first root.

When $x=1$: $(1)^3-8(1)^2+22(1)-20 = 1-8+22-20= -5 \neq 0$

When $x=2$: $(2)^3-8(2)^2+22(2)-20= 8 -32+44-20 = 0$

Therefore $x=2$ is a root.

Use polynomial division to factorise the cubic using $(x-2)$:

\[x^3-8x^2+22x-20\] can be written as \[x^2(x-2)-6x(x-2)+10(x-2)\]

Dividing this by $(x-2)$ gives


So we have $(x-2)(x^2-6x+10)=0$.

Now, use the quadratic formula to solve the quadratic $x^2-6x+10=0$ and find the last two roots.

\begin{align} x &= \frac{6 \pm \sqrt{(-6)^2-(4\times1\times 10)} }{2\times1}\\\\ &=\frac{6 \pm \sqrt{36 -40} }{2}\\\\ &=\frac{6 \pm \sqrt{-4} }{2}\\\\ &=\frac{6 \pm 2i}{2}\\ &=3 \pm i \end{align}

So $x= 3+i$ or $x=3-i$

Therefore, the roots of this cubic are $x_1=2$, $x_2=3+i$, $x_3= 3-i$.

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