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.
Find the roots of the polynomial x2−4x+29.
x=4±√(−4)2−(4×1×29)2×1=4±√16−1162=4±√−1002=4±10i2=2±5i
Therefore the roots of the polynomial are x1=2+5i and x2=2−5i.
Find the roots of the polynomial x3−8x2+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≠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):
x3−8x2+22x−20 can be written as x2(x−2)−6x(x−2)+10(x−2)
Dividing this by (x−2) gives
x2−6x+10
So we have (x−2)(x2−6x+10)=0.
Now, use the quadratic formula to solve the quadratic x2−6x+10=0 and find the last two roots.
x=6±√(−6)2−(4×1×10)2×1=6±√36−402=6±√−42=6±2i2=3±i
So x=3+i or x=3−i
Therefore, the roots of this cubic are x1=2, x2=3+i, x3=3−i.