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Polynomials with Complex Roots

The Fundamental Theorem of Algebra assures us that any polynomial with real number coefficients can be factored completely over the field of complex numbers .

In the case of quadratic polynomials , the roots are complex when the discriminant is negative.

Example 1:

Factor completely, using complex numbers.

x 3 + 10 x 2 + 169 x

First, factor out an x .

x 3 + 10 x 2 + 169 x = x ( x 2 + 10 x + 169 )

Now use the quadratic formula for the expression in parentheses, to find the values of x for which x 2 + 10 x + 169 = 0 .

Here a = 1 , b = 10 and c = 169 .

x = − b   ±   b 2   −   4 a c 2 a

x = − 10   ±   10 2   −   4 ( 1 ) ( 169 ) 2 ( 1 )         = − 10   ±   100   −   676 2         = − 10   ±   − 576 2

Write the square root using imaginary numbers.

x = − 10   ±   24 i 2         = − 5 ± 12 i

We now know that the values of x for which the expression

x 2 + 10 x + 169

equals 0 are x = − 5 + 12 i   and   x = − 5 − 12 i .

So, the original polynomial can be factored as

x 3 + 10 x 2 + 169 x = x ( x − [ − 5 + 12 i ] ) ( x − [ − 5 − 12 i ] )

You can verify this using FOIL .

Sometimes, you can factor a polynomial using complex numbers without using the quadratic formula. For instance, the difference of squares rule:

x 2 − a 2 = ( x + a ) ( x − a )

This can also be used with complex numbers when a 2 is negative, as follows:

x 2 + 25 = ( x + 5 i ) ( x − 5 i )

Example 2:

Factor completely, using complex numbers.

9 x 2 y + 64 y

First, factor out y .

9 x 2 y + 64 y = y ( 9 x 2 + 64 )

Now, use the difference of square rule to factor 9 x 2 + 64 .

9 x 2 + 64 = 9 x 2 − ( − 64 )                                         = ( 3 x ) 2 − ( 8 i ) 2                                         = ( 3 x + 8 i ) ( 3 x − 8 i )

Therefore, 9 x 2 y + 64 y = y ( 3 x + 8 i ) ( 3 x − 8 i ) .