Use a given zero to write P(x) as a product of linear and irreducible quadratic factors

Question

The polynomial in question is:

$x^4 - 8x^3 - 19x^2 + 288x - 612$

and the zero is $4 - i$.

What I don't understand is how to go from the given zero to factorizing, especially as it's imaginary.

Google's autocomplete showed others had searched for this problem before -- but there was nothing forthcoming. So I figured it might be useful to others if I asked it here.

Answer 1

Taking a polynomial for factorization such as

$$x^4 - 8x^3 - 19x^2 + 288x - 612$$

and with a given root $4-i$, it is clear that the polynomial has only real coefficients, and therefore there is a second root $4+i$. So we take these roots together like so:

$$(x-4-i)(x-4+i)=(x-4)^2-i^2=x^2-8x+17$$

Then we have a quadratic that we can use as a divisor on the original:

$$x^2-8x+17\mid x^4 - 8x^3 - 19x^2 + 288x - 612$$

I get

$$(x^2-8x+17)(x^2-36)=x^4 - 8x^3 - 19x^2 + 288x - 612$$

which means that the remaining factors are $(x-6),(x+6)$.

Answer 2

If you have only real coefficient in a polynomial, $a + ib$ is a zero, then also $a - ib$ is also a zero. So you can divide your polynomial for $((x-a)^2+b^2)$.

In this case you can divide your polynomial by $(x^2-8x+17)$