# The polynomial in question is: x^4−8x^3−19x^2+288x−612 and the zero is 4−i.

The polynomial in question is:
${x}^{4}-8{x}^{3}-19{x}^{2}+288x-612$
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Farbwolkenw
Taking a polynomial for factorization such as
${x}^{4}-8{x}^{3}-19{x}^{2}+288x-612$
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:
$\left(x-4-i\right)\left(x-4+i\right)=\left(x-4{\right)}^{2}-{i}^{2}={x}^{2}-8x+17$
Then we have a quadratic that we can use as a divisor on the original:
${x}^{2}-8x+17\mid {x}^{4}-8{x}^{3}-19{x}^{2}+288x-612$
I get
$\left({x}^{2}-8x+17\right)\left({x}^{2}-36\right)={x}^{4}-8{x}^{3}-19{x}^{2}+288x-612$
which means that the remaining factors are (x−6),(x+6) .