The polynomial in question is:

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

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

aphathalo
2022-10-08
Answered

The polynomial in question is:

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

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

You can still ask an expert for help

Farbwolkenw

Answered 2022-10-09
Author has **6** answers

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:

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

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

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

I get

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

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

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

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:

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

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

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

I get

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

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

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