When a polynomial has real coefficients, each nonreal root comes with its conjugate , and Vieta’s formulas then pin down the remaining roots and coefficients.
Problem 10.15.1. is a polynomial with root , and .
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The other root must be , since it must follow the property of polynomials. Then for the sum of the roots, since ,,,:
Now
We know , so , and .
So , so , so ,.
(Margin note: should be 1, not 9!)
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. So
and
So .
Example 10.15.2 (Quartic with nonreal roots)., with and for all . Also and . What is ?
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Grouping :
which together make ; the first is .
, so
That is,
Naming roots and their conjugates: .
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and .
Now since and , for each there is . So since is complex, , and since ,. So let and , equivalently ,. So and
If , as the problem posits, then must be such that their imaginary parts are equal. shows this clearly. Using :
where is real. Then, since ,
(the middle terms being ), so
Other routes tried: let , so
and likewise, with ,.
Or , giving
Side work: ;
?;
,,,; then
Also:
so . And , so , and , so .
Example 10.15.3. If are complex roots of the equation , where and , find in terms of and .
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The roots are , and , and
Expanding,
Also
So
Example 10.15.4. A cubic has leading coefficient , all its coefficients are real, and . If , find .
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Write , with roots , and . By Vieta,
Therefore
Test, dividing by :
leaves , whose roots are