I am trying to find all the zeros of the following polynomial:
$P(x) = 2x^4-5x^3+21x^2+11x+91$
But I am already given one zero, that being $x=2-3i$.
Normally, this question would not be an issue, but the complex is throwing me right off, and I have no idea where to start. I am pretty sure all I need is a bump in the right direction and I would be able to solve the remaining problem.
$\endgroup$4 Answers
$\begingroup$Complex roots always come in pairs, so $x=2+3i$ must also be a zero and hence the quadratic $x^2-4x+13$ is a factor. Factoring gives $P(x) = (x^2-4x+13)(2x^2+3x+7)$. One can find the roots of $2x^2+3x+7$ using the quadratic formula.
$\endgroup$ 4 $\begingroup$Hint. If a polynomial with real coefficients has a complex root then the complex conjugate is also a root. Then you have two roots and a quadratic real factor.
$\endgroup$ 2 $\begingroup$One idea you could use is that if a complex number is a root of a polynomial with real coefficients, then the complex conjugate is also a root to the polynomial. This means that 2+3i is another root to the polynomial. You can now attempt to factorize the polynomial.
$\endgroup$ $\begingroup$You actually have two zeroes: $2 + 3i$ and $2 - 3i$ because complex zeros always come in a pair of complex conjugates. Which means, you now have:
$$[x - (2 - 3i)][x - (2 + 3i)]$$
Expand this you get
$$(x^2 -4x + 13)$$
Then use this as a divisor to your original polynomial.
Side note: By Descarte’s Rule of Signs, both $P(x)$ and $P(-x)$ have two sign changes. How many zeros do you think are there?
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