While learning about polynomials, I came across this problem

$$x^4 + px^3 + qx^2 + rx + s = 0$$ is a polynomial with rational coefficients and $4$ complex roots. If two of the roots add up to $3 + 4i$ and the other two have a product of $13−i$, compute the value of $q$.

I let $m_1+m_2=3+4i$ and $m_3m_4=13-i$.

Since the coefficients are rational, we can apply the Complex Conjugate Roots Theorem. Because $(a+bi)(a-bi)$ always results in a real number, we know $m_3, m_4$ cannot be conjugates of each other. Hence, the conjugates are not together in the two equations.

I replaced the $m$'s with $a+bi, a-bi, c+di, c-di$ and attempted to solve for the variables. However, this led to an unwieldy system.

Is there a better way to solve this? Thanks!


By Vieta's formulas we have $$m_1 + m_2 + m_3 + m_4 = -p$$ and $$m_1m_2 + m_1m_3 + m_1m_4 + m_2m_3 + m_2m_4 + m_3m_4 = q$$ and $$m_1m_2m_3m_4 = s$$.

Use these formulas to compute $m_1m_2$, $m_3+m_4$, and note that $$q = m_1m_2 + (m_1 + m_2)(m_3 + m_4) + m_3m_4$$


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