The Fundamental Theorem of Algebra and Complex Roots
Advanced
Delves into the Fundamental Theorem of Algebra, which guarantees the existence of complex roots. Covers the corollary about the number of roots and the Complex Conjugate Root Theorem.
The Fundamental Theorem of Algebra and Complex Roots
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What is the statement of the Fundamental Theorem of Algebra?
Every non-constant single-variable polynomial with complex coefficients has at least one complex root. This means there is always at least one solution to \(P(x) = 0\).
What is the important corollary to the Fundamental Theorem of Algebra regarding the number of roots?
A polynomial of degree \(n > 0\) has exactly \(n\) roots in the complex number system. These roots may be real or complex, and some roots may be repeated (counted with multiplicity).
A polynomial has a degree of \(7\). How many roots does it have?
It has exactly \(7\) roots, counting multiplicities. These roots can be any combination of real and complex numbers.
What does the Complex Conjugate Root Theorem state?
If a polynomial has REAL coefficients, and if a complex number \(a + bi\) (where \(b \neq 0\)) is a root, then its complex conjugate, \(a - bi\), must also be a root.
What crucial condition must be met for the Complex Conjugate Root Theorem to apply?
The polynomial must have exclusively REAL coefficients. The theorem does not hold if any coefficient is complex or imaginary.
A polynomial with real coefficients has \(x = 2 - 5i\) as one of its roots. What must another root be?
According to the Complex Conjugate Root Theorem, the conjugate \(2 + 5i\) must also be a root.
A cubic polynomial with real coefficients is known to have roots at \(x = -3\) and \(x = 4i\). What is the third root?
Since the coefficients are real, complex roots must come in conjugate pairs. The conjugate of \(4i\) (or \(0+4i\)) is \(-4i\). The third root is \(x = -4i\).
If a polynomial has roots \(3+i\) and \(3-i\), what real quadratic factor corresponds to this pair of roots?
The factors are \((x - (3+i))\) and \((x - (3-i))\). Multiplying them gives \((x-3)-i\) and \((x-3)+i\), which is a difference of squares: \((x-3)^2 - i^2 = x^2 - 6x + 9 - (-1) = x^2 - 6x + 10\).
Can a cubic polynomial with real coefficients have exactly two complex roots and one real root?
Yes. The two complex roots must be a conjugate pair, and the third root would be real. For example, roots could be \(2+i\), \(2-i\), and \(5\).
Can a quartic (degree 4) polynomial with real coefficients have three complex roots and one real root?
No. Complex roots of a real-coefficient polynomial must come in conjugate pairs. This means there must be an even number of non-real complex roots. It could have \(0\), \(2\), or \(4\) complex roots, but not \(3\).
Find a third-degree polynomial with real coefficients that has zeros at \(x=1\) and \(x=3i\).
If \(3i\) is a zero, its conjugate \(-3i\) must also be a zero. The factors are \((x-1)\), \((x-3i)\), and \((x+3i)\).
\(P(x) = (x-1)(x-3i)(x+3i) = (x-1)(x^2 - (3i)^2) = (x-1)(x^2+9) = x^3 - x^2 + 9x - 9\).
How can any polynomial with real coefficients be factored?
It can be factored into a product of linear factors (corresponding to real roots) and irreducible quadratic factors with real coefficients (corresponding to complex conjugate root pairs).
