Question

Find a polynomial function ƒ(x) of least degree having only real coefficients and zeros as given. Assume multiplicity 1 unless otherwise stated. 2-i, 3, and -1

Accepted Answer

Identify the given zeros of the polynomial: $$2 - i$$, $$3$$, and $$-1. $$Since the polynomial must have only real coefficients, the complex zeros must come in conjugate pairs. Therefore, include the conjugate of $$2 - i$$, which is $$2 + i$$, as a zero as well. Write the factors corresponding to each zero. For zeros $$3$$ and $$-1$$, the factors are $$(x - 3)$$ and $$(x + 1)$$ respectively. For the complex zeros $$2 - i$$ and $$2 + i$$, the factors are $$(x - (2 - i))$$ and $$(x - (2 + i)). $$Multiply the complex conjugate factors to get a quadratic with real coefficients: 
$$ (x - (2 - i))(x - (2 + i)) = ((x - 2) + i)((x - 2) - i) = (x - 2)^2 - (i)^2 $$ Simplify the expression from the previous step using $$i^2 = -1$: 
 (x - 2)^2 - (-1) = (x - 2)^2 + 1$$ $$ Write the polynomial function f(x$$)$$ as the product of the quadratic from step 4 and the linear factors from step 2: 
 f(x) = ((x - 2)^2 + 1)(x - 3)(x + 1$$) $$

Find a polynomial function ƒ(x) of least degree having only real coefficients and zeros as given. Assume multiplicity 1 unless otherwise stated. 2-i, 3, and -1

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Key Concepts

Complex Conjugate Root Theorem

Constructing Polynomials from Roots

Multiplicity of Roots