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Polynomial function that has the zeros -5 and 4i

User Moe Matar
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1 Answer

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The polynomial function is:


\boxed{f(x)=a(x^3+5x^2+16x+80)}

Step-by-step explanation:

In this exercise, we will use two important theorems:

  • Fundamental Theorem of Algebra: We can factor completely any polynomial with real number coefficients over the field of complex numbers.

  • Complex Conjugate Root Theorem: If a polynomial in one variable has real coefficients, and
    a+bi is a root of that polynomial being
    a \ and \ b real numbers, then its complex conjugate
    a-bi is also a root of the polynomial function.

In this case we have the following roots:


x_(1)=-5 \ and \ x=4i

According to the Complex Conjugate Root Theorem:


x_(3)=-4i \\ \\ Is \ also \ a \ root \ of \ the \ polynomial \ function

Hence, we can write the function as:


f(x)=a(x-(-5))(x-4i)(x+4i) \\ \\ f(x)=a(x+5)(x-4i)(x+4i) \\ \\ \\ By \ distributive \ property: \\ \\ f(x)=a(x+5)(x^2+4xi-4xi-(4i)^2) \\ \\ f(x)=a(x+5)(x^2-16i^2) \\ \\ \\ i=√(-1) \\ \\  i^2=-1 \\ \\ \\ f(x)=a(x+5)(x^2+16) \\ \\ By \ distributive \ property: \\ \\ \boxed{f(x)=a(x^3+5x^2+16x+80)} \\ \\ For \ any \ real \ leading \ coefficient \ a

User Rob Starling
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