Write a polynomial function of least degree with integral coefficients that has the given zeros: `4` , `3i`
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Well, in order to have a polynomial with Real (let alone Integral) coefficients, you need to get rid of those complex numbers!
It turns out that the only way for you to have a polynomial without imaginary terms is to ensure that every complex root has its complex conjugate as another root.
How do I get the complex conjugate? Well, every complex number is of the form:
where `a` is the "real part" and `b` is the "imarginary part." In order to construct the complex conjugate, we simply take the additive inverse (negative) of the imaginary part! For example, the complex conjugate to `a+bi` would be:
How do these two help you get rid of imaginary numbers? Let me show you in general what happens to two roots that are complex conjugates:
`(x-(a+bi))(x-(a-bi)) = x^2 - ax + xbi -ax +a^2 - abi -xbi + abi -b^2i^2`
Now, let's simplify (remember, `i^2 = -1`) by putting all the real terms on one side, and all the imaginary terms on the other:
`x^2 - 2ax + a^2 + b^2 + i(-xb-ab+xb+ab)=x^2-2ax+a^2+b^2`
Notice that the imaginary side simply becomes zero! This would not happen if the roots were not complex conjugates.
So, how does all this apply to those two roots? Well, we are given one real root, and one imaginary root:
`x=4, x = 3i`
Well, remember how whenever we have an imaginary root, we need to take the negative of the imaginary part to make it a complex conjugate. Also, this complex conjugate is the only way to make the coefficients real. Therefore, we have to add one more root to this set of two in order to get real, integral coefficients. That additional root will be `-3i`.
So, our polynomial will become:
`(x-4)(x-3i)(x+3i) = (x-4)(x^2 + 3ix -3ix - 3^2i^2)`
Simplifying (again, make sure you account for `3^2i^2 = -9` :
`(x-4)(x^2+9) = x^3 -4x^2 +9x -36`
So there's our final polynomial. We have also found the minimum degree necessary to have integral coefficients: 3.
I hope that helps!
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