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opus
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As a preface to this theorem stated in my text, it states that:
"If all the coefficients of a polynomial ##P(x)## are real, then ##P## is a function that transforms real numbers into other real numbers, and consequently, ##P## can be graphed in the Cartesian Coordinate Plane."
It then goes on to state The Conjugate Roots Theorem:
Let ##P(x)## be a polynomial with only real coefficients. If the complex number ##a+bi## is a zero of ##P##, then so is the complex number ##a-bi##. In terms of linear factors of ##P##, this means that if ##x-(a+bi)## is a factor of ##P##, then so is ##x-(a-bi)##.
Now I understand the theorem, but what I've highlighted in bold is confusing me as it seems to be a contradiction. In the first statement, what I understand is that if we have real coefficients in a polynomial, then we'll have real numbers (and have no non-reals come up). In the theorem, it states the same given of only real coefficients, but now states that we can have non-real numbers come up. We then would not be able to graph all the solutions in the Cartesian Coordinate Plane.
Where am I thinking incorrectly here?
"If all the coefficients of a polynomial ##P(x)## are real, then ##P## is a function that transforms real numbers into other real numbers, and consequently, ##P## can be graphed in the Cartesian Coordinate Plane."
It then goes on to state The Conjugate Roots Theorem:
Let ##P(x)## be a polynomial with only real coefficients. If the complex number ##a+bi## is a zero of ##P##, then so is the complex number ##a-bi##. In terms of linear factors of ##P##, this means that if ##x-(a+bi)## is a factor of ##P##, then so is ##x-(a-bi)##.
Now I understand the theorem, but what I've highlighted in bold is confusing me as it seems to be a contradiction. In the first statement, what I understand is that if we have real coefficients in a polynomial, then we'll have real numbers (and have no non-reals come up). In the theorem, it states the same given of only real coefficients, but now states that we can have non-real numbers come up. We then would not be able to graph all the solutions in the Cartesian Coordinate Plane.
Where am I thinking incorrectly here?