Find a polynomial function ƒ(x) of least degree having only real ...

Question

Find a polynomial function ƒ(x) of least degree having only real coefficients and zeros as given. Assume multiplicity 1 unless otherwise stated. See Examples 4–6. 
5+i and 5-i

Accepted Answer

Hello, everyone using the zeros, 11 plus two, I and 11 minus two. I write the equation of a polynomial function of the least degree. The coefficient of the polynomial must be only real. Our answer choices are A F of X equals X squared minus 22 X minus B F of X equals X squared plus X plus 125 C F of X equals X squared minus 22 X plus 125 D F of X equals X squared plus 22 X minus 125. Knowing our zeros are 11 plus two, I and 11 minus two. I and those are conjugates of each other. I know that these are all the possible zeros that we need to work with. Had they not been conjugates of each other. We would have to find their conjugate, meaning the same terms with the opposite sign in the middle. So we can set up our F of X function. So F of X equals a multiplied by the factor. So our factor is gonna be X minus the zero we are given. So X minus in parentheses. 11 plus two. I close parentheses twice. And then our next factor is going to be X minus the next zero we're given. So in parentheses, 11 minus two, I closing both sets of parentheses. The A is there as a coefficient that we may or may not need. We don't know if there is a coefficient that canceled out while we were factoring to get these zeros. Originally, next, I'm going to distribute the negative into the complex number. So F of X equals A and then open parentheses, X minus 11 minus two. I closed parentheses and then open parentheses, X minus 11 plus two. I closed parentheses. From here, I'm going to multiply the two trinomial together. So F of X equals A and then in parentheses, I'm gonna distribute all the way through each term. So first X times X is X squared X multiplied by negative 11 is negative 11. X X multiplied by two. I is plus two I X. Next negative 11 multiplied by X is negative 11. X negative 11 multiplied by negative 11 is a positive 121. Negative 11 multiplied by two. I is negative 22. I next negative two, I multiplied by X is negative two I X negative two, I multiplied by negative 11 is a positive 22 I and finally negative two, I multiplied by positive two, I it's negative four I squared closed parentheses. I want to simplify this. I wanna combine my like terms before I do anything else? Because that's a lot of terms to look at. So F of X equals A and then in my parentheses, I'm starting with my highest term which is X squared and then I wanna see what else I need to combine. So looking for my X terms, I have negative 11 X and negative 11 X. So I'm gonna combine those two, negative 22 X and I'm honestly gonna put a little check mark above them once I use them. So I know I've used them next. I'm gonna deal with the imaginary numbers next to X. So I have a positive two I X and a negative two I X. Those will cancel out. So that gets to be out of the way. Looking for my constance, I have 121. So it's a positive 121 next I have minus or negative 22 I and a positive 22 I, so those will cancel out. So I'm gonna cross those out and then I'm left with my minus or negative four I squared. Recall that I squared is the same as negative one. So simplifying negative four I squared is like doing negative four multiplied by negative one, which is a positive four. So that simplified nicely. Um Right now I have F of X equals a multiplied by the parentheses X squared minus 22 X plus 121 plus four, I can combine one more set of like terms in there. So F of X equals a multiplied by the parentheses X squared minus 22 X plus 125. At this step, we are not told anything about the A, we don't have any extra values to be able to solve for it. So we're going to assume it's one, this means our final equation is F of X equals X squared minus 22 X plus 125. And that is our solution and it matches answer choice C have a nice day.

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Find a polynomial function ƒ(x) of least degree having only real coefficients and zeros as given. Assume multiplicity 1 unless otherwise stated. See Examples 4–6. 5+i and 5-i