In Exercises 25–32, find an nth-degree polynomial function with r...

Question

In Exercises 25–32, find an nth-degree polynomial function with real coefficients satisfying the given conditions. If you are using a graphing utility, use it to graph the function and verify the real zeros and the given function value. n=4; -2, 5, and 3+2i are zeros; f(1) = -96

Accepted Answer

Hey everyone here we are asked to identify the degree polynomial function with real coefficients that satisfies the given conditions. So we are first told that N is equal to three. So we know that we have a third degree polynomial. And with this we need to recall the linear factors. Nation theorem for a third degree polynomial, which tells us that F of X is equal to a sub N times the quantity of x minus C, sub one times the quantity of X minus C, sub two times the quantity of x minus C sub three. And next we just need to identify each of these zeros. So we're first told we have a zero of negative two. So we can let C sub one be equal to negative two. And next we are also told that two plus five I is a zero. So we can let it be equal to c sub two. And next, since two plus five I is a zero and our polynomial function has real coefficients. We know that the the conjugate to minus five I is also a zero. So we can let it be equal to C sub three. And next we just need to implement each of these zeros into the formula. So we have F of X is equal to a sub n, times the quantity of x minus c. Sub one, which is negative two times the quantity of x minus c sub two, which is two plus five, I times the quantity of x minus C, sub three, which is two minus five I. And now from here we just need to begin simplifying. So we have F of X is equal to a sub n times the quantity of x minus negative two gives us plus two. And for these last two sets of parentheses we can start foiling them out. So we have first X times X which is X squared. We then have X times negative quantity of two minus five. I we then have negative quantity of two plus five I times X. And then we have negative quantity of two plus five I times negative quantity of two minus five I giving us positive quantity two plus five I times quantity two minus five I. And now from here we just have to continue to simplify and first factor in these X is here. So we have F of X is equal to a sub n times the quantity of X plus two times the quantity of x squared, negative x times two, gives us negative two X negative x times negative five I gives us positive five X. I then we have negative two times X giving us negative two X. Then negative five I times X gives us negative five X. I. And now here we have these complex contra kits and so we know and we multiply them that the middle values will cancel out. And we also need to recall that I squared is equal to negative one. So here we can simply square both terms. So we can say plus the quantity of two squared and then we also square the coefficient of I here. So we have plus five squared and now simplifying further, we have F of X is equal to a sub N times the quantity of X plus two times the quantity of X squared minus two X plus minus two X. Gives us negative four X. Then we see that the five X. I's cancel out. And then we have plus two squared which is four plus five squared, which is 25. So we have plus 29. And now to finish simplifying we just have to again multiply out these two sets of parentheses. So we have F of X is equal to a sub N. Then we have X times X squared, giving us x cute X times negative four X gives us negative four X squared. Then X times 29 gives us plus 29 X. Then two times X squared gives us plus two X squared. Two times negative four X gives us negative eight X. And finally two times 49 gives us plus 58. Now we just have to simply simplify these terms here. So we have F of X is equal to a sub n times the quantity of X cubed. We then have negative four X squared plus two X squared giving us negative two X squared. Then we just have 29 X minus eight X giving us plus 21 X. And finally just bringing down this 58. So we have plus 58 and now we have a fully simplified function. So we want to start solving for a sub bend. And to do this we need to recall one of our given piece of information here, which is that F of one is equal to 78. So now we can set the function, we have to 78 we can substitute one in for X. So doing so we have 78 is equal to a sub N times a quantity of one q minus two times one squared plus 21 times one plus 58. And now with this we can just simplify and solve for M or a sub N. And so doing so we have 78 is equal to a sub n times 78. And now to get a sub N alone we divide both sides by which shows us that a sub N is equal to one. So now to get our final function here, we just have to substitute in this value for a sub N. So here we now have F of x is equal to one times the quantity of X cubed minus two X squared plus 21 X plus 58. And now just finally simplifying, we have an answer of F of x is equal to X cubed minus two X squared plus 21 X plus 58. And now if we were to graph this function, we would see hear that we see our real zero at negative 20. And we also see that at F of one, which is when X is one, we do have an output of 78. Therefore our function is correct and we are left with answer choice A with a function of X cubed minus two, X squared plus 21 X plus 58. Thanks for watching. I hope you found this video helpful and I'll see you again next time.

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In Exercises 25–32, find an nth-degree polynomial function with real coefficients satisfying the given conditions. If you are using a graphing utility, use it to graph the function and verify the real zeros and the given function value. n=4; -2, 5, and 3+2i are zeros; f(1) = -96