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 4–6.2-i, 3, and -1

Accepted Answer

Hello, everyone using the zeros six minus I four and negative three, we will write the equation of a polynomial function of the least degree. The coefficient of the polynomial must be only real and take multiplicity one. Unless otherwise stated, we're given four answer choices that are four slightly different polynomials. The first thing I'm going to do is I'm gonna start by writing my function. So F of X equals, and we need to put an A here to account for a coefficient that we might be able to find later. We might not be able to find later. And then I'm gonna open a set of parentheses for each factor that I know based on these zeros. So the first er, we're given is six minus I. So that means our first factor is X minus. The zero were given which in parentheses I put six minus I closed parentheses. My next factor is gonna be X again minus and this one is going to be the conjugate of six minus I, which is six plus I. This happens because the conjugate root theorem says that if one complex number is a root of a polynomial, its conjugate is also a root. So we call that a conjugate is the same terms. But with the opposite sign in between my next factor, I'm gonna have X minus four, which is one of our zeros. And then my next factor is gonna be X minus the zero, negative three. Now that I have that written out, I want to simplify what I can within my parentheses before I start my multiplication. So the F of X part does not change F of X equals A and that's gonna be multiplied by, in my first set of parentheses, I'm distributing a negative sign into the complex number. So X minus six plus I closed parentheses. Same thing is happening in my second set of parentheses. I'm distributing a negative sign into the complex number. So I have X minus six minus I closed parentheses. The parentheses X minus four does not change. So that stays put. And then in the parentheses, X minus negative three, I'm gonna rewrite that as X plus three oh three. So now I have F of X equals a multiplied by the factor X minus six plus I multiplied by the factor X minus six minus I multiplied by the factor X minus four multiplied by our factor X plus three. Now I need to multiply this all together, I'm gonna multiply in pairs. So I'm gonna multiply the first two sets of parentheses together and then the second two. So the F of X equals a part does not change multiply together X minus six plus I by X minus six minus I, I'm going to distribute each term from the first trinomial into the second. So X times X is X squared X multiplied by negative six is negative six, X X multiplied by negative I is negative and I'm gonna go I X and then negative six multiplied by X is negative six, X negative six multiplied by negative six is plus 36 negative six, multiplied by negative I is plus six I. And final term there I multiplied by X is plus I X, I multiplied by negative six is negative six. I, I multiplied by negative I is negative I squared. And I'm gonna just rewrite my X minus four times X plus three for the time being until I'm done simplifying the first set of multiplication. So now that I have this really long polynomial in this parentheses, I want to simplify what I can. So I'm gonna combine as many like terms as I possibly can. It looks like I only have one X squared term. And then when I look at my normal X terms without an exponent, I have a negative six X and a negative six X which is negative 12 X next uh minus or negative X I X and positive I X will cancel out. I've already used these. So I'm just putting a little check mark above them to keep track of it for myself. Then I have a positive 36 I have a positive six I and a negative six I. So those will cancel out then minus negative or sorry minus I squared. So those 10 plus terms or nineish terms are now simplified down to four. That works a lot better for me. So next to it, I'm going to foil out X minus four, multiplied by X plus three. And when I do this, I get X squared minus four, X plus three, X minus 12. So both sets of parentheses have a little bit more simplification that can be done. So F of X stays put equals a and then in my first set of parentheses, I recall that negative I squared is the same thing as plus one. So I'm going to, instead of keeping negative I squared there add one to my constant of 36. So this parenthesis now has X squared minus 12, X plus 37 close parentheses. And in my second set of parentheses, I have X terms that I can combine. So I have X squared minus X minus 12. All right. One more set of multiplication to do. And this one's going to be a little long. So bear with me. But we are going to multiply together our two trinomial. So we're going to multiply X squared minus 12 X M plus 37 by X squared minus X minus 12. And I'm gonna start it over here. Just so I have a little bit more space. So F of X equals a multiplied by and I'm gonna go step by step. So X squared multiplied by X squared is XX to the fourth X squared multiplied by negative X is negative X to the third or cubed, X squared multiplied by negative 12 is negative 12 X squared. Next negative 12 X multiplied by X squared is negative 12, X cubed, negative 12, X multiplied by negative X is plus 12, X squared, negative 12, X multiplied by negative 12 is a positive 144 X and then 37 multiplied by X squared is X squared. 37 multiplied by negative X is minus 37 X and 37 multiplied by negative 12 is a negative 444. All right. So we did all the multiplication. Now we need to simplify it. So F of X equals a multiplied by parentheses. And I'm gonna to simplify what I can. So I have X to the fourth and then looking for X cubes, I have a negative one, X cubed and a negative 12 X cubed. So that gives me a negative 13 X cubed. I'm just putting little checkmarks next to them when I use them. So I know they're used looking at my X squared terms, I have a negative 12 X squared and a positive 12 X squared. So those will cancel. But I also have a positive 37 X squared. So that gets brought down into my simplification. Looking at my X terms, I have 144 X and negative 37 X which combined to a positive 107 X and then minus 444. All right. So we've gotten this all down to F of X equals a multiplied by X raised to the fourth power minus 13 X cubed plus 37 X squared plus 107 X minus 444. As we're given no information to be able to find that leading coefficient A, we are gonna leave this as if the leading coefficient is one. So F of X will equal X rays to the fourth power minus 13 X cubed plus 37 X squared plus 107 X minus 444. That is our polynomial. And after all of that multiplication it matches answer choice. A have a nice day.

Struggling with College Algebra?

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 4–6.2-i, 3, and -1

Zeros of Polynomial Functions practice set