Find a polynomial function ƒ(x) of degree 3 with real coefficient...

Question

Find a polynomial function ƒ(x) of degree 3 with real coefficients that satisfies the given conditions. See Example 4. 
Zeros of -3, 1, and 4; ƒ(2)=30

Accepted Answer

Hello, everyone. We've been asked to provide the polynomial function with zeros at negative 52 and three. And that F of four equals 54. The degree of the polynomial function is three. Our answer choices are A F of X equals three, X cubed minus nine X squared minus X plus 90 B F of X equals three X cubed minus 57 X plus 90 C F of X equals three X cubed minus nine X squared plus 57 X plus 90 D F of X equals three X cubed plus 57 X plus 90. The zeroes we are given were negative 52 and three. This means that we found that X equals negative five, X equals two and X equals three. We now need to work backwards from there to find the factor that gave us these values. Recall that to find the value of X, we set the factor equal to zero. So we're gonna work backwards. So for X equals negative five, I'm gonna add five to both sides. So I get that X plus five equals zero. I now know X plus five is one of my factors. Doing the same with X equals two. I subtract two from both sides. I get X minus two equals zero, X minus two will also be one of our factors. And then X equals three. I subtract three from both sides and I get X minus three equals zero. So X minus three is also a factor. Now, I'm gonna set this up to be able to find my full equation or function. So I know that the function we're looking for is F of X that's gonna equal, we're gonna put a for any possible coefficient that's in there, that may have just gotten canceled out and not represented directly in these factors. And then we're going to multiply that by each factor. So a multiplied by the factor X plus five multiplied by the factor X minus two multiplied by the factor X minus three. Now I'm gonna do all the multiplication. So F of X is gonna stay put F of X equals A and I'm going to foil X plus five multiplied by X minus two. So when I do that X times X is X squared negative two times X is negative two. X inner gives me five multiplied by X which is five, X last five times negative two, which is negative 10 closed parentheses. And next, we'll still need to multiply that by the factor X minus three before I multiply it by X minus three, I'm gonna combine my like terms. So I get F of X equals a multiplied by the parentheses X squared plus three X because I combined my negative two X and my five X minus 10 closed parentheses. Now multiplied by the factor X minus three. I'm going to multiply X minus three times the trinomial X squared plus three X minus 10. So it's still F of X equals A and I'm going to keep this in a set of parentheses. X squared multiplied by X would be X to the third X squared multiplied by negative three is negative three, X squared. Then three, X multiplied by X give me plus three, X squared three, X multiplied by negative three is negative nine X. Next negative 10, multiplied by X is negative 10, X and negative 10 multiplied by negative three is plus 30. Close parentheses. Before I go anywhere near my next step, I want to combine all my like terms and see what I have to work with. So F of X equals a then open parentheses, combining my like terms. I still have X to the third power X cubed, negative three, X squared and positive three X squared. Cancel out negative nine, X minus 10. X will give me negative 19 X plus 30. So we've simplified all of our factors and this is where the fact that F of four equals 54 comes into play, we are gonna put four in for X in this function and we're gonna let F of X equal 54. And we're gonna solve for a, so 54 equals a multiplied by. There'd be four cubed minus 19 times four plus 30. So 54 equals a then in my parentheses, four cubed is 64. Negative 19 times four is minus or negative plus 30. Combining what's in my parentheses? 54 equals a multiplied by and 64 minus 76 is negative 12 plus 30. Gives us a positive 18. So we have 54 equals a times 18, dividing both sides by 18. A equals three. So now that we know a equals three, we can plug that in for a, distribute it through the trinomial and find our final function. So putting that over here, I'm gonna use this function. So F of X equals a times three X cubed minus 19 X plus 30. But now I'm going to put three in for a. So F of X equals three multiplied by the parentheses X cubed minus 19 X plus 30 closed parentheses, distributing that in F of X equals three X cubed minus 57 X plus 90. And that should be all of our simplification. Our polynomial is um the degree of three because it's an X cubed. So F of X equals three X cubed minus X plus 90. And that is answer choice B have a nice day.

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Find a polynomial function ƒ(x) of degree 3 with real coefficients that satisfies the given conditions. See Example 4. Zeros of -3, 1, and 4; ƒ(2)=30