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Let $f(x)$ be a polynomial function of second degree. If $f(1)=f(-1)$ and $a_{1}, a_{2}, a_{3}$ are in A. P., then $f^{prime}left(a_{1}ight)$, $f^{prime}left(a_{2}i...

Question

Let $f(x)$ be a polynomial function of second degree. If $f(1)=f(-1)$ and $a_{1}, a_{2}, a_{3}$ are in A. P., then $f^{prime}left(a_{1}ight)$, $f^{prime}left(a_{2}ight), f^{prime}left(a_{3}ight)$ are in(A) A.P.(B) G.P.(C) H.P.(D) None of these

Let $f(x)$ be a polynomial function of second degree. If $f(1)=f(-1)$ and $a_{1}, a_{2}, a_{3}$ are in A. P., then $f^{prime}left(a_{1} ight)$, $f^{prime}left(a_{2} ight), f^{prime}left(a_{3} ight)$ are in (A) A.P. (B) G.P. (C) H.P. (D) None of these



Answers

If $x=a$ is a zero of a polynomial function $f,$ then the following three statements are true.
(a) $x=a$ is a _____ of the polynomial equation $f(x)=0$
(b)
_____ is a factor of the polynomial $f(x)$
(c) $(a, 0)$ is an _____ of the graph of $f$

We were given 1/5 degree polynomial X to the fifth plus ball above. A problem doesn't stress by that, but it does specify that when you plug to you get zero. When you plug in one, you get zero. Basically, two X equals one and two are zeros of the pollen, all zeros, so that means X minus one and x plus X minus. Two are factors of F of X, and the problem has what other function. What other function of polynomial divides into ever backs evenly? Well, if X minus one divides into it and X minus two divides into it than their product about it into it evenly as well. So let's find this product. If you do, X minus one times X minus two You will get X squared minus three x What is after a sorry X squared minus two Yanks plus two. And that's gonna be your answer. And that corresponds to It's a choice D

Okay, so we have our falling function f of X, and we will to, for part a solve for F evaluated that negative one. So this is going to give us. And they would have won two part two minus three times negative one but in plus for sabbatical to a one plus a three plus a force That's for Post War, which is an eight and then for F evaluated at two, we had a two squared minus three times to and then plus four. So that gives us a four minus six plus four. That's an eight minus six, which is equal to two.

So we have our falling function after back and report A. We want to have always tough violated at negative one. So this is three times negative one number two plus the negative one minus a five. So this gives us a three minus six, which is equal to negative three, and I followed it a two fer part being. We get three in times to three part two plus a two on minus five. The service here is equal to a three times a four and then to minus five after minus three. So we get 12 Ministry, which is equal to nine.

Okay, so we're giving her full function and we went to for part a evaluated if it's negative one. So this is negative. Negative one to mark two plus two. I'm thinking of 1353 minus AIDS. Really have a negative one and then minus two minus eight. Fanatical. 89 10 11. That's minus 11. And then we went on to part B. We have minus two to part two, plus two times +22 or three minus eight. So we at minus warn and in plus two trying 2 to 3. Car three. We just 16 minus its 2 16 minus eight minus four That's you put too positive for.


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