
New Member
Form a polynomial whose degrees and zeros are given
Zeros: 3, multiplicity 2; 3, multiplicity 2; degree 4

New Member
I not only need help with this problem. I need to understand how to work it. I do not even know where to start with this problem.

Elite Member
Before we dive in, it might help if you specified the part(s) you don't understand. Do you know what a polynomial's "degree" is? Do you know what "zeroes" are? Do you know what "multiplicity" means?
Thank you.
Eliz.

New Member
I know what zeros are and I know what a polynomial is but i do not understand how to take this information and make a polynomial out of it. I do not understand what multiplicity is either and I do not know why there is 3 of them. Also i thought that the degree told you how many zeros there were... so does that mean there is a double zero somewhere?

Elite Member
I'm not sure what you mean by "there is 3 of them"...?
The "zeroes" are the values for x that make the polynomial equal to zero. If they are real numbers, then they are also where the polynomial crosses or touches the xaxis.
The "multiplicity" is "how often they occur". For instance, (x  3)<sup>2</sup> has a zero of x = 3, with multiplicity two, because the factor x  3 occurs twice.
Note that, if x = a is a zero, then you solved x  a = 0 to find it, which means that x  a was a factor. Use this to work backwards from the zeroes to the factors. Use the multiplity information to find out how many times the factors appear. Then multiply together all the factors to find the polynomial.
Eliz.

New Member
Wait a minute... the zeros are already in the question? If so then i think im reading the problem wrong....
Zeros: 3, multiplicity 2; 3, multiplicity 2; degree 4
So what im getting now is that 3 is a zero and it shows up 2 times in the polynomial as (x3)^2. Another zero is 3 and that shows up as (x+3)^2
So my polynomial is f(x)=(x3)^2(x+3)^2
Right?

Elite Member
Yup. Except they might want you to multiply all that out. And they might want you to answer "in full generality", which means you have to account for the fact that nonzeroable factors might also be included. The "general" form would be:
. . . . .y = a (x  3)<sup>2</sup> (x + 3)<sup>2</sup>
...but maybe all multiplied out. Because the real number "a" doesn't affect the degree (in this case, 4) of the polynomial and, since it's never equal to zero, it doesn't affect the zeroes of the polynomial. So the polynomial could be y = 6 (x  3)<sup>2</sup> (x + 3)<sup>2</sup> or y = 47 (x  3)<sup>2</sup> (x + 3)<sup>2</sup> or whatever.
Eliz.

New Member
So the next problem is f(x)=1/4x^2 (x^23)(x+3)
so 1/4 would be considered a....
one of the zeros would be 3 and one would be positive 3 and there is no multiplicity.... correct?

Elite Member
Please post new questions as new threads, rather than as replies to old threads, where they tend to be overlooked.
Where are you getting that x = 3 is a zero? I see no "x  3" factor. Also, if something is a zero, it must occur, so it must have multiplicity of 1 or more. You can't have a zero that occurs ("x = 3 is a zero") and yet doesn't ("with no multiplicity").
Also, what happened to the "x<sup>2</sup>" factor? Can't you get a zero (with multiplicity 2, obviously) from that?
Eliz.
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