# Finding a polynomial with degree, zeroes, and multiplicity

• Nov 25th 2008, 10:07 AM
mwok
Finding a polynomial with degree, zeroes, and multiplicity
Question:
Find a polynomial f(x) of degree 6 such that 0,3 are zeroes of multiplicity 3 and f(2) = -24.

Can someone explain this step-by-step? what exactly is "multiplicity"?
• Nov 25th 2008, 10:12 AM
Moo
Hello,
Quote:

Originally Posted by mwok
Question:
Find a polynomial f(x) of degree 6 such that 0,3 are zeroes of multiplicity 3 and f(2) = -24.

Can someone explain this step-by-step? what exactly is "multiplicity"?

If a number a has a multiplicity b in a polynomial $f(x)$, it means that $(x-a)^b$ divides this polynomial.

So here, $f(x)=k (x-3)^3 (x-0)^3$
that's how it works.

Now, substitute x=2 to find k ! :)
• Nov 25th 2008, 10:20 AM
mwok
I see.
Find a polynomial f(x) of degree 5 such that 0,3 are zeroes of multiplicity 3 and f(2) = -24. (Notice the degree is now 5)

How would you solve this one?
• Nov 25th 2008, 10:27 AM
Moo
Quote:

Originally Posted by mwok
I see.
Find a polynomial f(x) of degree 5 such that 0,3 are zeroes of multiplicity 3 and f(2) = -24. (Notice the degree is now 5)

How would you solve this one?

It is not possible.

Note that if the degree of a polynomial is n, there are at most 5 roots for it.

If a root a has a multiplicity of b, it's like you have b roots that are all equal to a.

In fact, if a has a multiplicity of b, you have $f(a)=f'(a)=f''(a)=\ldots=f^{(b-1)}(a)=0$

$f^{(b-1)}(x)$ denots the (b-1)th derivative of f.

I replied to someone here : http://www.mathhelpforum.com/math-he...tiplicity.html dunno if it can give you another view on it