Find a polynomial with zeros:3, multiplicity 2;-3, multiplicity 2; and degree 4

Question

Help me find this out

Answer 1

Breaking it down:

 

Zeros at 3, with multiplicity 2 implies (x-3)²

Zeros at -3, with multiplicity 2 implies (x+3)²

 

Putting these terms together into a polynomial 

 

f(x) = (x-3)²(x+3)² = (x²-9)²

 

It can be determined from inspection that this will be a polynomial of degree 4, but let's expand it out 

 

f(x) = (x-3)²(x+3)² = x⁴-18x²+81

Answer 2

Alright, when a problem states that it has multiplicity of 2, that means that the final polynomial has a zero occur at that location twice. This of course can be extended to multiplicites of more than two, which would just mean it happened the multiplicities times at that location, i.e., if it had a multiplicity of 3 it would have a zero at that location 3 times, so on and so forth. Anyways, I digress.

 

With a multiplicity of 2 for the zero at 3, that would imply that we have x-3 as a factor of the polynomial twice, or part of the polynomial can be written as :

 

p(x) = (x-3)²q(x)

 

where p(x) is the polynomial we are trying to determine and q(x) is the remaining factors that we have yet to determine.

 

We also know that p(x) has a zero at -3 with a multiplicity of 2 as well, meaning that the polynomial can be written as:

 

p(x) = (x-3)²*(x+3)²q(x)

 

Since the first two factors are going to give a polynomial of degree 4, we assume that there are no further factors of p(x) and that means that q(x) must equal 1.

 

Thus, p(x) = (x-3)²(x+3)² or it can also be written as p(x) = [(x-3)(x+3)]² = (x²-9)² = x⁴ - 18x²  + 81.