Zeros & Multiplicities of Polynomial

Question

Given the polynomial f(x)=-.0029(x+12)(x+5)2(x-9)3
find the zeros and their corresponding multiplicities.

I know that in order to find the zeros you must factor and set the equation equal to zero and solve for x but I am unsure what to do for -.0029 and (x+5)2. Any help is greatly appreciated.

Thank you!

Answer 1

 f(x)=-.0029(x+12)(x+5)²(x-9)³

 

Michael is correct that the multiplicity has to do with the exponents.

If you write this out all the way your expanded polynomial is:

 

f(x) = -.0029(x+12)(x+5)(x+5)(x-9)(x-9)(x-9)

 

The zero at x=-12 appears 1 time so has a multiplicity of 1.

You can see that the zero at x=-5 appears 2 times and has multiplicity 2.

What about the other zero?

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NOTE:

If this was multiplied out you would have a polynomial of degree 6,

meaning the highest exponent on a variable would be on the x⁶.  Based

on that you would expect there to be 6 zeroes.  Here we see only 3 ...

{-12, -5, 9} ... With this polynomial we have 6 zeroes in actuality, but

3 of the zeroes are "repeats" due to the multiplicity of the zeroes at

x=-5 and x=9.

 

Answer 2

You don't have to do anything about -0.0029.  The factor (x+5)² is a factor the yields x=-5 as one of the polynomial's zeros.  Since the factor has an exponent of 2, the multiplicity of the zero x=-5 is 2.  This means that the zero x=-5 appears twice, and when the polynomial is graphed, the curve is tangential to the x-axis but never crossing the x-axis.

 

For the remaining, these are the factors you need to focus on:

 

(x + 12)

 

(x - 9)³