Find the zeros and give the multiplicity of each for $f(x)=x^{2}(x-8)^{2}(x+7)^{4}$.
Note: to be counted as correct, you must get all answers correct.
- One zero is $x=\square$ and has a multiplicity of $\square$ help (numbers)
- Another zero is $x=\square$ and has a multiplicity of: $\square$ help (numbers)
- The last zero is $x=\square$ and has a multiplicity of: $\square$
Answer
So, the answers are: One zero is \(x=\boxed{0}\) and has a multiplicity of \(\boxed{2}\). Another zero is \(x=\boxed{8}\) and has a multiplicity of \(\boxed{2}\). The last zero is \(x=\boxed{-7}\) and has a multiplicity of \(\boxed{4}\).
Step 1 ：The zeros of a polynomial are the values of x that make the polynomial equal to zero. The multiplicity of a zero is the number of times that zero appears as a root of the polynomial.
Step 2 ：In this case, the zeros are given by the factors of the polynomial. The multiplicity of each zero is given by the exponent of its corresponding factor. So, we can directly read off the zeros and their multiplicities from the given polynomial.
Step 3 ：The zeros of the polynomial \(f(x)=x^{2}(x-8)^{2}(x+7)^{4}\) are \(x=0\), \(x=8\), and \(x=-7\). The multiplicities of these zeros are 2, 2, and 4 respectively.
Step 4 ：So, the answers are: One zero is \(x=\boxed{0}\) and has a multiplicity of \(\boxed{2}\). Another zero is \(x=\boxed{8}\) and has a multiplicity of \(\boxed{2}\). The last zero is \(x=\boxed{-7}\) and has a multiplicity of \(\boxed{4}\).
