Given:
The function is [tex]y=x(x+2)^2[/tex].
To find:
The zeros and their multiplicity.
Solution:
A polynomial is defined as
[tex]y=a(x-c_1)^{m_1}(x-c_2)^{m_2}...(x-c_n)^{m_n}[/tex] ...(i)
where, a is a constant, [tex]c_1,c_2,...,c_n[/tex] are zeros with multiplicity [tex]m_1,m_2,...,m_n[/tex] respectively.
We have,
[tex]y=x(x+2)^2[/tex]
It can be written as
[tex]y=(x-0)^1(x-(-2))^2[/tex] ...(ii)
On comparing (i) and (ii), we get
[tex]c_1=0,m_1=1[/tex]
[tex]c_2=-2,m_2=2[/tex]
Therefore, the zeros of the given function are 0 and -2 with multiplicity 1 and 2 respectively.
