find the zeros of this function. state the multiplicity of multiple zeros

y(3x+2)^3(x-5)^5

Question

find the zeros of this function. state the multiplicity of multiple zeros 

y(3x+2)^3(x-5)^5

anonymous · Answer

So our function is: $y(x)=(3x+2)^3(x-5)^5$. As we know, a zero of $y(x)$ a value $x$ such that $y(x) = 0$. Since we've factored out our function into a product, at least one factor must be $0$ for the product (i.e. the result of the function) to be $0$; thus, to find the zeroes, we only need to solve for the $x$s that could result in one of the factors being $0$.$$(3x+2)^3=0\3x+2=0\3x=-2\x=-\frac23\\ $x-5)^5=0\x-5=0\x=5$$The multiplicity of a zero \(x$ refers to the number of factors that using said $x$ would cause to equal $0$.
Since $(3x+2)^3$ is really just $(3x+2)(3x+2)(3x+2)$, i.e. $(3x+2)$ multiplied with itself thrice, the zero $-\frac23$ has multiplicity $3$.
Similarly, the zero $5$ has multiplicity 5 (since $(x-5)^5=(x-5)(x-5)(x-5)(x-5)(x-5)$).