Answer
a. $x=-5$.
b. The multiplicity for zero $x=-5$ is $2$ and is even.
c. The maximum possible number of turning points is $1$.
d. See graph
Work Step by Step
a.
Set $f(x)=0$:
$$0=x^2+10x+25$$ $$0=(x+5)^2$$ $$x+5=0$$ $$x=-5$$ Thus, the real zero is $x=-5$.
b.
$$f(x)=(x+5)^2$$ From the power of the factor $(x+5)^2$ which is $2$, the multiplicity for zero $x=-5$ is $2$ and is even.
c.
Since the power of the leading coefficient of the polynomial $x^2+10x+25$ is $2$, the maximum possible number of turning points is: $$n-1=2-1=1$$
d. The graph of the function is as shown and it shows the zero is $x=-5$, the multiplicity of zero $x=-5$ is even since the graph bounces off the $x$-axis, and number of turning points is $1$.
