Answer
a. $x=1-\sqrt2$, $x=0$ and $x=1+\sqrt2$.
b. The multiplicity for zero $x=0$ is $1$ and is odd.
The multiplicity for zero $x=1-\sqrt2$ is $1$ and is odd.
The multiplicity for zero $x=1+\sqrt2$ is $1$ and is odd.
c. The maximum possible number of turning points is $2$.
d. See graph
Work Step by Step
a.
Set $f(x)=0$:
$$0=5x(x^2-2x-1)$$ $$0=x$$ $$x=0$$ $$0=x^2-2x-1$$ $$x=\frac{-(-2)\pm\sqrt{(-2)^2-4(1)(-1)}}{2(1)}=1\pm\sqrt2$$ $$x_1=1-\sqrt2$$ $$x_2=1+\sqrt2$$ Thus, the real zeros are $x=1-\sqrt2$, $x=0$ and $x=1+\sqrt2$.
b.
$$f(x)=5x^1\left(x-(1-\sqrt2)\right)^1\left(x-(1+\sqrt2)\right)^1$$
From the power of the factor $x^1$ which is $1$, the multiplicity for zero $x=0$ is $1$ and is odd.
From the power of the factor $(x-(1-\sqrt2))^1$ which is $1$, the multiplicity for zero $x=1-\sqrt2$ is $1$ and is odd.
From the power of the factor $(x-(1+\sqrt2))^1$ which is $1$, the multiplicity for zero $x=1+\sqrt2$ is $1$ and is odd.
c.
Since the degree of the polynomial $5x(x^2-2x-1)=5x^3-10x^2-5x$ is $3$, the maximum possible number of turning points is: $$n-1=3-1=2$$
d. The graph of the function is as shown and it shows the zeros are $x=1-\sqrt2$, $x=0$ and $x=1+\sqrt2$, the multiplicity of zero $x=1-\sqrt2$ is odd since the graph crosses the $x$-axis, the multiplicity of zero $x=0$ is odd since the graph crosses the $x$-axis and the multiplicity of zero $x=1+\sqrt2$ is odd since the graph crosses also the $x$-axis, and the number of turning points is $1$.
