Answer
a. $x=\frac{-5-\sqrt{37}}{2}$ and $x=\frac{-5+\sqrt{37}}{2}$.
b. The multiplicity for zero $x=\frac{-5-\sqrt{37}}{2}$ is $1$ and is odd.
The multiplicity for zero $x=\frac{-5+\sqrt{37}}{2}$ is $1$ and is odd.
c. The maximum possible number of turning points is $1$.
d. See graph
Work Step by Step
a.
Factoring out $\frac{1}{2}$ of the polynomial:
$$f(x)=\frac{1}{2}(x^2+5x-3)$$
Set $f(x)=0$:
$$0=\frac{1}{2}(x^2+5x-3)$$ $$0=x^2+5x-3$$ $$x=\frac{-5\pm\sqrt{5^2-4(1)(-3)}}{2(1)}=\frac{-5\pm\sqrt{37}}{2}$$ $$x_1=\frac{-5-\sqrt{37}}{2}$$ $$x_2=\frac{-5+\sqrt{37}}{2}$$
Thus, the real zeros are $x=\frac{-5-\sqrt{37}}{2}$ and $x=\frac{-5+\sqrt{37}}{2}$.
b.
$$f(x)=\frac{1}{2}\left(x+\frac{-5-\sqrt{37}}{2}\right)^1\left(x-\frac{-5+\sqrt{37}}{2}\right)^1$$
From the power of the factor $\left(x+\frac{-5-\sqrt{37}}{2}\right)^1$ which is $1$, the multiplicity for zero $x=\frac{-5-\sqrt{37}}{2}$ is $1$ and is odd.
From the power of the factor $\left(x-\frac{-5+\sqrt{37}}{2}\right)^1$ which is $1$, the multiplicity for zero $x=\frac{-5+\sqrt{37}}{2}$ is $1$ and is odd.
c.
Since the degree of the polynomial $\frac{1}{2}x^2+\frac{5}{2}x-\frac{3}{2}$ 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 zeros are $x=\frac{-5-\sqrt{37}}{2}$ and $x=\frac{-5+\sqrt{37}}{2}$, the multiplicity of zero $x=\frac{-5-\sqrt{37}}{2}$ is odd since the graph crosses the $x$-axis and the multiplicity of zero $x=\frac{-5+\sqrt{37}}{2}$ is odd since the graph crosses also the $x$-axis, and the number of turning points is $1$.
