Calculating the number of multiplications necessary to evaluate a polynomial
I am not sure what the slide was intended for, so I just gave a
straight answer to your question in a comment, viz that the slide
really intended 3 multiplications for $4x^3=4\times x\times x\times
x$, and 4 additions to add the five terms.

Now of course, there are better ways to do that. Various powers of $x$
can be obtained in multiplying other powers of $x$ together using the
rule $x^m\times x^n=x^{m+n}$.

The method of repeated squaring presente by David Richerby in his
answer is a systematic way to get high powers of $x$ as fast as
possible. However, when you need a few specific powers, you can try to
adapt it to compute just these powers with as few multiplications as
possible.

So you can compute, for example:

$\begin{align}
x^2&amp;=x\times x \\
x^3&amp;=x^2\times x \\
x^6&amp;=x^3\times x^3 \\
x^{12}&amp;=x^6\times x^6 \\
x^{24}&amp;=x^{12}\times x^{12} \\
x^{25}&amp;=x^{24}\times x \\
x^{49}&amp;=x^{24}\times x^{25} \\
x^{50}&amp;=x^{25}\times x^{25} \\
x^{99}&amp;=x^{49}\times x^{50}
\end{align}$

Hence you get all needed powers of $x$ with only 9 multiplications of
powers of $x$, to which you have to add 4 multiplications for the
coefficients, which makes a total of 13 multiplications. The number of
additions remains at 4.

But you can do a bit better using a technique that is most
efficient when all powers of $x$ are being used, rather than having a sparse
polynomial. The technique, known as Horner's method, consists in
factorizing some powers of the variable, when it helps the
computation.

In the case of your sparse polynomial, you may notice that
$7x^{25}+6x^{99}=x^{25}(7+6x^{74})$. Hence rather than computing
$x^{50}$ and $x^{99}$, you compute only

$x^{74}=x^{49}\times x^{25}$ and then $x^{25}(7+6x^{74})$.

This saves one multiplication, because both $7x^{25}+6x^{99}$ and
$x^{25}(7+6x^{74})$ take exactly 2 multiplications and one addition,
assuming the powers of $x$ are already computed, but computing the
needed powers takes one less multiplication.

So you are down to a total of 12 multiplications and 4 additions.

I do not see any way to do better on this example, but I have no
systematic algorithm to check that. However, I am rather confident
that someone must have worked on it.

BTW, what was the theme of the set of slides?