Source : Moses Richardson, College Algebra (1947) , ยง111, page $241$.
Note : Richardson is also the author of a ( to my eyes) admirable introductory book called Fundamentals Of Mathematics ( available at Archive.org).
The theorem reads as follows:
If $r$ is a positive root of the [polynomial] equation $f(x)=0$ with real
coefficients , then the quotient $Q(x)$ obtained by dividing $f(x)$ by
$(x-r)$ has at least one less variation in sign than $f(x)$ has.
The author places himself in the case where the leading coefficient is positive ( which can be obtained without changing the solution set of the equation by multiplying $f(x)$ by $-1$).
He only gives an " outline of the proof" , using as a paradigmatic example the equation $ f(x)= x^{8}+x^{6}+2x^{5}-4x^{4}-80x^{2}+x+2$
He operates the synthetic division of $f(x)$ by $2$ and reasons on the coefficients of the quotient $Q(x)$ , that is, on the third line of the division tableau.
(1) He first explains that the first coefficient of $Q(x)$ will be positive and will not change to negative before $f(x)$ changes in sign. So, $Q(x)$ will not have , so far, more change of signs than $f(x)$.
(2) Once the coefficients of $Q(x)$ have ( possibly) become negative, they will remain so until $f(x)$ changes from negetive to positive . Same consequence as in (1) .
Points (1) and (2) show clearly that $Q(x)$ cannot have more changes of signs than $f(x)$.
The third point aims at establishing the " not more,and even at least one less" part of the theorem.
Here is the argument :
" Since the last number in the third line of synthetic division must
be $0$ the last coefficient of $Q(x)$ must have the opposite sign from
the last coefficient of $f(x)$ [ this I understand well] . Hence the
largest number of variations $Q(x)$ can have is one less than the
number of variations of $f(x)$ and it may have fewer than that."
I cannot see how what precedes " hence" entails what comes after it.
I suppose the logical link is quite obvious, since Richardson is a lucid and reader friendly author, but I miss something here.
Best Answer
The leading coefficients of $Q(x)$ and $f(x)$ have the same sign, and the last coefficients of $Q(x)$ and $f(x)$ have opposite signs; hence, $Q(x)$ and $f(x)$ have a different number of sign changes. Since $Q(x)$ does not have more sign changes than $f(x)$, it must have fewer.