How many real roots does $f(x) = 2x^5 - 3x^4 + x^3 - 4x^2 - 6x + 4$ have? A multiple choice questions asks me to find the number of real roots of $f(x) = 2x^5 - 3x^4 + x^3 - 4x^2 - 6x + 4$.
A brute force approach involving the factor theorem and algebraic long division gives $3$ real roots.
Is there a quicker way?

I was thinking Descartes rule of signs tells us that since there are $4$ sign changes, there are at most 4 positive roots. We can also use the rule on $f(-x)$ to say there is at most one negative root. However, I can't see how to continue along this path.
I tried the derivative and deduced that there at least two turning points on the curve $y=f'(x) = 10x^4 - 12x^3 + 3x^2 - 8x - 6$, since $10 &gt; 0$ and the $y$ intercept is negative. But that doesn't rule out anything since the turning points could be above the axis.
Of course I could work this out in more detail, but I am looking for efficiency.
We only need to use Descartes' rule of signs.
When $x\le0$ you have worked out that $f(x)$ has one negative root.
As $f(x+1)=2x^5+7x^4+9x^3+x^2-13x-6$, there is one sign change so $f(x)$ has one positive root when $x\ge1$.
Consider $g(x)=x^5f(1/x)$. As $g(x+1)=4x^5+14x^4+12x^3-7x^2-17x-6$, there is one sign change so $g(x)$ has one positive root when $x\ge1$. Hence $f(x)$ has one positive root when $0&lt;x&lt;1$.
Therefore $f(x)$ has three real roots.
Using  factorization it's very quickly here:
$$2x^5 - 3x^4 + x^3 - 4x^2 - 6x + 4=$$
$$=2x^5+2x^4-5x^4-5x^3+6x^3+6x^2-10x^2-10x+4x+4=$$
$$=(x+1)(2x^4-5x^3+6x^2-10x+4)=$$
$$=(x+1)(2x^4-4x^3-x^3+2x^2+4x^2-8x-2x+4)=$$
$$=(x+1)(x-2)(2x^3-x^2+4x-2)=(x+1)(x-2)(2x-1)(x^2+2),$$
which gives three real roots.