I want to find the number of real roots of the polynomial $x^3+7x^2+6x+5$.
Using Descartes rule, this polynomial has either only one real root or 3 real roots (all are negetive). How will we conclude one answer without doing some long process?
Find number of real roots of the polynomial $x^3+7x^2+6x+5$.
calculuselementary-functionspolynomialsreal-analysis
Related Solutions
Let $$f(x)=x^4-4x^3+12x^2+x-1\;,$$ Then $$f'(x) = 4x^3-12x^2+24x$$
and $$f''(x)=12x^2-24x+24 = 12(x^2-2x+2)=12[(x-1)^2+1]>0\;\forall \;x\in \mathbb{R}$$
Now Using Rolles Theorem If $f(x)=0$ has $n$ real Roots, Then $f'(x)=0$ has at least $n-1$ real roots
and $f''(x)=0$ has at least $n-2$ real roots.
Or in other words If $f''(x)=0$ has $n$ roots, Then $f'(x)=0$ has at most $n+1$ roots
and $f''(x)=0$ has at most $n+1$ roots.
Now here $f''(x)=0$ has no real roots, Then $f'(x)=0$ has at most one real root and $f(x)=0$
has at most $2$ real roots.
Now Here $f(0)=-1$ and $f(1)=9$. So one root lie between $(0,1)$
and $f(-1)=15$ and $f(0) = -1$. So other root lie between $(-1,0)$
So $$f(x)=x^4-4x^3+12x^2+x-1 =0$$ has exactly two real roots.
The best way to solve this is to use Sturm's theorem. This gives an algorithm for computing the number of distinct real roots of any polynomial. The Wikipedia page is quite good, but I'll outline the method here.
Let $f(x)$ be a polynomial. We define a sequence as follows: $$P_0=f$$ $$P_1=f'$$ $$P_{n+2}=-P_{n}\text{ mod }P_{n+1}$$ where $f'$ is the derivative of the polynomial and, for polynomials $P$ and $Q$, we define $P\text{ mod }Q$ to be the remainder of dividing $P$ by $Q$ - that is, the unique polynomial $R$ of degree less than $\deg Q$ such that $P=cQ+R$ for some other polynomial $c$. (This is also just the result you get by polynomial long division)
For instance, suppose we want to know how many roots $f(x)=x^3+2x+1$ has using this method - of course, we know the answer is $1$, but we should check. We get the following chain: $$P_0=x^3+2x+1$$ $$P_1=3x^2+2$$ $$P_2=-\frac{4}3x-1$$ $$P_3=\frac{-59}{16}.$$
For any real number $a$, we define $V(a)$ to be the number of sign changes in the sequence $P_0(a),P_1(a),P_2(a),P_3(a)$, where we ignore any zeros. Assuming neither $a$ or $b$ are themselves roots, Sturm's theorem states that $V(a)-V(b)$ is the number of real roots between $a$ and $b$.
Note that $V(-\infty)=\lim_{a\rightarrow-\infty}V(a)$ or $V(\infty)=\lim_{b\rightarrow\infty}V(b)$ are easy to compute by looking at the leading terms of each polynomial. For instance, here we have that $V(-\infty)=2$ since, towards $-\infty$ we have that $P_0$ tends to $-\infty$, $P_1$ to $\infty$, $P_2$ to $\infty$ and $P_3$ is negative - so two sign changes. Then $V(\infty)=1$ because $P_0$ and $P_1$ are positive near $\infty$ and $P_2$ and $P_3$ are negative. This polynomial has $V(-\infty)-V(\infty)=1$ roots, as expected, since it is an increasing function.
This can be a bit laborious to do by hand, but it always works for any polynomial.
The only trick to proving this, at least in the square-free case, is to consider what happens to sign changes in this sequence as one moves along the real line: The number of sign changes can only change near a root of one of the polynomials. However, note that, for some polynomial $c$, we have the following relationship: $$P_{n}=cP_{n+1}-P_{n+2}$$ Note that if $P_{n+1}$ has a root at a place where $P_n$ doesn't, then near that root, $P_n$ and $P_{n+2}$ must have opposite signs, since $P_n=-P_{n+2}$ at the root. So long as $P_0$ was squarefree (i.e. has no multiple roots), we can note that no consecutive terms share a root, so this always happens. As a result, the zero of $P_{n+1}$ does not affect the number of sign changes. However, if $P_0$ has a root, then the number of sign changes decreases by one there, since, near that root, $f$ and $f'$ have opposite signs prior to the root and equal signs after.
Best Answer
I note that this is "close" to $$(x+5)(x+1)(x+1)=x^3+7x^2+11x+5$$ which has a repeated root at $-1$, and another root at $-5$. The repeated root at $-1$ is a local minimum, considering the general shape of a cubic with positive leading coefficient.
So you have $$(x+5)(x+1)(x+1)-5x$$
Adding that $-5x$ is going to push the local minimum upward, since $-5x$ is positive near $-1$. The doubled root will be perturbed into two non-real complex conjugate roots. And only the perturbed root near $-5$ will still be real.