0
$\begingroup$

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?

$\endgroup$
2
  • 1
    $\begingroup$ Any real roots are negative as the polynomial is $5$ when $x=0$ and increasing above that. $\endgroup$ – Ross Millikan Jan 8 '20 at 1:34
  • 1
    $\begingroup$ You could use the cubic discriminant. If it is positive, three real roots. Negative, one real root and two non-real complex conjugate roots. $\endgroup$ – alex.jordan Jan 8 '20 at 1:36
3
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ want to use that, at -1, the graph will be just touching the x-axis. right? $\endgroup$ – sabeelmsk Jan 8 '20 at 4:21
  • $\begingroup$ @sabeelmsk At $-1$, the graph of $(x+5)(x+1)(x+1)$ touches the $x$-axis and turns around. Then adding $-5x$ pushes that low point up (and a little to the right). So now the local low point is above the $y$-axis, and the only real root is where the graph crosses the $x$-axis off to the far left. $\endgroup$ – alex.jordan Jan 8 '20 at 4:26
2
$\begingroup$

If there are three real roots, the value of the function must be of opposite signs at the points the derivative is zero.

$\endgroup$
2
  • $\begingroup$ this is not easy to check $\endgroup$ – sabeelmsk Jan 8 '20 at 4:13
  • 2
    $\begingroup$ Yes it is. Take the derivative, which gives a quadratic. Use the quadratic formula. That gives you two $x$ values to plug into your original cubic. It is some work, but is rigorous and will work for any cubic. If you see an easier route, use it. $\endgroup$ – Ross Millikan Jan 8 '20 at 4:37
0
$\begingroup$

If you do not want to use the cubic discriminant, the easy solution has been proposed by Ross Millikan.

Consider that you lookf for the zero's of function $$f(x)=x^3+7x^2+6x+5$$ for which $$f'(x)=3x^2+14x+6 \qquad \text{and} \qquad f''(x)=6x+14$$ The first derivative cancels for $$x_\pm=-\frac{1}{3} \left(7\pm\sqrt{31}\right)$$ and you have $$f''(x_+)=2 \sqrt{31}>0\qquad \text{and} \qquad f''(x_-)=-2 \sqrt{31}<0$$ So, by the second derivative test, $x_+$ corresponds to a local minimum and $x_-$ to a local maximum.

Now $$f(x_+)=\frac{443-62 \sqrt{31}}{27} >0\qquad \text{and} \qquad f(x_-)=\frac{443+62 \sqrt{31}}{27} >0$$ Since $f(x_-)>0$ there is only one real root which will be on the left of $x_-$.

Based on this simply acquired information, we can even obtain an estimate of the solution writing $$f(x)=f(x_-)+\frac 12 f''(x_-)(x-x_-)^2\implies x =x_--\sqrt{-2\frac {f(x_-)}{f''(x_-)}}$$ $$x=-\frac{1}{3} \left(7+\sqrt{31}\right)-\sqrt{\frac{443+62 \sqrt{31}}{27 \sqrt{31}}}\approx -6.47905$$ while the exact solution is $$x=-\frac{1}{3} \left(7+2 \sqrt{31} \cosh \left(\frac{1}{3} \cosh ^{-1}\left(\frac{443}{62 \sqrt{31}}\right)\right)\right)\approx -6.15745$$

$\endgroup$
0
$\begingroup$

If you take $$f(x)=x^3+7x^2+6x+5$$ and subtract $5$, you have $$g(x)=(x+6)(x+1)x$$ a cubic with roots at $-6$, $-1$, and $0$. Note the graph goes down through the root at $-1$ and up through the root at $0$.

To recover the original, you need to add $5$ to $g$. It's hard to believe that adding $5$ will fail to pull the entire segment of the graph between $-1$ and $0$ up above the $x$-axis. Indeed, if $x$ is in $[-1,0]$, then $$\lvert g(x)\rvert=\lvert (x+6)(x+1)x\rvert >5\lvert(x+1)x\rvert\geq5\left(\frac14\right)$$ So $g$'s local minimum is greater than $-\frac54$. Adding $5$ definitely takes the local minimum up higher than the $x$-axis. This leaves only one root for $f$.

$\endgroup$
0
$\begingroup$

The first derivative is $f'(x)=3x^2+14x+6$. We observe that this is negative when $x=-1$: $f'(-1)=-5$. From this we consider the tangent line $y=-5(x+1)+5$.

There is another tangent line where the graph crosses the $y$-axis: $y=6x+5$.

We can look at the second derivative to see that the curve's inflection point happens when $x=-\frac{7}{3}$. The value doesn't matter, just note this is to the left of $-1$.

After solving the system of equations from the two lines, we can find they cross at a point whose $y$-value is $\frac{11}{25}$, which is positive. It follows that the entirety of the curve to the right of the inflection point is positive. And it follows from that that there can only be one real root.

enter image description here

$\endgroup$
0
$\begingroup$

Use the Discriminant formula for a 3rd ordered polynomial.Click here!

N.B The discriminant is zero if and only if at least two roots are equal. If the coefficients are real numbers, and the discriminant is not zero, the discriminant is positive if the roots are three distinct real numbers, and negative if there is one real root and two complex conjugate roots

Coming to the ans of your question as discriminant is coming negative , hence there is one real root and two complex conjugate roots

$\endgroup$
-1
$\begingroup$

If you actually want the roots of a cubic, you can use the rational root test to find one of the roots. After you have one of the roots you can easily do a long division to find the resulting second degree polynomial from there you just apply the quadratic formula. This will yield the exact roots, but if you want to just know if you have three real or one real and two complex that is a different scenario.

$\endgroup$
2
  • 3
    $\begingroup$ The cubic could have real roots without having any rational roots. $\endgroup$ – alex.jordan Jan 8 '20 at 1:41
  • $\begingroup$ you are correct, I was looking only for rational roots, we must have some requirements on the polynomial to expect only rational roots. $\endgroup$ – Kori Jan 8 '20 at 1:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.