The minimum value of the polynomial $x(x+1)(x+2)(x+3)$ is:
a) $0$
b) $9/16$
c) $-1$
d) $-3/2$

Please answer in detail.


closed as off-topic by Gibbs, Namaste, Key Flex, Theoretical Economist, Micah Oct 7 '18 at 21:20

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  • $\begingroup$ Instead of asking us to answer us in detail, you should in detail tell us what you have tried. $\endgroup$ – Kezer Oct 7 '18 at 10:21
  • $\begingroup$ Welcome to MSE! One suggestions-Whenever you ask any question on MSE, please give your thought process or working so as the person answering will know how much to tell. So please give your workings for this question also and edit it accordingly. $\endgroup$ – jayant98 Oct 7 '18 at 10:24
  • $\begingroup$ Sir I got 3rd degree equation what to do $\endgroup$ – Mark 7 Oct 7 '18 at 10:26
  • $\begingroup$ Sir they are not x values they are functions value@smcc $\endgroup$ – Mark 7 Oct 7 '18 at 10:27
  • $\begingroup$ Then type it and edit the question. Give your workings! $\endgroup$ – jayant98 Oct 7 '18 at 10:27



  • $\begingroup$ * Thank you sir. * $\endgroup$ – Mark 7 Oct 7 '18 at 10:30
  • $\begingroup$ You only need the zeroes of $x^2+3x+1$. $\endgroup$ – Henno Brandsma Oct 7 '18 at 10:33
  • $\begingroup$ No need sir as square function at minimum can be 0 polynomial min value can be -1 $\endgroup$ – Mark 7 Oct 7 '18 at 10:36
  • $\begingroup$ @Mark7 You only need the value, so you're right; I was thinking to compute at what $x$ it is assumed. You do need to remark that $0$ is assumed at all in that square term. $\endgroup$ – Henno Brandsma Oct 7 '18 at 10:38

Hint: If $$f(x)=uvw$$ then we get $$f'(x)=u'uw+uv'w+uvw'$$ $$f'(x)=\left( x+1 \right) \left( x+2 \right) \left( x+3 \right) +x \left( x+2 \right) \left( x+3 \right) +x \left( x+1 \right) \left( x+3 \right) +x \left( x+1 \right) \left( x+2 \right) $$

  • $\begingroup$ Yeah but we get 3rd degree equation $\endgroup$ – Mark 7 Oct 7 '18 at 10:22
  • $\begingroup$ How solve it sir $\endgroup$ – Mark 7 Oct 7 '18 at 10:23

As you know, To find the minimum value of any function we differentiate it and then equate it to $0$. The reason for this is that:

The graph has horizontal tangent where the function has maximum or minimum value. So, in order to find the points where the function maximum or minimum values, first you need to find the derivative of the function and set the derivative is equal to zero and solve the equation for the points called critical points.

Now, to find the minimum value of the polynomial of your function, you can write it as:

$f(x) = x^4 + 6x^3 + 11x^2 +6x$

Differentiate with respect to $x$

$f'(x) = 4x^3 + 18x^2 + 22x +6$

Which can be written better as:

$f'(x) = (2x+3)(2x^2+6x+2)$

Now you have to equate $f'(x)$ in order to get minimum value.

$f'(x) = (2x+3)(2x^2+6x+2) = 0$

This means either $2x+3 = 0$ or $ 2x^2+6x+2 = 0$

Either $x = -3/2$ or $x=(-3 + √5)/2$ or $x=(-3 - √5)/2$

Now you can check values of $f(x)$ with these critical points and the one with the minimum value, will get you your answer.

  • $\begingroup$ $-1$ and $-2$ are not roots of $2x^2+6x+2$.... $\endgroup$ – Henno Brandsma Oct 7 '18 at 10:36
  • $\begingroup$ Oh, my mistake sorry for that! Thank you for pointing that out. I am editing it now. $\endgroup$ – PradyumanDixit Oct 7 '18 at 10:38

The maxima/minima can be found by finding the derivative of the function and setting it to equal $0$. $$f(x) = x(x+1)(x+2)(x+3) \implies f(x) = x^4+6x^3+11x^2+6x$$ $$f’x = 4x^3+18x^2+22x+6$$ The derivative should be factored so it could be solved. $$\implies 2(2x^3+9x^2+11x+3)$$ $$\implies 2[(2x^3+3x^2)+(6x^2+9x)+(2x+3)]$$ $$\implies 2[x^2(2x+3)+3x(2x+3)+1(2x+3)]$$ $$\implies 2(2x+3)(x^2+3x+1)$$

Set $f’x = 0$

$$f’x = 0$$ $$2(2x+3)(x^2+3x+1) = 0$$ $$2x+3 = 0 \implies 2x+1 = -3 \implies \boxed{x = -1.5}$$ $$OR$$ $$x^2+3x+1 = 0 \implies \boxed{x = \frac{-3\pm\sqrt{5}}{2}}$$ The final step is carrying out the second derivative test and plugging in $x$. $$f’’x = 12x^2+36x+22$$ The question asks for a minima, so $f’’x > 0$, as the slope becomes positive past the inflection point.

$$f’’-1.5 = -5$$ $$f’’\frac{-3-\sqrt{5}}{2} = 10$$ $$f’’\frac{-3+\sqrt{5}}{2} = 10$$

$\frac{-3\pm\sqrt{5}}{2}$ meets this criterion, substitute it in place of x in the original function. (Both of them give the same answer.)

$$y_{minimum} = \frac{-3\pm\sqrt{5}}{2}\biggr(\frac{-3\pm\sqrt{5}}{2}+1\biggr)\biggr(\frac{-3\pm\sqrt{5}}{2}+2\biggr)\biggr(\frac{-3\pm\sqrt{5}}{2}+3\biggr) = -1$$

So, the minimum value of the polynomial is $-1$, when $x = \frac{-3+\sqrt{5}}{2}$ or when $x = \frac{-3-\sqrt{5}}{2}$.


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