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No. of real solutions in polynomial equation


Well-known member
Apr 5, 2012
The no. of real solution of the equation $\displaystyle 1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!} = 0$


Well-known member
MHB Math Helper
Mar 22, 2013
Re: no. of real solution in polynomial equation

Analysis of the discriminant shows that the discriminant is 1/576 which is obviously > 0. So, either there are two pairs of complex conjugate roots or none at all.

So, check if the additional discriminant (\(\displaystyle 8ac - 3b^2\)) is greater or smaller than 0. In this case, it is -1/3 < 0, and hence all the roots are distinct and real.
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MHB Oldtimer
Staff member
Feb 7, 2012
Re: no. of real solution in polynomial equation

$ 1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!} = \frac1{4!}(x^4 + 4x^3 + 12x^2 + 24x + 24) = \frac1{4!}\bigl((x^2+2x+2)^2 + 4(x+2)^2 + 4\bigr) >0$ for all real $x$, so the equation has no real solutions.