How Many Real Roots Does This Equation Have?

[anemone]

Here is this week's POTW:

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Find the number of real roots of the equation $1+\dfrac{x}{1!}+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\dfrac{x^4}{4!}+\dfrac{x^5}{5!}+\dfrac{x^6}{6!}=0$.

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[anemone]

Congratulations to the following members for their correct solution!(Cool)

1. lfdahl
2. Opalg

Solution from Opalg:

Spoiler

Let $P(x) = 1+\dfrac{x}{1!}+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\dfrac{x^4}{4!}+\dfrac{x^5}{5!}+\dfrac{x^6}{6!}$. Then
$$\begin{aligned}720P(x) &= x^6 + 6x^5 + 30x^4 + 120x^3 + 360x^2 + 720x + 720 \\ &= (x^3 + 3x^2 + 6x + 15)^2 + 9x^4 + 54x^3 + 234x^2 + 540x + 495 \\ &= (x^3 + 3x^2 + 6x + 15)^2 + 9(x^4 + 6x^3 + 26x^2 + 60x + 55) \\ &= (x^3 + 3x^2 + 6x + 15)^2 + 9\bigl((x^2 + 3x + 4)^2 + 9x^2 + 36x + 39\bigr) \\ &= (x^3 + 3x^2 + 6x + 15)^2 + 9\bigl((x^2 + 3x + 4)^2 + 9(x + 2)^2 + 3\bigr) \\ &= (x^3 + 3x^2 + 6x + 15)^2 + 9(x^2 + 3x + 4)^2 + 81(x + 2)^2 + 27.\end{aligned}$$ That is a sum of squares, so it is always positive (with minimum value at least $27$). Therefore $P(x)$ is always positive and so the equation $P(x) = 0$ has no real roots.