$P(x) = 2x^{98} + 5x^{97} + 5x^{96} + 5x^{95} +\cdots + 5x + 3 = 0$

If the equation above holds true, which of the following is true?

-2 is a root of
$P(x)=0$
.
$-\dfrac 32$
is a root of
$P(x)=0$
.
The number of rational roots is 3.

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$\begin{aligned} P(x) & = 2x^{98}+5x^{97}+5x^{96}+5x^{95}+\cdots + 5x + 3 \\ & = 5\sum_{k=0}^{98} x^k - 3x^{98} - 2 \\ & = 5\left(\frac {x^{99}-1}{x-1}\right) - 3x^{98} - 2 = 0 \end{aligned}$

Therefore,

$\begin{aligned} 5\left(\frac {x^{99}-1}{x-1}\right) & = 3x^{98} + 2 \\ 5x^{99}-5 & = (3x^{98} + 2)(x-1) \\ 5x^{99}-5 & = 3x^{99} + 2x - 3x^{98} - 2 \\ 2x^{99} + 3x^{98} - 2x - 3 & = 0 \\ (x^{98}-1)(2x+3) & = 0 \end{aligned}$

$\implies x = \begin{cases} -1 \\ - \frac 32 \end{cases}$ . Note that $x=1$ is not a solution. There are two rational root and one of them is $-\frac 32$ .