Real Roots

$P(x)=x^9-37x^8-2x^7+74x^6+x^4-37x^3-2x^2+74x$

First of all factor out an $x$

$P(x)=x(x^8-37x^7-2x^6+74x^5+x^3-37x^2-2x^1+74$

Notice that inside the parenthesis the first $4$ terms are just the last $4$ terms multiplyed by $x^5$ , hence

$P(x)=x(x^5+1)(x^3-37x^2-2x+74)$

With a similar reasoning

$\begin{aligned} P(x)&=x(x^5+1)[x^2(x-37)-2(x-37)]\\ &=x(x^5+1)(x^2-2)(x-37)\\ &=x(x^5+1)(x+\sqrt{2})(x-\sqrt{2})(x-37) \end{aligned}$

Thus there are only $\boxed{5}$ real roots, namely

$0,-1,\sqrt{2},-\sqrt{2},37$

The other $4$ roots are the imaginary fifth roots of $-1$ :

$x_1=\dfrac{1+\sqrt{5}}{4}+i\sqrt{\dfrac{5-\sqrt{5}}{8}}$

$x_2=\dfrac{1+\sqrt{5}}{4}-i\sqrt{\dfrac{5-\sqrt{5}}{8}}$

$x_3=\dfrac{1-\sqrt{5}}{4}+i\sqrt{\dfrac{5+\sqrt{5}}{8}}$

$x_4=\dfrac{1-\sqrt{5}}{4}-i\sqrt{\dfrac{5+\sqrt{5}}{8}}$

$\begin{aligned} x^9 -37x^8-2x^7+74x^6 + x^4 -37x^3-2x^2+74x & = 0 \\ x^9 + x^4 -37x^8-37x^3-2x^7-2x^2+74x^6 +74x & = 0 \\ \left(x^5 + 1\right)\left(x^4 -37x^3-2x^2+74x\right) & = 0 \\ x\left(x^5 + 1\right)\left(x^3 -37x^2-2x+74\right) & = 0 \\ x\left(x^5 + 1\right)\left(x^2(x-37)-2(x-37)\right) & = 0 \\ x\left(x^5 + 1\right)\left(x-37\right)\left(x^2 - 2\right) & = 0 \\ x\left(x^5 + 1\right)\left(x-37\right)\left(x - \sqrt 2 \right)\left(x + \sqrt 2\right) & = 0 \end{aligned}$

Therefore, there are $\boxed{5}$ real numbers $\{0, -1, -37, \sqrt 2, - \sqrt 2\}$ that are roots to the polynomial.