If $2x^3+4x^2+6x+8=2468$ and $x$ is a positive number, what is the value of $x^3+9x^2+8x+8$ If $2x^3+4x^2+6x+8=2468$ and $x$ is a positive number, what is the value of $x^3+9x^2+8x+8$
I'm trying to find $x$:
this is what i did so far:
$2x^3+4x^2+6x+8=2468$
Divide by $2$
$x^3+2x^2+3x+4=1234$
Minus $4$, $x^3+2x^2+3x=1230$
factor out an $x$
$x(x^2+2x+3)=1230$
and now, i'm stuck.

If you notice that $2,4,6,8$ correspond to the coefficients of the polynomial $2x^{3} + 4x^{2} + 6x + 8$, then $2x^{3} + 4x^{2} + 6x + 8 = 2468$ is implied by $x =$ ?
Using this root and the factor theorem, we can then arrive at $(x-10)(x^{2} + 12x + 123) = 0$, which shows that $x=10$ is the only (real; for complex numbers cannot be ordered like the reals) solution.
