Divisible by this year? (Part 8: Get JINXED! Again.)

Let $f(x) = x^4 + 1989x^3 -8162747x^2+202809800x$

also $g(x) = e^{2014x}-1$

and $h(x) = \ln { (x-10069) }$

We must equate all these factors to $0$ .

Factoring $f(x)$ we will get

$f(x) = x(x - 25)(x - 2014)(x + 4028) = 0$

Thus the roots of $f(x)$ are $x = 0, 25, 2014, -4028$

Equating $g(x) = 0$ gives

$g(x) = 0$

$e^{2014x}-1 = 0$

$e^{2014x} = 1$

$e^{2014x} = e^0$

$2014x = 0$

$x = 0$

Equating $h(x)=0$ gives

$h(x) = 0$

$\ln { (x-10069) } = 0$

$\ln { (x-10069) } = \ln1$

$x - 10069 = 1$

$x = 10070$ .

These are not yet the roots of $p(x)$ . To get the roots of $p(x)$ , we must SUBSTITUTE them back to the original equation.

Only $x = 10070$ satisfies $p(x) = 0$ since the other values of $x$ makes the natural logarithm part undefined.

Since $10070$ is divisible by $2014$ , the probability will be $100$ %