If $9^{x^2+2x-3}-4\left(3^{x^2+2x-2}\right)+27=0$  then the product of all possible values of x is

The correct answer is 20.

Let us solve the given exponential equation step-by-step:

$9^{x^2+2x-3}-4\left(3^{x^2+2x-2}\right)+27=0$

First, notice that the exponents share a common structure. Let us define a new variable to simplify the expressions. We can express the exponents in terms of $y=x^2+2x-3$. Notice that the exponent in the second term is $x^2+2x-2$, which is simply $y+1$.

Substituting $y$ into our original equation, we obtain:

$9^y-4\left(3^{y+1}\right)+27=0$

Next, we write the terms with a common base of 3. We know that $9^y={\left(3^2\right)}^y={\left(3^y\right)}^2$ and $3^{y+1}=3^y·3^1=3·3^y$. Let us plug these rewrites into our equation:

${\left(3^y\right)}^2-4·3·3^y+27=0$

${\left(3^y\right)}^2-12\left(3^y\right)+27=0$

This is now a quadratic equation in terms of $3^y$. Let $u=3^y$. Substituting $u$ gives:

$u^2-12u+27=0$

We can solve for $u$ by factoring this quadratic equation. We are looking for two numbers that multiply to 27 and add to -12. These numbers are -9 and -3.

$(u-9)(u-3)=0$

Thus, we have two possible values for $u$: $u=9$ and $u=3$. We must analyze both cases to find all possible values of $x$.

Case 1: $u=9$
Since $u=3^y$, we have $3^y=9$. Because $9=3^2$, it follows that $y=2$.

Recall that $y=x^2+2x-3$. Substituting $y=2$ yields:

$x^2+2x-3=2$

$x^2+2x-5=0$

Let the roots of this quadratic equation be $x_1$ and $x_2$. According to Vieta's formulas, for any quadratic equation $a{x}^2+bx+c=0$, the product of the roots is $\frac{c}{a}$. Therefore, the product of roots for this equation is $x_1·x_2=\frac{-5}{1}=-5$.

Case 2: $u=3$
Since $u=3^y$, we have $3^y=3$. Because $3=3^1$, it follows that $y=1$.

Again, using $y=x^2+2x-3$ and substituting $y=1$ yields:

$x^2+2x-3=1$

$x^2+2x-4=0$

Let the roots of this quadratic equation be $x_3$ and $x_4$. Using Vieta's formulas, the product of roots for this equation is $x_3·x_4=\frac{-4}{1}=-4$.

The problem asks for the product of all possible values of $x$. This means we must multiply the products of the roots from both cases together:

$\text{Total Product}=(x_1·x_2)·(x_3·x_4)$

$\text{Total Product}=(-5)·(-4)=20$

Therefore, the product of all possible values of $x$ is 20.