After substituting $x=1$, carefully compute $(1+3)^2$ and $(1-5)$, and use $p(1)=192$ to write an equation involving only $k$ and constants.

Once you know $k$, substitute $x=0$ into $p(x)$ and simplify $(0+3)^2$ and $(0-5)$, then multiply everything together.

Both $(1-5)$ and $(0-5)$ are negative; check that you keep track of these signs when solving for $k$ and when finding $p(0)$.

In Desmos, type the expression k = 192 / ((1+3)^2 * (1-5)). Desmos will compute the numerical value of $k$ for you; note this value for the next step.

Now type the expression p0 = k * (0+3)^2 * (0-5) using the same $k$ value from Step 1. The numeric result that Desmos shows for p0 is the value of $p(0)$.

Start by substituting $x=1$ into the definition of $p(x)$:

You are told that $p(1)=192$, so set this equal to 192:

Now simplify the numerical parts $(1+3)$ and $(1-5)$.

Compute the numerical expressions:

• $1+3 = 4$, so $(1+3)^2 = 4^2 = 16$.
• $1-5 = -4$.

So the equation becomes:

Multiply $16$ and $-4$:

Now solve for $k$ by dividing both sides by $-64$:

Simplify this fraction to get the value of $k$.

Once you have $k$, substitute $x=0$ into the original function:

Simplify the numerical parts:

• $0+3 = 3$, so $(0+3)^2 = 9$.
• $0-5 = -5$.

So

Now plug in your value of $k$ from Step 2 into $-45k$ and simplify. This calculation gives $p(0)=135$.