Since $P$ is cubic and you know two roots from the table, think of writing $P(x)$ as $a(x-1)(x-3)$ times one more linear factor.

After you write $P(x)=a(x-1)(x-3)(x-r)$, plug in $x=0$ and $x=2$ using the table values to create two equations involving $a$ and $r$.

Once you have determined $a$ and $r$ and have a complete expression for $P(x)$, substitute $x=4$ to find $P(4)$. Do your sign arithmetic carefully.

In a Desmos expression line, type P(x) = -4/3*(x-1)*(x-3)*(x+1) to define the polynomial. You should see its graph appear.

Add the points (0,-4), (1,0), (2,4), and (3,0) to Desmos (for example, by typing them as separate expressions). Check that each point lies on the graph of $y=P(x)$, confirming the polynomial is correct.

In a new expression line, type P(4) and let Desmos compute the value, or move the cursor along the graph to $x=4$ and read off the corresponding y-value. That y-value is $P(4)$.

From the table, $P(1)=0$ and $P(3)=0$.

That means $x=1$ and $x=3$ are roots of the polynomial, so $(x-1)$ and $(x-3)$ are factors of $P(x)$. Since $P$ is a cubic (degree 3), it must have the form

for some constants $a$ and $r$ (where $r$ is the third root, which we do not know yet).

Plug $x=0$ into $P(x)=a(x-1)(x-3)(x-r)$ and use the table value $P(0)=-4$:

So

This gives a relationship between $a$ and $r$. We will use another point to find their exact values.

Now plug $x=2$ into the factored form and use $P(2)=4$ from the table:

This simplifies to

But $P(2)=4$, so

From the earlier step we have $ar = \tfrac{4}{3}$, so $a = \tfrac{4}{3r}$. Substitute this into $a(r-2)=4$:

Divide both sides by $4$:

So

Then from $ar=\tfrac{4}{3}$,

So the full polynomial is

Now plug $x=4$ into the polynomial we found:

Compute the factors inside the parentheses:

• $4-1=3$
• $4-3=1$
• $4+1=5$

So

The $3$ in the numerator cancels with the $3$ in the denominator, giving

Therefore, $P(4) = -20$.