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Question 142·Hard·Math

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A cubic polynomial $P$ satisfies the following table of values.

$x$	$0$	$1$	$2$	$3$
$P(x)$	$-4$	$0$	$4$	$0$

What is the value of $P(4)$ ?

Your answer

When you see a polynomial of known degree with several function values in a table, first look for x-values where the polynomial equals zero; these give you linear factors immediately. For a cubic, two distinct zeros tell you it has the form $a(x-r_1)(x-r_2)(x-r_3)$ . Use one or two additional table entries to solve for the remaining factor and the leading coefficient by plugging in the corresponding x-values and forming equations. Once you have the factored form, substitute the requested x-value and compute carefully, watching signs to avoid simple arithmetic errors.

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Look for zeros in the table

Which x-values in the table give $P(x)=0$ ? What does that tell you about factors of the polynomial?

Use the fact that P is a cubic

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.

Use given values to find the unknowns

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 $r$ .

Evaluate after finding the formula

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.

Desmos Guide

Enter the polynomial using the factored form

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.

Verify it matches the table points

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.

Find P(4) from Desmos

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 $P(4)$ .

Step-by-step Explanation

Use the zeros from the table to factor P(x) partly

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

P(x) = a(x-1)(x-3)(x-r)

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

Use P(0)=-4 to relate a and r

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

P(0) = a(0-1)(0-3)(0-r) = a(-1)(-3)(-r) = a(-3r) = -4.

-3ar = -4 \quad\Rightarrow\quad ar = \frac{4}{3}.

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

Use P(2)=4 to solve for a and r

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

P(2) = a(2-1)(2-3)(2-r) = a(1)(-1)(2-r).

This simplifies to

P(2) = a(r-2).

But $P(2)=4$ , so

a(r-2) = 4.

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

\frac{4}{3r}(r-2) = 4.

Divide both sides by $4$ :

\frac{r-2}{3r} = 1.

r - 2 = 3r \quad\Rightarrow\quad -2 = 2r \quad\Rightarrow\quad r = -1.

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

a(-1) = \frac{4}{3} \quad\Rightarrow\quad a = -\frac{4}{3}.

So the full polynomial is

P(x) = -\frac{4}{3}(x-1)(x-3)(x+1).

Evaluate P(4) using the factored form

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

P(4) = -\frac{4}{3}(4-1)(4-3)(4+1).

Compute the factors inside the parentheses:

$4-1=3$
$4-3=1$
$4+1=5$

P(4) = -\frac{4}{3} \cdot 3 \cdot 1 \cdot 5.

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

P(4) = -4 \cdot 5 = -20.

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

Question 142·Hard·Math

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Step-by-step Explanation

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Strategy

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Step-by-step Explanation