Question 70·Hard·Math

For a polynomial, what factor do you get if you know that $p(a)=0$? Think about an expression like $(x-a)$ and how it behaves when $x=a$.

Once you know all the $x$-values from the table where $p(x)=0$, ask: which answer choice represents factors that would give exactly those zeros?

For any choice that includes a factor $(x-c)$, that would force $p(c)=0$. Compare each candidate's implied zeros with the $p(x)$ values in the table to see which one must be true.

In Desmos, add a new table (using the "+" button, then "Table"). In the first column (for $x$), enter $-2$, $-1$, $0$, $1$, and $2$. In the second column (for $p(x)$), enter $5$, $0$, $-3$, $-1$, and $0$ respectively so you have the points $(-2,5)$, $(-1,0)$, $(0,-3)$, $(1,-1)$, and $(2,0)$ plotted.

Look at the plotted points and find which ones lie on the $x$-axis (where the $y$-value is $0$). The $x$-coordinates of those points are the zeros of $p(x)$, which correspond to linear factors of the form $(x-a)$.

Using the zeros you observed, think about which answer choice has factors that would produce exactly those zeros (via expressions of the form $(x-a)$). Pick the choice whose factors align with the $x$-coordinates of the points lying on the $x$-axis in your Desmos plot.

Look at the $p(x)$ column in the table and identify which $x$-values give $p(x)=0$.

From the table:

• $p(-1)=0$
• $p(2)=0$

So $x=-1$ and $x=2$ are zeros (roots) of the polynomial $p$.

For any polynomial $p(x)$, the Factor Theorem says:

• If $p(a)=0$, then $(x-a)$ is a factor of $p(x)$.
• Equivalently, each zero $x=a$ of $p$ gives a linear factor $(x-a)$.

So each zero you found in the table corresponds to a linear factor of $p(x)$.

Let the two zeros be $a=-1$ and $b=2$.

By the Factor Theorem, the polynomial has linear factors $(x-a)$ and $(x-b)$.

A factor of $p(x)$ that has both zeros built in is the product of these two linear factors, $(x-a)(x-b)$.

Now you just need to substitute the actual values of $a$ and $b$ and compare with the answer choices.

We have $a=-1$ and $b=2$.

• The factor from $a=-1$ is $(x-(-1)) = (x+1)$.
• The factor from $b=2$ is $(x-2)$.

Combining both zeros into one factor gives $(x+1)(x-2)$.

Among the answer choices, this matches choice D, so $(x+1)(x-2)$ must be a factor of $p$ based on the table.