Question

if $f(y) = 11y^2 - 31y + 1$, use synthetic division to find $f(3)$.

Explanation:

Step1: Set up synthetic division

We are dividing \(11y^{2}-31y + 1\) by \(y - 3\) (since we want to find \(f(3)\), we use \(c = 3\) for synthetic division). The coefficients of the polynomial are \(11\), \(-31\), and \(1\).
We write the coefficients and the value \(3\) as follows:
\[

$$\begin{array}{r|rrr} 3 & 11 & -31 & 1 \\ & & & \\ \hline & 11 & & \\ \end{array}$$

\]

Step2: Bring down the leading coefficient

Bring down the \(11\) to the bottom row:
\[

$$\begin{array}{r|rrr} 3 & 11 & -31 & 1 \\ & & & \\ \hline & 11 & & \\ \end{array}$$

\]

Step3: Multiply and add

Multiply the number we just brought down (\(11\)) by \(3\): \(11\times3 = 33\). Add this to the next coefficient (\(-31\)): \(-31+33 = 2\). Write this result in the bottom row:
\[

$$\begin{array}{r|rrr} 3 & 11 & -31 & 1 \\ & & 33 & \\ \hline & 11 & 2 & \\ \end{array}$$

\]

Step4: Multiply and add again

Multiply the new number in the bottom row (\(2\)) by \(3\): \(2\times3=6\). Add this to the last coefficient (\(1\)): \(1 + 6=7\). Write this result in the bottom row:
\[

$$\begin{array}{r|rrr} 3 & 11 & -31 & 1 \\ & & 33 & 6 \\ \hline & 11 & 2 & 7 \\ \end{array}$$

\]

The last number in the bottom row is the value of \(f(3)\) when we use synthetic division (by the Remainder Theorem, the remainder when dividing by \(y - c\) is \(f(c)\)).

Answer:

\(7\)