Question

differentiated task #1
problem: determine if $(x - 3)$ is a factor of $f(x) = x^3 - 2x^2 - 5x + 6$.
step a: the setup

• what number will go outside of the box? (inverse of -3): _______
• list the coefficients: __, , , __

step b: the table

Explanation:

Step1: Find root for synthetic division

For factor $(x-3)$, set $x-3=0$, so $x=3$.

Step2: List polynomial coefficients

For $f(x)=x^3 - 2x^2 -5x +6$, coefficients are $1, -2, -5, 6$.

Step3: Perform synthetic division

$$\begin{array}{r|rrrr} 3 & 1 & -2 & -5 & 6 \\ & & 3 & 3 & -6 \\ \hline & 1 & 1 & -2 & 0 \\ \end{array}$$

Bring down 1. Multiply $1\times3=3$, add to $-2$: $-2+3=1$. Multiply $1\times3=3$, add to $-5$: $-5+3=-2$. Multiply $-2\times3=-6$, add to $6$: $6+(-6)=0$.

Step4: Interpret remainder

A remainder of 0 means $(x-3)$ is a factor.

Answer:

• Number outside the box: $3$
• Coefficients: $[1, -2, -5, 6]$
• Synthetic division result: Remainder = $0$, so $(x-3)$ is a factor of $f(x)=x^3 - 2x^2 -5x +6$