Algebra 2 Common Core

Equat$x-3,$ to zero then solve for $x,$: \begin{align*} x-3&=0 \\ x&=3 \end{align*} Substituting $x= 3$ in the given expression, $x^3+4x^2+x-6 ,$ results to \begin{align*} & (3)^3+4(3)^2+(3)-6 \\&= 27+4(9)+3-6 \\&= 27+36+3-6 \\&= 60 .\end{align*} Since the substitution above is NOT equal to zero, then the remainder is NOT zero. By the Factor Theorem, $x-3$ is not a factor of $x^3+4x^2+x-6$.