Question
Show that $(x + 2)$ is a factor of $f(x) = x^3 + 4x^2 + x - 6.$
✓
Answer
Given: $f(x) = x^3 + 4x^2 + x - 6$
Now, $f(-2) = (-2)^3 + 4(-2)^2 + (-2) - 6$
$= -8 + 16 - 2 - 6$
$= 0$
$\therefore$ $(x + 2)$ is a factor of $f(x) = x^3 + 4x^2 + x - 6.$
Now, $f(-2) = (-2)^3 + 4(-2)^2 + (-2) - 6$
$= -8 + 16 - 2 - 6$
$= 0$
$\therefore$ $(x + 2)$ is a factor of $f(x) = x^3 + 4x^2 + x - 6.$
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