Question
Factorise $x^3 + 6x^2 + 11x + 6$ completely using factor theorem.
✓
Answer
Let $f(x)=x^3+6 x^2+11 x+6$
$f \text { or }=x=-1$
$f(-1)=(-1)^3+6(-1)^2+11(-1)+6$
$=-1+6-11+6=12-12=0$
Hence, $(x+1)$ is a factory of $f(x)$
$\therefore x^3+6 x^2+11 x+6=(x+1)\left(x^2+5 x+6\right)$
$=(x+1)\left(x^2+2 x+3 x+6\right)$
$=(x+1)[x(x+2)+3(x+2)]$
$=(x+1)(x+2)(x+3)$
$f \text { or }=x=-1$
$f(-1)=(-1)^3+6(-1)^2+11(-1)+6$
$=-1+6-11+6=12-12=0$
Hence, $(x+1)$ is a factory of $f(x)$
$\therefore x^3+6 x^2+11 x+6=(x+1)\left(x^2+5 x+6\right)$
$=(x+1)\left(x^2+2 x+3 x+6\right)$
$=(x+1)[x(x+2)+3(x+2)]$
$=(x+1)(x+2)(x+3)$
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