Question
Factorise: $x^3 - 6x^2 + 11x - 6$
✓
Answer
Let $p(x)=x^3-6 x^2+11 x-6$ Constant term of $p(x)=-6$
$\therefore$ Factors of $-6$ are $\pm 1, \pm 2, \pm 3, \pm 5, \pm 6$ By trial,
we find that $p(1)=0$, so $(x-1)$ is a factor of $p(x)$.
$\left[\therefore(1)^3-6(1)^2+11(1)-6=1-6+11-6=0\right]$
Now, we see that $x^3-6 x^2+11 x-6=x^3-x^2-5 x^2+5 x+6 x-6$
$=x^2(x-1)-5 x(x-1)+(6 x-1)=(x-1)\left(x^2-5 x+6\right)[$ taking $(x-1)$ common factor $]$
Now, $\left(x^2-5 x+6\right)=x^2-3 x-2 x+6[$ by spliting middle term]
$= x(x - 3) -2(x - 3) = (x - 3)(x - 2) $
$\therefore x^3 - 6x^2 + 11x - 6 = (x - 1)(x - 2)(x - 3)$
$\therefore$ Factors of $-6$ are $\pm 1, \pm 2, \pm 3, \pm 5, \pm 6$ By trial,
we find that $p(1)=0$, so $(x-1)$ is a factor of $p(x)$.
$\left[\therefore(1)^3-6(1)^2+11(1)-6=1-6+11-6=0\right]$
Now, we see that $x^3-6 x^2+11 x-6=x^3-x^2-5 x^2+5 x+6 x-6$
$=x^2(x-1)-5 x(x-1)+(6 x-1)=(x-1)\left(x^2-5 x+6\right)[$ taking $(x-1)$ common factor $]$
Now, $\left(x^2-5 x+6\right)=x^2-3 x-2 x+6[$ by spliting middle term]
$= x(x - 3) -2(x - 3) = (x - 3)(x - 2) $
$\therefore x^3 - 6x^2 + 11x - 6 = (x - 1)(x - 2)(x - 3)$
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