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