MCQ
The factors of $x^3-7 x-6$ are.
- A$x(x - 6) (x - 1)$
- B$(x - 1) (x - 3) (x + 2)$
- ✓$(x + 1) (x + 2) (x - 3)$
- D$\left(x^2-6\right)(x-1)$
✓
Answer
Correct option: C.
$(x + 1) (x + 2) (x - 3)$
The given expression to be factorized is $x^3-7 x+6$
This can be written in the form
$x^3-7 x+6=x^3-(1+6) x+6$
$=x^3-x-6 x+6$
Take common $x$ from the first two terms and $-6$ from the last two terms.
Then we have,
$x^3+7 x+6=x\left(x^2-1\right)-6(x-1)$
$=x\left\{(x)^2-(1)^2\right\}-6\{x-1\}$
$=x(x+1)(x-1)-6(x-1)$
Finally, take common $(x-1)$ from the above expression,
$x^3-7 x+6=(x-1)\{(x+1)-6\}$
$=(x-1)\left(x^2+x-6\right)$
$=(x-1)\left(x^2+3 x-2 x-6\right)$
$=(x-1)\{x(x+3)-2(x+3)\}$
$=(x-1)(x+3)(x-2)$
This can be written in the form
$x^3-7 x+6=x^3-(1+6) x+6$
$=x^3-x-6 x+6$
Take common $x$ from the first two terms and $-6$ from the last two terms.
Then we have,
$x^3+7 x+6=x\left(x^2-1\right)-6(x-1)$
$=x\left\{(x)^2-(1)^2\right\}-6\{x-1\}$
$=x(x+1)(x-1)-6(x-1)$
Finally, take common $(x-1)$ from the above expression,
$x^3-7 x+6=(x-1)\{(x+1)-6\}$
$=(x-1)\left(x^2+x-6\right)$
$=(x-1)\left(x^2+3 x-2 x-6\right)$
$=(x-1)\{x(x+3)-2(x+3)\}$
$=(x-1)(x+3)(x-2)$
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