Question

Factorise : $2x^3-3x^2-3x+2$

• A. $(x+1)(x-2)(2x-1)$
• B. $(x-1)(x-2)(2x-1)$
• C. $(x+1)(x+2)(2x-1)$
• D. $(x+1)(x-2)(x-1)$

Answer 1

The correct option is A $(x+1)(x-2)(2x-1)$

$2x^3-3x^2-3x+2$
When the coefficient of highest power is more than one, we have to divide the constant by respective coefficient.
$\frac{2}{2}=1=$ factors can $\pm 2,\pm 1/2,\pm 1$
When, $x=-1\Rightarrow 2(-1{)}^3-3(-1{)}^2-3(-1)+2=0$
When, $x=2\Rightarrow 2(2{)}^3-3(2{)}^2-3\left(2\right)+2=0$
When, $x=1/2\Rightarrow 2(1/2{)}^3-3(1/2{)}^2-3\left(1/2\right)+2=0$
For other factors $f\left(n\right)\ne 0$
Hence, $f\left(x\right)=(x+1)(x-2)(x-1/2)$
or
$f\left(x\right)=(x+1)(x-2)(2x-1)$
So, the correct answer is option (a).