Question

Factorise $2x^7-128x$

• A. $2x\left[\right(x-2\left)\right(x^2+2x+4\left)\right(x-2\left)\right(x^2-2x+4\left)\right]$
• B. $x\left[\right(x-2\left)\right(x^2+2x+4\left)\right(x+2\left)\right(x^2-2x+4\left)\right]$
• C. $=2x\left[\right(x-2\left)\right(x^2+2x+4\left)\right(x+2\left)\right(x^2-2x+4\left)\right]$
• D. $2x\left[\right(x-2\left)\right(x^2+2x+4\left)\right(x+2\left)\right(x^3-2x+4\left)\right]$

Answer 1

The correct option is C $=2x\left[\right(x-2\left)\right(x^2+2x+4\left)\right(x+2\left)\right(x^2-2x+4\left)\right]$

$2x^7-128x$
$=2x(x^6-64)$
$=2x\left[\right(x^3{)}^2-8^2]$
We know that,
{ $a^2-b^2=(a+b)(a-b)$}
$=2x\left[\right(x^3-8\left)\right(x^3+8\left)\right]$
We know that,
{$x^3-y^3=(x-y)(x^2+xy+y^2)$ }
{$x^3+y^3=(x+y)(x^2-xy+y^2)$ }
$=2x\left[\right(x^3-2^3\left)\right(x^3+2^3\left)\right]$
$=2x\left[\right(x-2\left)\right(x^2+2x+4\left)\right(x+2\left)\right(x^2-2x+4\left)\right]$