Question

$\text{Factorise:}{x}^4-(x-4{)}^4$

• A. $16[x-2][x^2-4x+8]$
• B. $16[x+2][x^2-4x+8]$
• C. $16[x-2][x^2+4x+8]$
• D. $32[x-2][x^2-4x+8]$

Answer 1

The correct option is A $16[x-2][x^2-4x+8]$

$x^4-(x-4{)}^4=(x^2{)}^2-\left[\right(x-4{)}^2{]}^2$
$\text{[Using the identity:}{a}^2-b^2=(a+b)(a-b)]$

$=[x^2+(x-4{)}^2\left]\right[x^2-(x-4{)}^2]$

$\text{[Using the identity:}(a-b{)}^2=a^2-2ab+b^2]$

$=[x^2+(x^2-8x+16\left)\right][x^2-(x^2-8x+16\left)\right]$

$=[x^2+x^2-8x+16][x^2-x^2+8x-16]$

$=[2x^2-8x+16][8x-16]$

Taking '2' common from the first group and '8' common from the second group, we get
$=\left(2\right)\left(8\right)[x^2-4x+8][x-2)]$
$=16[x-2][x^2-4x+8]$