Question
Factorize: $x(x - 2)(x - 4) + 4x - 8$
✓
Answer
$x(x - 2)(x - 4) + 4x - 8$
$= x(x - 2)(x - 4) + 4(x - 2)$
Taking $(x - 2) $common in both the terms
$=(x - 2){x(x - 4) + 4}$
$=(x - 2){x^2 - 4x + 4}$
Now splitting the middle term of $x^2 - 4x + 4$
$= (x - 2){x^2 - 2x - 2x + 4}$
$= (x - 2){x( x - 2) -2(x - 2)}$
$= (x - 2){(x - 2)(x - 2)}$
$= (x - 2)(x - 2)(x - 2)$
$= (x - 2)^3$
$\therefore x(x - 2)(x - 4) + 4x - 8$
$ = (x - 2)^3$
$= x(x - 2)(x - 4) + 4(x - 2)$
Taking $(x - 2) $common in both the terms
$=(x - 2){x(x - 4) + 4}$
$=(x - 2){x^2 - 4x + 4}$
Now splitting the middle term of $x^2 - 4x + 4$
$= (x - 2){x^2 - 2x - 2x + 4}$
$= (x - 2){x( x - 2) -2(x - 2)}$
$= (x - 2){(x - 2)(x - 2)}$
$= (x - 2)(x - 2)(x - 2)$
$= (x - 2)^3$
$\therefore x(x - 2)(x - 4) + 4x - 8$
$ = (x - 2)^3$
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