Question

factorise 2(x-3)²-32​

Answer 1

Answer:

2(x²+9-6x)-32

2x²-12x+18-32

2x²-12x-14

2[x²-6x-7]

2[x²-7x+x-7]

2[x(x-7)+1(x-7)]

2(x+1)(x-7)

pls marks brainliest

Answer 2

$Given \: 2(x-3)^{2} - 32$

$= 2[ (x-3)^{2} - 16 ]$

$= 2[ (x-3)^{2} - 4^{2} ]$

/* By algebraic identity */

$\boxed{ \pink{ a^{2} - b^{2} = ( a+b)(a-b)}}$

$= 2[ (x-3)+4] [ (x-3) - 4 ]$

$= 2( x - 3 + 4 )( x - 3 - 4 )$

$= 2( x + 1 )( x - 7 )$

Therefore.,

$\red{ Factors \:of \: 2(x-3)^{2} - 32}$

$\green {= 2( x + 1 )( x - 7 )}$

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