Math, asked by manojsamad629, 6 months ago

factorise 2(x-3)²-32​

Answers

Answered by sarthaknitinagarkar
5

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

Answered by mysticd
8

 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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