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\[\frac{ k+3 }{ 4k-2 }\times \left( 12k ^{2} +2k-4\right)\]
Try to factorise 12k^2+2k-4 and tell me vat u get
no clue
i am no good at this stuff
well can u tell me two factors of -48 such that their sum is 2?
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-8 and 6
well it is other way round. i mean the numbers are 8 and -6. Nw 2x can be written as 8x-6x. agree?
sure
\(12k^2+2k-4=12k^2+8k-6k-4\). Nw can u factorise \(12k^2+2k\)?
still dont get it
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\(12k^2+2k=(2*6*k*k)+(2*k)\) What is common to both \(12k^2 and 2k\)?
1 and 2 ?
2k is common
ok
take it out. then what remains?
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\[12k ^{2}\]
|dw:1368552600854:dw|
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