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Question

Factorise:
6(2x3x)22(2x3x)20

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Solution

6(2x3x)22(2x3x)20Put (2x3x)=vEquation becomes 6v22v206v212v+10v203v(2v4)+5(2v4)(3v+5)(2v5)Put the value back(3(2x3x)+5)(2(2x3x)5)(6x9x+5)(4x6x5)


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