Find he possible dimensions for the cuboid whose volume is V=6kt2-21kt-12k.

 6kt2 - 21kt - 12k

= k(6t2 - 21t - 12)

= k(6t2 - 24t + 3t - 12)

= k{(6t(t - 4) + 3(t - 4)}

= k{(6t+3)(t - 4)}

= 3k (2t+1)(t - 4)

Thus, the dimensions are 3k, (2t+1), (t - 4).

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