Answer:
Factorising \[15\text{ }(y+3)({{y}^{2}}16),\] We get, \[5\times 3\times (y+3)(y4)(y+4)\] On factorising, \[5({{y}^{2}}y12),\]we get \[5({{y}^{2}}4y+3y12)\] \[=5\text{ }[y(y4)+3(y4)]\] \[=5(y4)(y+3)\] Therefore, on dividing the first expression by the second expression, we get \[\frac{15(y+3)({{y}^{2}}-16)}{5({{y}^{2}}-y-12)}\] \[\frac{5\times 3\times (y+3)(y-4)(y+4)}{5\times (y-4)(y+3)}\] \[=3(y+4)\]
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