question_answer

Divide: \[15\left( y+3 \right)\left( {{y}^{2}}-16 \right)by\text{ }5\left( {{y}^{2}}-y-12 \right).\]

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 \[\frac{15(y+3)({{y}^{2}}-16)}{5({{y}^{2}}-y-12)}\]

\[\frac{5\times 3\times (y+3)(y-4)(y+4)}{5\times (y-4)(y+3)}\]

\[=3(y+4)\]