Question
Divide $21(y+3)\left(y^2-16\right)$ by $7\left(y^2-y-12\right)$.
✓
Answer
We have, $21(y+3)\left(y^2-16\right)+7\left(y^2-y-12\right)$
$
\begin{aligned}
\text { Factorising } & 21(y+3)\left(y^2-16\right) \\
= & 21(y+3)\left[(y)^2-4 \times 4\right] \\
= & 21(y+3)\left[(y)^2-(4)^2\right]
\end{aligned}
$
$
\begin{array}{l}
=21(y+3)(y+4)(y-4) \\
=3 \times 7(y+3)(y+4)(y-4)
\end{array}
$
Now, factorising $7\left(y^2-y-12\right)$
$
\text { Since, } \begin{array}{ll}
& 4 \times(-3)=-12 \\
& 4+(-3)=1 \text { or } 4-3=1
\end{array}
$
On putting these values in given expression, we get
$
\begin{array}{l}
7\left(y^2-y-12\right)=7\left[y^2-(4-3) y-12\right] \\
=7\left(y^2-4 y+3 y-12\right) \\
=7[(y(y-4)+3(y-4)] \\
=7[(y-4)(y+3)]=7(y-4)(y+3) \\
\therefore \frac{21(y+3)\left(y^2-16\right)}{7\left(y^2-y-12\right)} \\
=\frac{3 \times 7(y+3)(y+4)(y-4)}{7(y-4)(y+3)}=3(y+4)
\end{array}
$
$
\begin{aligned}
\text { Factorising } & 21(y+3)\left(y^2-16\right) \\
= & 21(y+3)\left[(y)^2-4 \times 4\right] \\
= & 21(y+3)\left[(y)^2-(4)^2\right]
\end{aligned}
$
$
\begin{array}{l}
=21(y+3)(y+4)(y-4) \\
=3 \times 7(y+3)(y+4)(y-4)
\end{array}
$
Now, factorising $7\left(y^2-y-12\right)$
$
\text { Since, } \begin{array}{ll}
& 4 \times(-3)=-12 \\
& 4+(-3)=1 \text { or } 4-3=1
\end{array}
$
On putting these values in given expression, we get
$
\begin{array}{l}
7\left(y^2-y-12\right)=7\left[y^2-(4-3) y-12\right] \\
=7\left(y^2-4 y+3 y-12\right) \\
=7[(y(y-4)+3(y-4)] \\
=7[(y-4)(y+3)]=7(y-4)(y+3) \\
\therefore \frac{21(y+3)\left(y^2-16\right)}{7\left(y^2-y-12\right)} \\
=\frac{3 \times 7(y+3)(y+4)(y-4)}{7(y-4)(y+3)}=3(y+4)
\end{array}
$
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