Question

Factorise the expression and divide them as directed: $39y^3\left(50y^2-98\right)÷26y^2\left(5y+7\right)$.

Answer 1

Step 1: Factorize the expression $39y^3\left(50y^2-98\right)$.

$\begin{array}{rcl}& \Rightarrow & 39y^3\left(2\times 25y^2-2\times 49\right)=39y^3\times 2\left(25y^2-49\right)\\ & =& 78y^3\left({\left(5y\right)}^2-{\left(7\right)}^2\right)\end{array}$

Using identity $a^2-b^2=\left(a+b\right)\left(a-b\right)$, expand the above expression.

$78y^3\left(5y+7\right)\left(5y-7\right)$

Step 2: Divide the above expression by $26y^2\left(5y+7\right)$.

$\Rightarrow \frac{78y^3\left(5y+7\right)\left(5y-7\right)}{26y^2\left(5y+7\right)}=3y\left(5y-7\right)$

Hence, the required final answer is $3y\left(5y-7\right)$.