Question

$\text{Divide}39y^2(25y^2-49)\text{by}13y(5y+7).$

• A. $3y(5y-14)$
• B. $3y(5y-17)$
• C. $3y(5y-7)$
• D. $3(5y-8)$
  

Answer 1

The correct option is C $3y(5y-7)$

Given, the expression is
$\frac{39y^2(25y^2-49)}{13y(5y+7)}.$

Now, the numerator can be factorized as,
$39y^2(25y^2-49)=39y^2\left[\right(5y{)}^2-\left(7{)}^2\right]$

Using the identity: $a^2-b^2=(a+b)(a-b)$
$39y^2\left[\right(5y{)}^2-\left(7{)}^2\right]$
$=3\times 13\times y\times y\times \left[\right(5y+7\left)\right(5y-7\left)\right]$

Therefore,
$\frac{39y^2(25y^2-49)}{13y(5y+7)}=\frac{3\times 13\times y\times y\times (5y+7)(5y-7)}{13\times y\times (5y+7)}$

After canceling the common terms, we get,
$\frac{39y^2(25y^2-49)}{13y(5y+7)}=3\times y\times (5y-7)=3y(5y-7)$