Question

Find the value of $39{{y}^{2}}\left( 50{{y}^{2}}-98 \right)\div 26{{y}^{2}}\left( 5y+7 \right)$ .

Answer 1

Hint: We have to write the given expression in the form of numerator and denominator. Then, we have to cancel the common terms from the numerator and the denominator. We have to factorize 50 and 98 such that one factor of the both are similar and the other must be a perfect square. Then, we have to apply algebraic identity and simplify the expression further.

Complete step by step answer:
We have to find the value of $39{{y}^{2}}\left( 50{{y}^{2}}-98 \right)\div 26{{y}^{2}}\left( 5y+7 \right)$ . Let us write this expression as follows.
$= \dfrac{39{{y}^{2}}\left( 50{{y}^{2}}-98 \right)}{26{{y}^{2}}\left( 5y+7 \right)}$
We can cancel ${{y}^{2}}$ from the numerator and the denominator.
$= \dfrac{39\require{cancel}\cancel{{{y}^{2}}}\left( 50{{y}^{2}}-98 \right)}{26\require{cancel}\cancel{{{y}^{2}}}\left( 5y+7 \right)}$
We will get the result of the above simplification as follows
$= \dfrac{39\left( 50{{y}^{2}}-98 \right)}{26\left( 5y+7 \right)}$
Let us factorize 50 and 98.
$= \dfrac{39\left( 2\times 25{{y}^{2}}-2\times 49 \right)}{26\left( 5y+7 \right)}$
We have to take the common term outside of the bracket.
$= \dfrac{39\times 2\left( 25{{y}^{2}}-49 \right)}{26\left( 5y+7 \right)}$
We can write 25y and 49 in terms of squares of 5y and 9.
$= \dfrac{39\times 2\left( {{\left( 5y \right)}^{2}}-{{7}^{2}} \right)}{26\left( 5y+7 \right)}$
We know that ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ . Therefore, the above expression can be written as
$= \dfrac{39\times 2\left( \left( 5y+7 \right)\left( 5y-7 \right) \right)}{26\left( 5y+7 \right)}$
Let us cancel the common factors and $5y-7$ from the numerator and the denominator.
\[= \dfrac{39\times \require{cancel}\cancel{2}\require{cancel}\cancel{\left( 5y+7 \right)}\left( 5y-7 \right)}{{{\require{cancel}\cancel{26}}^{13}}\require{cancel}\cancel{\left( 5y+7 \right)}}\]
We will get the result of the above simplification as follows
\[= \dfrac{39\left( 5y-7 \right)}{13}\]
Let us cancel the common factor 13 from the numerator and the denominator.
\[\begin{align}
  & = \dfrac{{{\require{cancel}\cancel{39}}^{3}}\left( 5y-7 \right)}{\require{cancel}\cancel{13}} \\
 & = 3\left( 5y-7 \right) \\
\end{align}\]
We have to apply the distributive property on the above expression.
$= 15y-27$
Hence, the value of $39{{y}^{2}}\left( 50{{y}^{2}}-98 \right)\div 26{{y}^{2}}\left( 5y+7 \right)$ is \[3\left( 5y-7 \right)\] or $15y-27$ .

Note: Students must deeply understand and learn the concept of algebra and algebraic identities. They must also be thorough with the properties of algebraic expressions including commutative, associative and distributive properties. In questions of this type, instead of going for direct division by long division method, students must simplify the expression as much as possible.