Question

How do you simplify the expression \[\dfrac{2{{k}^{2}}+11kp+12{{p}^{2}}}{9{{k}^{2}}p-33k{{p}^{2}}+28{{p}^{3}}}\div \dfrac{6{{k}^{2}}+17kp+12{{p}^{2}}}{9{{k}^{2}}p-16{{p}^{3}}}\]?

Answer 1

Hint: In order to find the solution of the given question that is to simplify \[\dfrac{2{{k}^{2}}+11kp+12{{p}^{2}}}{9{{k}^{2}}p-33k{{p}^{2}}+28{{p}^{3}}}\div \dfrac{6{{k}^{2}}+17kp+12{{p}^{2}}}{9{{k}^{2}}p-16{{p}^{3}}}\] convert the division form between the two-given fractions into the multiplication form between two fractions by using the formula \[\dfrac{\dfrac{a}{b}}{\dfrac{c}{d}}=\dfrac{a}{b}\times \dfrac{d}{c}\]. After that factorise the expression in the numerator and denominator of both the fraction and then simplify the expression.

Complete step by step solution:
According to the question, given expression in the question is as follows:
\[\dfrac{2{{k}^{2}}+11kp+12{{p}^{2}}}{9{{k}^{2}}p-33k{{p}^{2}}+28{{p}^{3}}}\div \dfrac{6{{k}^{2}}+17kp+12{{p}^{2}}}{9{{k}^{2}}p-16{{p}^{3}}}\]
We can rewrite the above expression as follows:
\[\Rightarrow \dfrac{\dfrac{2{{k}^{2}}+11kp+12{{p}^{2}}}{9{{k}^{2}}p-33k{{p}^{2}}+28{{p}^{3}}}}{\dfrac{6{{k}^{2}}+17kp+12{{p}^{2}}}{9{{k}^{2}}p-16{{p}^{3}}}}\]
Applying the formula \[\dfrac{\dfrac{a}{b}}{\dfrac{c}{d}}=\dfrac{a}{b}\times \dfrac{d}{c}\] in the above expression, we will have:
\[\Rightarrow \dfrac{2{{k}^{2}}+11kp+12{{p}^{2}}}{9{{k}^{2}}p-33k{{p}^{2}}+28{{p}^{3}}}\times \dfrac{9{{k}^{2}}p-16{{p}^{3}}}{6{{k}^{2}}+17kp+12{{p}^{2}}}\]
Now take p common from the denominator of the first fraction and from the numerator of the second fraction in the above expression, we will have:
\[\Rightarrow \dfrac{2{{k}^{2}}+11kp+12{{p}^{2}}}{p\left( 9{{k}^{2}}-33kp+28{{p}^{2}} \right)}\times \dfrac{p\left( 9{{k}^{2}}-16{{p}^{2}} \right)}{6{{k}^{2}}+17kp+12{{p}^{2}}}\]
After this factorise the denominator of the first fraction using the method of splitting the middle term, we will have:
\[\Rightarrow \dfrac{2{{k}^{2}}+11kp+12{{p}^{2}}}{p\left( 9{{k}^{2}}-12kp-21kp+28{{p}^{2}} \right)}\times \dfrac{p\left( 9{{k}^{2}}-16{{p}^{2}} \right)}{6{{k}^{2}}+17kp+12{{p}^{2}}}\]
Now take the terms in common, we will have:
\[\Rightarrow \dfrac{2{{k}^{2}}+11kp+12{{p}^{2}}}{p\left( 3k\left( 3k-4p \right)-7p\left( 3k-4p \right) \right)}\times \dfrac{p\left( 3k-4p \right)\left( 3k+4p \right)}{6{{k}^{2}}+17kp+12{{p}^{2}}}\]
After simplifying the above expression further, we will have:
\[\Rightarrow \dfrac{2{{k}^{2}}+11kp+12{{p}^{2}}}{p\left( 3k-7p \right)\left( 3k-4p \right)}\times \dfrac{p\left( 3k-4p \right)\left( 3k+4p \right)}{6{{k}^{2}}+17kp+12{{p}^{2}}}\]
\[\Rightarrow \dfrac{2{{k}^{2}}+11kp+12{{p}^{2}}}{\left( 3k-7p \right)}\times \dfrac{\left( 3k+4p \right)}{6{{k}^{2}}+17kp+12{{p}^{2}}}\]
Now factorise the numerator of the first fraction and denominator of the second expression using the method of splitting the middle term, we will have:
\[\Rightarrow \dfrac{2{{k}^{2}}+3kp+8kp+12{{p}^{2}}}{\left( 3k-7p \right)}\times \dfrac{\left( 3k+4p \right)}{6{{k}^{2}}+9kp+8kp+12{{p}^{2}}}\]
After this take the terms in common, we will have:
\[\Rightarrow \dfrac{k\left( 2k+3p \right)+4p\left( 2k+3p \right)}{\left( 3k-7p \right)}\times \dfrac{\left( 3k+4p \right)}{3k\left( 2k+3p \right)+4p\left( 2k+3p \right)}\]
Simplifying the above expression further, we will have:
\[\Rightarrow \dfrac{\left( k+4p \right)\left( 2k+3p \right)}{\left( 3k-7p \right)}\times \dfrac{\left( 3k+4p \right)}{\left( 3k+4p \right)\left( 2k+3p \right)}\]
Clearly, we can see some terms under the bracket are there in both numerator and denominator, therefore they will get cancelled and we can rewrite the above expression as follows:
\[\Rightarrow \dfrac{\left( k+4p \right)}{\left( 3k-7p \right)}\]
\[\Rightarrow \dfrac{2{{k}^{2}}+11kp+12{{p}^{2}}}{9{{k}^{2}}p-33k{{p}^{2}}+28{{p}^{3}}}\div \dfrac{6{{k}^{2}}+17kp+12{{p}^{2}}}{9{{k}^{2}}p-16{{p}^{3}}}=\dfrac{\left( k+4p \right)}{\left( 3k-7p \right)}\]
Therefore, after simplification of the given expression \[\dfrac{2{{k}^{2}}+11kp+12{{p}^{2}}}{9{{k}^{2}}p-33k{{p}^{2}}+28{{p}^{3}}}\div \dfrac{6{{k}^{2}}+17kp+12{{p}^{2}}}{9{{k}^{2}}p-16{{p}^{3}}}\] we get the final answer as \[\dfrac{\left( k+4p \right)}{\left( 3k-7p \right)}\].

Note: Student make mistakes while using the wrong formula which convert the division form between the two-given fractions into the multiplication form between two fractions that is they sometimes use \[\dfrac{\dfrac{a}{b}}{\dfrac{c}{d}}=\dfrac{a}{b}\times \dfrac{c}{d}\] instead of the correct formula \[\dfrac{\dfrac{a}{b}}{\dfrac{c}{d}}=\dfrac{a}{b}\times \dfrac{d}{c}\], which is completely wrong and leads to the wrong answer.