
How will you multiply \[\dfrac{{2{x^2} + 5xy + 2{y^2}}}{{4{x^2} - {y^2}}} \div \dfrac{{{x^2} + xy - 2{y^2}}}{{2{x^2} + xy - {y^2}}}\] ?
\[
A) \dfrac{{x + {y^2}}}{{x - y}} \\
B) \dfrac{{x + y}}{{x - y}} \\
C) \dfrac{{x + y}}{{{x^2} + y}} \\
D) \dfrac{{x - y}}{{x - y}} \\
\]
Answer
548.4k+ views
Hint: In order to divide the rational expressions the same method for dividing numerical fractions will be used which means that we need to flip-and- multiply. The basic point of solving the equation is factorization wherever required.
Formula: In the above equation we need to remember that dividing by a fraction is equal to multiplying by the reciprocal of fraction which is as followed
\[\dfrac{a}{b} \div \dfrac{c}{d} = \dfrac{a}{b} \times \dfrac{d}{c}\]
Complete step by step solution:
Firstly we will factorize the algebraic fractions wherever it will be possible
\[\dfrac{{2{x^2} + 5xy + 2{y^2}}}{{4{x^2} - {y^2}}} \div \dfrac{{{x^2} + xy - 2{y^2}}}{{2{x^2} + xy - {y^2}}}\]
So here the above equation can be factorized as followed
\[
\Rightarrow \dfrac{{quadratic{\text{ trinomial}}}}{{differences{\text{ of squares}}}} \div \dfrac{{quadratic{\text{ trinomial}}}}{{quadratic{\text{ trinomial}}}} \\
\Rightarrow \dfrac{{(2x + y)(x + 2y)}}{{(2x + y)(2x - y)}} \times \dfrac{{(x + 2y)(x - y)}}{{(2x - y)(x + y)}} \\
\]
Now in order for dividing by fraction we need to multiply by the reciprocal and during multiplication we need to cancel the like factors
\[
\Rightarrow \dfrac{{(2x + y)(x + 2y)}}{{(2x + y)(2x - y)}} \times \dfrac{{(2x - y)(x + y)}}{{(x + 2y)(x - y)}} \\
\Rightarrow \dfrac{{x + y}}{{x - y}} \\
\]
So the answer is option ‘B’ \[\dfrac{{x + y}}{{x - y}}\]
Additional information: We need to keep in mind that the numerator is the product of the numerators of the given fractions and denominator is the product of the denominators of the given fractions while finding the product of two given fractions.
Notes: We will find while solving the above equation that typically, rational expressions will not be given in factored form but herein this equation, first we need to factorize all numerators and denominators completely. Later on we will multiply and if there are any common factors then we will cancel it. It is considered as best practice to leave answers in factored form while solving such equations.
Formula: In the above equation we need to remember that dividing by a fraction is equal to multiplying by the reciprocal of fraction which is as followed
\[\dfrac{a}{b} \div \dfrac{c}{d} = \dfrac{a}{b} \times \dfrac{d}{c}\]
Complete step by step solution:
Firstly we will factorize the algebraic fractions wherever it will be possible
\[\dfrac{{2{x^2} + 5xy + 2{y^2}}}{{4{x^2} - {y^2}}} \div \dfrac{{{x^2} + xy - 2{y^2}}}{{2{x^2} + xy - {y^2}}}\]
So here the above equation can be factorized as followed
\[
\Rightarrow \dfrac{{quadratic{\text{ trinomial}}}}{{differences{\text{ of squares}}}} \div \dfrac{{quadratic{\text{ trinomial}}}}{{quadratic{\text{ trinomial}}}} \\
\Rightarrow \dfrac{{(2x + y)(x + 2y)}}{{(2x + y)(2x - y)}} \times \dfrac{{(x + 2y)(x - y)}}{{(2x - y)(x + y)}} \\
\]
Now in order for dividing by fraction we need to multiply by the reciprocal and during multiplication we need to cancel the like factors
\[
\Rightarrow \dfrac{{(2x + y)(x + 2y)}}{{(2x + y)(2x - y)}} \times \dfrac{{(2x - y)(x + y)}}{{(x + 2y)(x - y)}} \\
\Rightarrow \dfrac{{x + y}}{{x - y}} \\
\]
So the answer is option ‘B’ \[\dfrac{{x + y}}{{x - y}}\]
Additional information: We need to keep in mind that the numerator is the product of the numerators of the given fractions and denominator is the product of the denominators of the given fractions while finding the product of two given fractions.
Notes: We will find while solving the above equation that typically, rational expressions will not be given in factored form but herein this equation, first we need to factorize all numerators and denominators completely. Later on we will multiply and if there are any common factors then we will cancel it. It is considered as best practice to leave answers in factored form while solving such equations.
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