Question

Divide the following and write your answer in lowest terms:
\[\dfrac{2{{x}^{2}}+13x+15}{{{x}^{2}}+3x-10}\div \dfrac{2{{x}^{2}}-x-6}{{{x}^{2}}-4x+4}\]

Answer 1

Hint: First, rewrite the given expression by changing the second term. This can be done by multiplying the reciprocal of the second term by the first term instead of dividing the second term itself by the first term. Then find the roots by expanding the middle terms of all the bigger terms. In the end, all the terms will get cancelled out and you will arrive at the final answer.

Complete step-by-step answer:
In this question, we need to find the value of \[\dfrac{2{{x}^{2}}+13x+15}{{{x}^{2}}+3x-10}\div \dfrac{2{{x}^{2}}-x-6}{{{x}^{2}}-4x+4}\] in the lowest terms.
The given expression is: \[\dfrac{2{{x}^{2}}+13x+15}{{{x}^{2}}+3x-10}\div \dfrac{2{{x}^{2}}-x-6}{{{x}^{2}}-4x+4}\]

We can rewrite it by changing the second term. This can be done by multiplying the reciprocal of the second term by the first term instead of dividing the second term itself by the first term.
Using the above method, we can rewrite the given expression as the following:
\[\dfrac{2{{x}^{2}}+13x+15}{{{x}^{2}}+3x-10}\times \dfrac{{{x}^{2}}-4x+4}{2{{x}^{2}}-x-6}\]

Now, we will expand the middle terms of the numerators and denominators of both the terms in order to form groups in the numerator and the denominators.
Using the above method, we can rewrite the above expression as the following:
\[\dfrac{2{{x}^{2}}+10x+3x+15}{{{x}^{2}}+5x-2x-10}\times \dfrac{{{x}^{2}}-2x-2x+4}{2{{x}^{2}}-4x+3x-6}\]

Now, we will take (x + 5) common from the first two terms and the last two terms separately of the numerator and the denominator of the first term and we will take (x - 2) common from the first two terms and the last two terms separately of the numerator and the denominator of the second term.

Using the above method, we can rewrite the above expression as the following:
\[\dfrac{2{{x}^{2}}+13x+15}{{{x}^{2}}+3x-10}\div \dfrac{2{{x}^{2}}-x-6}{{{x}^{2}}-4x+4}=1\]

Now, we will take (x + 5) common from the numerator and the denominator of the first term and we will take (x - 2) common from the numerator and the denominator of the second term.
Using the above method, we can rewrite the above expression as the following:
\[\dfrac{\left( 2x+3 \right)\left( x+5 \right)}{\left( x-2 \right)\left( x+5 \right)}\times \dfrac{\left( x-2 \right)\left( x-2 \right)}{\left( 2x+3 \right)\left( x-2 \right)}\]
\[\dfrac{\left( 2x+3 \right)\left( x+5 \right)\left( x-2 \right)\left( x-2 \right)}{\left( x-2 \right)\left( x+5 \right)\left( 2x+3 \right)\left( x-2 \right)}=1\]

Hence, \[\dfrac{2{{x}^{2}}+13x+15}{{{x}^{2}}+3x-10}\div \dfrac{2{{x}^{2}}-x-6}{{{x}^{2}}-4x+4}=1\]
This is our final answer.

Note: In this question, it is very important to know that we can rewrite the given expression by changing the second term. This can be done by multiplying the reciprocal of the second term by the first term instead of dividing the second term itself by the first term.