# Simplify the given Expression : $\dfrac{{{a}^{2}}+10a+21}{{{a}^{2}}+6a-7}\times \dfrac{{{a}^{2}}-1}{a+3}$.

Last updated date: 22nd Mar 2023

•

Total views: 304.5k

•

Views today: 5.85k

Answer

Verified

304.5k+ views

Hint:The given problem is related to simplification by factorization. Express each term as a product of its factors and then simplify the expression.

Complete step-by-step answer:

We will proceed with the solution by taking each term, factoring it, and then substituting the factored form in the given expression. Then we will simplify the given expression.

The given expression is $\dfrac{{{a}^{2}}+10a+21}{{{a}^{2}}+6a-7}\times \dfrac{{{a}^{2}}-1}{a+3}$. The first term of the expression is ${{a}^{2}}+10a+21$. It is a quadratic in $a$ . We will use middle term splitting to factorize the quadratic. We have to split $10a$ into two terms, such as their sum is equal to $10a$ and their product is equal to $21{{a}^{2}}$ .

We can write $10a$ as $7a+3a$ . Here, the sum of $7a$ and $3a$ is $10a$ and the product of $7a$ and $3a$ is $21{{a}^{2}}$ . So, ${{a}^{2}}+10a+21$can be written as ${{a}^{2}}+7a+3a+21$ .

$\Rightarrow {{a}^{2}}+10a+21=a\left( a+7 \right)+3\left( a+7 \right)$

$\Rightarrow {{a}^{2}}+10a+21=\left( a+3 \right)\left( a+7 \right)$

So, we can write ${{a}^{2}}+10a+21$ as $\left( a+3 \right)\left( a+7 \right)$ .

Now, the second term is ${{a}^{2}}+6a-7$ . It is a quadratic in $a$ . We will use middle term splitting to factorize the quadratic. We have to split $6a$ into two terms, such as their sum is equal to $6a$ and their product is equal to \[-7{{a}^{2}}\] .

We can write $6a$ as $7a-a$ . Here, the sum of $7a$ and $-a$ is $6a$ and the product of $7a$ and $-a$ is $-7{{a}^{2}}$ . So, ${{a}^{2}}+6a-7$can be written as ${{a}^{2}}+7a-a-7$ .

$\Rightarrow {{a}^{2}}+6a-7=a\left( a+7 \right)-1\left( a+7 \right)$

$\Rightarrow {{a}^{2}}+6a-7=\left( a-1 \right)\left( a+7 \right)$

So, we can write ${{a}^{2}}+6a-7$ as $\left( a-1 \right)\left( a+7 \right)$ .

Now, the third term is ${{a}^{2}}-1$ . We know, we can write ${{a}^{2}}-1$ as ${{a}^{2}}-{{1}^{2}}$. So, ${{a}^{2}}-1={{a}^{2}}-{{1}^{2}}=\left( a+1 \right)\left( a-1 \right)$ . So, we can write ${{a}^{2}}-1$ as $\left( a+1 \right)\left( a-1 \right)$

Now, the fourth term is $\left( a+3 \right)$ . It is already in its simplest form. So, we don’t need to factorize it.

Now, the expression $\dfrac{{{a}^{2}}+10a+21}{{{a}^{2}}+6a-7}\times \dfrac{{{a}^{2}}-1}{a+3}$ can be written as $\dfrac{\left( a+7 \right)\left( a+3 \right)}{\left( a+7 \right)\left( a-1 \right)}\times \dfrac{\left( a-1 \right)\left( a+1 \right)}{\left( a+3 \right)}$ .

$\Rightarrow \dfrac{{{a}^{2}}+10a+21}{{{a}^{2}}+6a-7}\times \dfrac{{{a}^{2}}-1}{a+3}=a+1$

Hence, the simplified value of $\dfrac{{{a}^{2}}+10a+21}{{{a}^{2}}+6a-7}\times \dfrac{{{a}^{2}}-1}{a+3}$ is $a+1$ .

Note: While making substitutions, make sure that the substitutions are done correctly and no sign mistakes are present. Sign mistakes can cause the final answer to be wrong.

Complete step-by-step answer:

We will proceed with the solution by taking each term, factoring it, and then substituting the factored form in the given expression. Then we will simplify the given expression.

The given expression is $\dfrac{{{a}^{2}}+10a+21}{{{a}^{2}}+6a-7}\times \dfrac{{{a}^{2}}-1}{a+3}$. The first term of the expression is ${{a}^{2}}+10a+21$. It is a quadratic in $a$ . We will use middle term splitting to factorize the quadratic. We have to split $10a$ into two terms, such as their sum is equal to $10a$ and their product is equal to $21{{a}^{2}}$ .

We can write $10a$ as $7a+3a$ . Here, the sum of $7a$ and $3a$ is $10a$ and the product of $7a$ and $3a$ is $21{{a}^{2}}$ . So, ${{a}^{2}}+10a+21$can be written as ${{a}^{2}}+7a+3a+21$ .

$\Rightarrow {{a}^{2}}+10a+21=a\left( a+7 \right)+3\left( a+7 \right)$

$\Rightarrow {{a}^{2}}+10a+21=\left( a+3 \right)\left( a+7 \right)$

So, we can write ${{a}^{2}}+10a+21$ as $\left( a+3 \right)\left( a+7 \right)$ .

Now, the second term is ${{a}^{2}}+6a-7$ . It is a quadratic in $a$ . We will use middle term splitting to factorize the quadratic. We have to split $6a$ into two terms, such as their sum is equal to $6a$ and their product is equal to \[-7{{a}^{2}}\] .

We can write $6a$ as $7a-a$ . Here, the sum of $7a$ and $-a$ is $6a$ and the product of $7a$ and $-a$ is $-7{{a}^{2}}$ . So, ${{a}^{2}}+6a-7$can be written as ${{a}^{2}}+7a-a-7$ .

$\Rightarrow {{a}^{2}}+6a-7=a\left( a+7 \right)-1\left( a+7 \right)$

$\Rightarrow {{a}^{2}}+6a-7=\left( a-1 \right)\left( a+7 \right)$

So, we can write ${{a}^{2}}+6a-7$ as $\left( a-1 \right)\left( a+7 \right)$ .

Now, the third term is ${{a}^{2}}-1$ . We know, we can write ${{a}^{2}}-1$ as ${{a}^{2}}-{{1}^{2}}$. So, ${{a}^{2}}-1={{a}^{2}}-{{1}^{2}}=\left( a+1 \right)\left( a-1 \right)$ . So, we can write ${{a}^{2}}-1$ as $\left( a+1 \right)\left( a-1 \right)$

Now, the fourth term is $\left( a+3 \right)$ . It is already in its simplest form. So, we don’t need to factorize it.

Now, the expression $\dfrac{{{a}^{2}}+10a+21}{{{a}^{2}}+6a-7}\times \dfrac{{{a}^{2}}-1}{a+3}$ can be written as $\dfrac{\left( a+7 \right)\left( a+3 \right)}{\left( a+7 \right)\left( a-1 \right)}\times \dfrac{\left( a-1 \right)\left( a+1 \right)}{\left( a+3 \right)}$ .

$\Rightarrow \dfrac{{{a}^{2}}+10a+21}{{{a}^{2}}+6a-7}\times \dfrac{{{a}^{2}}-1}{a+3}=a+1$

Hence, the simplified value of $\dfrac{{{a}^{2}}+10a+21}{{{a}^{2}}+6a-7}\times \dfrac{{{a}^{2}}-1}{a+3}$ is $a+1$ .

Note: While making substitutions, make sure that the substitutions are done correctly and no sign mistakes are present. Sign mistakes can cause the final answer to be wrong.

Recently Updated Pages

Calculate the entropy change involved in the conversion class 11 chemistry JEE_Main

The law formulated by Dr Nernst is A First law of thermodynamics class 11 chemistry JEE_Main

For the reaction at rm0rm0rmC and normal pressure A class 11 chemistry JEE_Main

An engine operating between rm15rm0rm0rmCand rm2rm5rm0rmC class 11 chemistry JEE_Main

For the reaction rm2Clg to rmCrmlrm2rmg the signs of class 11 chemistry JEE_Main

The enthalpy change for the transition of liquid water class 11 chemistry JEE_Main

Trending doubts

Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Write a letter to the Principal of your school to plead class 10 english CBSE