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

Answer

Verified

381.9k+ 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

Which of the following would not be a valid reason class 11 biology CBSE

What is meant by monosporic development of female class 11 biology CBSE

Draw labelled diagram of the following i Gram seed class 11 biology CBSE

Explain with the suitable examples the different types class 11 biology CBSE

How is pinnately compound leaf different from palmately class 11 biology CBSE

Match the following Column I Column I A Chlamydomonas class 11 biology CBSE

Trending doubts

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Which one of the following places is unlikely to be class 8 physics CBSE

Select the word that is correctly spelled a Twelveth class 10 english CBSE

Difference Between Plant Cell and Animal Cell

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

What is the past tense of read class 10 english CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

Elucidate the structure of fructose class 12 chemistry CBSE

What is pollution? How many types of pollution? Define it