
How do you factor ${{a}^{2}}+2a+ab+2b$?
Answer
511.8k+ views
Hint: In this question we will group the similar terms in the given expression to get common terms. we will break down the expression into two parts and take common and then combine them. We will then take out the common multiple from both the terms to write it in the factored form. We will then cross check the answer by multiplying the factors.
Complete step-by-step solution:
We have the given expression as:
$\Rightarrow {{a}^{2}}+2a+ab+2b$
Since the given equation is not directly factorable, we will rearrange the equations to get the similar terms.
Now let’s split the terms into two separate equations,
We can make ${{a}^{2}}+2a\to \left( 1 \right)$ and $ab+2b\to \left( 2 \right)$, therefore the original equation will be $\left( 1 \right)+\left( 2 \right)$.
Now consider $\left( 1 \right)$,
$\Rightarrow {{a}^{2}}+2a$
Now since $a$ is common in both the terms, we can take it out as common and write it as:
$\Rightarrow a\left( a+2 \right)$
Now consider $\left( 2 \right)$,
$\Rightarrow ab+2b$
Now since $b$ is common in both the terms, we can take it out as common and write it as:
$\Rightarrow b\left( a+2 \right)$
Now since both the equations are simplified, we can join equation $\left( 1 \right)$ and $\left( 2 \right)$, and write it as:
$\Rightarrow a\left( a+2 \right)+b\left( a+2 \right)$
Now since the term $\left( a+2 \right)$ is common in both the terms we can take it out as common and write it as:
$\Rightarrow \left( a+2 \right)\left( a+b \right)$, which is the final factorized way of writing the given equation.
Note: To check whether the final factored equation is correct we can multiply the factors, and if we get the given equation back, the factors are correct.
We can multiply the factors as:
$a\times a+a\times b+2\times a+2\times b$
On simplifying we get:
${{a}^{2}}+ab+2a+2b$
On rearranging we get:
${{a}^{2}}+2a+ab+2b$, which is the original equation, therefore the factors are correct.
It is to be remembered that factors are the digits which make up a number, a factor should be indivisible by any number except $1$ and it should not be in the form of a number which is raised to a power.
Complete step-by-step solution:
We have the given expression as:
$\Rightarrow {{a}^{2}}+2a+ab+2b$
Since the given equation is not directly factorable, we will rearrange the equations to get the similar terms.
Now let’s split the terms into two separate equations,
We can make ${{a}^{2}}+2a\to \left( 1 \right)$ and $ab+2b\to \left( 2 \right)$, therefore the original equation will be $\left( 1 \right)+\left( 2 \right)$.
Now consider $\left( 1 \right)$,
$\Rightarrow {{a}^{2}}+2a$
Now since $a$ is common in both the terms, we can take it out as common and write it as:
$\Rightarrow a\left( a+2 \right)$
Now consider $\left( 2 \right)$,
$\Rightarrow ab+2b$
Now since $b$ is common in both the terms, we can take it out as common and write it as:
$\Rightarrow b\left( a+2 \right)$
Now since both the equations are simplified, we can join equation $\left( 1 \right)$ and $\left( 2 \right)$, and write it as:
$\Rightarrow a\left( a+2 \right)+b\left( a+2 \right)$
Now since the term $\left( a+2 \right)$ is common in both the terms we can take it out as common and write it as:
$\Rightarrow \left( a+2 \right)\left( a+b \right)$, which is the final factorized way of writing the given equation.
Note: To check whether the final factored equation is correct we can multiply the factors, and if we get the given equation back, the factors are correct.
We can multiply the factors as:
$a\times a+a\times b+2\times a+2\times b$
On simplifying we get:
${{a}^{2}}+ab+2a+2b$
On rearranging we get:
${{a}^{2}}+2a+ab+2b$, which is the original equation, therefore the factors are correct.
It is to be remembered that factors are the digits which make up a number, a factor should be indivisible by any number except $1$ and it should not be in the form of a number which is raised to a power.
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