Question

How do you factor \[2{{a}^{2}}+3a+1=0\]?

Answer 1

Hint: This type of problem is based on the concept of factoring and solving a. First, we have to consider the given equation with the variable a. Then, split the middle term of the equation in such a way that it has a common term in both the first and last term. Here, the middle term is 3a, so we can write 3a as the sum of 2a and a. Then, take the common terms from the obtained equation and represent the equation as a product of two functions of a. Thus, the two functions are the factors of the given equation which is the required answer.

Complete step by step solution:
According to the question, we are asked to find the factor of \[2{{a}^{2}}+3a+1=0\].
 We have been given the quadratic equation is \[2{{a}^{2}}+3a+1=0\]. ---------(1)
Here, the given quadratic equation is with variable a.
Let us now split the middle term in such a way that the sum of the two terms is equal to 3 and the product of the two terms is equal to 2.
We know that \[2\times 1=2\] and 2+1=3.
Let us substitute in equation (1).
We get
\[2{{a}^{2}}+\left( 2+1 \right)a+1=0\]
On using distributive property in the obtained equation, that is, \[a\left( b+c \right)=ab+ac\]
We get
\[2{{a}^{2}}+2a+1a+1=0\]
Let us now find the common term.
Here, 2a is the common term from the first two terms and 1 is common from the last two terms.
Therefore, we get
 \[2a\left( a+1 \right)+1\left( a+1 \right)=0\].
From the obtained equation, find that (a+1) is common.
On taking out (a+1) common from the two terms, we get
\[\left( a+1 \right)\left( 2a+1 \right)=0\]
Now, we have expressed the given equation as a product of two functions with variable a.
These two functions are the factors of the equation (1).
Therefore, the factors are (a+1) and (2a+1).

Note: Whenever we get such types of problems, we should make necessary calculations to the quadratic equation and then take out the common terms out of the bracket if any, to obtain the values of a. Also, avoid calculation mistakes based on the sign conventions. Here, we can solve this question by using a quadratic formula also.