Question

How do you factor \[2{{x}^{2}}+5x+2\]?

Answer 1

Hint: In the given question, we are given a quadratic equation and we are asked to factor it. We are given the quadratic equation in the form \[a{{x}^{2}}+bx+c\]. In order to factor the given equation, we will have to split the middle term and for this we will be applying the sum and the product rule. The sum of the middle terms should give \[b=5\] and the product should be \[ac=2\times 2=4\]. We will then have, \[\Rightarrow 2{{x}^{2}}+4x+x+2\]. We will then take the common terms out and we will get the required factors of the given quadratic equation.

Complete step-by-step solution:
According to the given equation, we are given a quadratic equation which we have to factor.
The given quadratic equation that we have is,
\[2{{x}^{2}}+5x+2\]----(1)
We can see that the given equation is of the form \[a{{x}^{2}}+bx+c\], where ‘a’, ‘b’ and ‘c’ have the corresponding values from the equation (1).
In order to factor the given quadratic equation, we will have to split the middle term. And we will be using the sum ad the product rule for the same.
The sum of the middle terms should be \[b=5\].
The product of the middle terms should be \[ac=2\times 2=4\].
And so we have,
\[= 2{{x}^{2}}+(4x+x)+2\]----(2)
We will open up the brackets and we get,
\[= 2{{x}^{2}}+4x+x+2\]-----(3)
We will now pair up the terms and we will take the common parts from each of the pair made and we get,
\[= 2x\left( x+2 \right)+1\left( x+2 \right)\]----(4)
From the equation (4), we will take \[\left( x+2 \right)\] out as common, we get,
\[= \left( x+2 \right)\left( 2x+1 \right)\]
Therefore, the factors of the given quadratic equation \[2{{x}^{2}}+5x+2=\left( x+2 \right)\left( 2x+1 \right)\].

Note: The factors of an expression simply refer to the monomial terms that when multiplied give the corresponding quadratic equation. So, we do not have to solve for the variable involved or solve in terms of the variable in the expression and also it would not be the required answer. While splitting the middle terms make sure that the terms add up and multiply to give the appropriate answers or not, else the answer will come out wrong.