How do you factor and solve \[3{x^2} + 6x + 3 = 0\] ?

Hint:Factoring reduces the higher degree equation into its linear equation. In the above given question, we need to reduce the quadratic equation into its simplest form in such a way that addition of products of the factors of first and last term should be equal to the middle term

Complete step by step solution:
 \[a{x^2} + bx + c\] is a general way of writing quadratic equations where a, b and c are the numbers.
In the above expression,
a=3, b=6, c=3
 \[3{x^2} + 6x + 3 = 0\]
We can reduce the above by dividing it by 3.
 \[{x^2} + 2x + 1 = 0\]
where now a=1 b=2 c=1
First step is by multiplying the coefficient of \[{x^2}\] and the constant term 1, we get ${x^2}$.
After this, factors of ${x^2}$should be calculated in such a way that their addition should be equal to $2x$.
Factors of 1 can be 1 and 1
where $x + x = 2x$.
So, further we write the equation by equating it with zero and splitting the middle term according to the factors.
 \[
\Rightarrow {x^2} + 2x + 1 = 0 \\
\Rightarrow {x^2} + x + x + 1 = 0 \\
\]
Now, by grouping the first two and last two terms we get common factors.
 \[
\Rightarrow x\left( {x + 1} \right) + 1\left( {x + 1} \right) = 0 \\
\Rightarrow \left( {x + 1} \right)(x + 1) = 0 \\
\]
Taking x common from the first group and 1 common from the second we get the above equation. We can further solve it and taking square root we get,
 \[
{\left( {x + 1} \right)^2} = 0 \\
x = - 1 \\
\]
Therefore, we get the above value for $x$.

Note: In quadratic equation, an alternative way of finding the factors is by directly solving the equation by using a formula which is given below:

 \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\] .By substituting the values of a=3, b=6 and c=3 we get the factors of x.
Therefore, the value we get is -1.