How do you factor $ 5{{x}^{2}}+6x+1 $ ?

Hint: When we factorize a quadratic equation $ a{{x}^{2}}+bx+c $ we write $ bx $ as $ mx+nx $ where $ m+n=b $ and $ mn=ac $ , then we can easily factorize the equation. We can split $ 6x $ to $ 5x $ and $ x $ to solve this question.

Complete step by step answer:
The given equation is $ 5{{x}^{2}}+6x+1 $ which a quadratic equation. if we compare the equation to standard quadratic equation $ a{{x}^{2}}+bx+c $ then $ a=5 $ , $ b=6 $ and $ c=1 $
To factor a quadratic equation we can find two numbers m and n such that the sum of m and n is equal to b and the product of m and n is $ ac $. Then we can spilt $ bx $ to $ mx+nx $ then we can factor the equation easily.
In our case
 $ \begin{align}
  & ac=5\times 1 \\
 & ac=5 \\
\end{align} $
and $ b=6 $
So pair of 2 numbers whose product is 6 are (1,5) , (2,6)
But there is only one pair whose sum is 5 that is (1,5)
We can $ 6x $ spilt to $ x+5x $
So
Taking x common in the first half of the equation and taking 1 common in second half of the equation.
 $ \Rightarrow 5{{x}^{2}}+x+5x+1=x\left( 5x+1 \right)+1\left( 5x+1 \right) $
Taking $ 5x+1 $ common
 $ \Rightarrow x\left( 5x+1 \right)+1\left( 5x+1 \right)=\left( x+1 \right)\left( 5x+1 \right) $
In this way, we can factor a quadratic equation another method is directly finding the roots of the equation.

Note:
While factorizing a quadratic equation we can’t always able to split $ bx $ such that their product is equal to ac because sometimes the roots can be an irrational number. So in that case we can try the complete square method to factorize the equation.