How do you factor $2{{y}^{2}}+7y+5$?

Hint: First multiply the coefficient of \[{{y}^{2}}\] with the constant term. Then find two numbers whose sum or difference is the coefficient of ‘y’ and multiplication is the previous result. Split the given equation in terms of multiplication of two factors. Do the necessary simplification to obtain the required result.

Complete step by step solution:
Factorization: It is simply the resolution of a polynomial into factors such that when multiplied together they will result in an original polynomial. In the factorization method, we can reduce any algebraic or quadratic equation into its simpler form, where the equations are represented as the product of factors.
The equation we have, $2{{y}^{2}}+7y+5$
Multiplying the coefficient of \[{{x}^{2}}\] with the constant term, we get $2\times 5=10$
Now, we have to find two numbers whose sum is ‘7’ and multiplication is ‘10’.
Thus, the numbers are ‘2’ and ‘5’.
Hence, our equation can be written as
$\begin{align}
  & 2{{y}^{2}}+7y+5 \\
 & \Rightarrow 2{{y}^{2}}+2y+5y+5 \\
\end{align}$
Taking common ‘2y’ from first two terms and ‘5’ from last two terms, we get
$\Rightarrow 2y\left( y+1 \right)+5\left( y+1 \right)$
Again taking common $\left( y+1 \right)$ from both the terms, we get
$\Rightarrow \left( y+1 \right)\left( 2y+5 \right)$
This is the required solution of the given question.

Note: Finding two numbers whose sum is 20 and multiplication is $3\times 12=36$, should be the first approach to the factorization. Then taking common terms out and grouping them together we can obtain the required factorization. We can also execute the factorization by completing the square method by reducing the coefficient of ‘\[{{y}^{2}}\]’.