How do you factor the trinomial $10{{x}^{2}}+11x+3$ ?

Hint: The trinomial given in the above problem is a quadratic expression and we are going to factorize this quadratic expression by multiplying the coefficient of ${{x}^{2}}$ and the constant and then we will write the factors for this result of multiplication. And then arrange the factors in such a way so that addition or subtraction will give the coefficient of x. After that we will find some common terms in the quadratic expression and in this way, we will find the factors of the given trinomial.

Complete step-by-step solution:
The equation given above which we have to solve is as follows:
$10{{x}^{2}}+11x+3$
As you can see that the highest power of this equation is 2 so the degree of the above equation is 2 and hence, the above equation is a quadratic equation.
Now, we are going to multiply the coefficient of ${{x}^{2}}$ (i.e. 10) and the constant term (i.e. 3) so the multiplication of these two terms is:
$\Rightarrow 10\times 3=30$
Now, we are going to write the factors of 30 which are equal to:
$\begin{align}
  & 30=1\times 30 \\
 & 30=2\times 15 \\
 & 30=3\times 10 \\
 & 30=5\times 6 \\
\end{align}$
If you look at the last factorization of 30 then you will find that on adding this factor (5 + 6) you will get the coefficient of x which is 11 so substituting $5+6$ in place of 11 in the above equation we get:
$\begin{align}
  & \Rightarrow 10{{x}^{2}}+\left( 5+6 \right)x+3 \\
 & \Rightarrow 10{{x}^{2}}+5x+6x+3 \\
\end{align}$
Taking $5x$ as common from the first two terms in the above expression and 3 as common from the last two terms and we get,
$\Rightarrow 5x\left( 2x+1 \right)+3\left( 2x+1 \right)$
Now, taking $\left( 2x+1 \right)$ as common from the above we get,
$\Rightarrow \left( 2x+1 \right)\left( 5x+3 \right)$
Hence, we have factorized the given trinomial to $\left( 2x+1 \right)\left( 5x+3 \right)$.

Note: You can check the above factors of the trinomial which we have solved solutions by equating each bracket to 0 and then find the values of x. After that we will substitute these values of x and see whether putting these values in the given trinomial will give you answer 0 or not.
Equating $\left( 2x+1 \right)$ to 0 we get,
$\begin{align}
  & \Rightarrow 2x+1=0 \\
 & \Rightarrow x=-\dfrac{1}{2} \\
\end{align}$
Equating $\left( 5x+3 \right)$ to 0 we get,
$\begin{align}
  & \Rightarrow 5x+3=0 \\
 & \Rightarrow x=-\dfrac{3}{5} \\
\end{align}$
Substituting the value of $x=-\dfrac{1}{2}$ in the above trinomial we get,
$\begin{align}
  & \Rightarrow 10{{\left( -\dfrac{1}{2} \right)}^{2}}+11\left( -\dfrac{1}{2} \right)+3 \\
 & \Rightarrow 10\left( \dfrac{1}{4} \right)-\dfrac{11}{2}+3 \\
 & \Rightarrow \dfrac{5}{2}-\dfrac{11}{2}+3 \\
 & \Rightarrow \dfrac{5-11}{2}+3 \\
 & \Rightarrow -\dfrac{6}{2}+3 \\
\end{align}$
In the above, 6 will be cancelled out by 2 three times so the above expression will become:
$\begin{align}
  & \Rightarrow -3+3 \\
 & =0 \\
\end{align}$
Hence, on substituting the value of $x=-\dfrac{1}{2}$ in the given expression we are getting 0. This means the factor $\left( 2x+1 \right)$ is the correct factor of the given trinomial.
Similarly, we can check the value of $x=-\dfrac{3}{5}$ and see whether $\left( 5x+3 \right)$ is the correct factor or not.