How do you factor \[5{{x}^{2}}+30x+50\]?

Hint: First take 5 common from all the terms present. Now, check if real factors can be formed by finding the discriminant value. Substitute the obtained quadratic expression equal to 0 and find the values of x and write them as x = m and x = n. Now, consider the product \[\left( x-m \right)\left( x-n \right)\] to get the required factored form of the polynomial.

Complete step by step answer:
Here, we have been provided with the quadratic polynomial: \[5{{x}^{2}}+30x+50\] and we are asked to factor it.
Taking 5 common from all the terms, we get,
\[\Rightarrow 5{{x}^{2}}+30x+50=5\left( {{x}^{2}}+6x+10 \right)\]
Now, we need to factorize \[\left( {{x}^{2}}+6x+10 \right)\]. First let us check if it can be factored into real factors or not. For this we need to find the discriminant value of this equation. If the discriminant value is greater than or equal to 0 then we can form real factors and if the discriminant value is greater than or equal to 0 then we can form real factors and if the discriminant value is less than 0 then we can form complex factors. So, let us check.
Considering \[{{x}^{2}}+6x+10\] and assuming the coefficient of \[{{x}^{2}}\] as ‘a’, coefficient of x as ‘b’ and the constant term as ‘c’, we have,
Applying the discriminant formula: -
\[\Rightarrow \] Discriminant (D) = \[{{b}^{2}}-4ac\]
\[\begin{align}
  & \Rightarrow D={{6}^{2}}-4\left( 1 \right)\left( 10 \right) \\
 & \Rightarrow D=36-40 \\
 & \Rightarrow D=-4 \\
 & \Rightarrow D<0 \\
\end{align}\]
Therefore, we have to form complex factors. To do this, we have to apply the quadratic formula to find the values of x. So, using the quadratic formula, we have,
\[\begin{align}
  & \Rightarrow x=\dfrac{-b\pm \sqrt{D}}{2a} \\
 & \Rightarrow x=\dfrac{-6\pm \sqrt{-4}}{2\times 1} \\
\end{align}\]
\[\begin{align}
  & \Rightarrow x=\dfrac{-6\pm 2i}{2} \\
 & \Rightarrow x=-3\pm i \\
\end{align}\]
Since, \[x=-3+i\] and \[x=-3-i\] are the roots of the given equation so \[\left[ x-\left( -3+i \right) \right]\left[ x-\left( -3-i \right) \right]\] are the factors of the equation.
So, we can write,
\[\Rightarrow 5{{x}^{2}}+30x+50=5\left( x+3-i \right)\left( x+3+i \right)\]
Hence, \[5\left( x+3-i \right)\left( x+3+i \right)\] is the factored form of the given quadratic polynomial.

Note:
One may note that we have not applied the middle term split method to factorize the polynomial because the roots are not real and it will be difficult for us to think of the factors like \[-3+i\] and \[-3-i\]. Remember the quadratic formula to solve the above question. Note that we have taken 5 common from all the terms at the initial step of the solution to make our calculation of the discriminant easier.