For all \[x\], $ {x^2} + 2ax + 10 - 3a > 0 $ , then the interval in which lies is
A. $ a < 5 $
B. $ - 5 < a < 2 $
C. $ a > 5 $
D. $ 2 < a < 5 $

Hint: For a quadratic inequality \[a{x^2} + bx + c > 0\] to be true, $ a $ should be greater than zero and the value of discriminant $ {b^2} - 4ac $ should be less than zero. If $ \left( {x + a} \right)\left( {x - b} \right) < 0 $ , then $ x $ belongs to the interval $ \left( { - b,a} \right) $ , this concept will help in deciding the required interval.

Complete step-by-step answer:
Factorisation of polynomials can also be said as the process of converting addition and subtraction of terms into the form of the product of the terms. Factorisation is necessary to convert the complex looking expressions into the product of linear terms. In 1793, Theodor von Schubert produced the first algorithm for the factorization of the polynomial. The linear factors are being deduced from the polynomials in order to factorise the polynomials.
As per the question, for all $ x $ , $ {x^2} + 2ax + 10 - 3a > 0 $ .
We know that, $ a{x^2} + bx + c > 0 $ only when $ a > 0 $ and $ {b^2} - 4ac < 0 $ .
Now, we can compare the inequality
$ {x^2} + 2ax + 10 - 3a > 0 $ with $ a{x^2} + bx + c > 0 $ .
On comparing, we get, $ a = 1 $ , $ b = 2a $ , $ c = \left( {10 - 3a} \right) $ .
Now, substitute 1 for $ a $ , $ 2a $ for $ b $ and $ 10 - 3a $ for $ c $ in the inequality $ {b^2} - 4ac < 0 $ .
After substituting, we get,
 $ \Rightarrow {\left( {2a} \right)^2} - 4\left( 1 \right)\left( {10 - 3a} \right) < 0 $ .
On simplifying the equation, we get,
 $
\Rightarrow 4{a^2} - 4\left( {10 - 3a} \right) < 0\\
4{a^2} - 40 + 12a < 0
 $
Now, we can divide the entire inequality by 4.
 $ {a^2} - 10 + 3a < 0 $
On rearranging the inequality, we get,
 $ \Rightarrow {a^2} + 3a - 10 < 0......\left( 1 \right) $
We can factorise $ {a^2} + 3a - 10 $ .
We know that, $ 5 - 2 = 3 $ and $ 5 \times \left( { - 2} \right) = - 10 $
So, $ {a^2} + 3a - 10 $ can be further factorised as
 $ {a^2} + 3a - 10 = {a^2} + 5a - 2a - 10 $
On factorising, we get,
 $
{a^2} + 3a - 10 = a\left( {a + 5} \right) - 2\left( {a + 5} \right)\\
 = \left( {a + 5} \right)\left( {a - 2} \right)......\left( 2 \right)
 $
From equations (1) and (2),
 $ \left( {a + 5} \right)\left( {a - 2} \right) < 0 $
This means, $ a \in \left( { - 5,2} \right) $ .

Note: Students often find it difficult to find the factors of the equation. So, it is important to keep in mind that for the equation of the form $ {x^2} - mx + n $ , if the roots are $ a,{\rm{ }}b $ then $ a + b = m $ and $ a \cdot b = n $ We need to just compare the given expression to $ {x^2} - mx + n $ and convert it in the form of $ \left( {x - a} \right)\left( {x - b} \right) $ . This can be an alternative method to solve this problem.