How do you factorize $10{{x}^{2}}+3x-4$?

Hint: We will be using the method of factorization to factor the given quadratic equation. We have the middle term as 3x, so we need to find two numbers such that their sum is 3x and the product is -40. Once we get the numbers, we can easily take out the common factors and get the answer.

Complete step-by-step solution:
The given equation is $10{{x}^{2}}+3x-4$ which is a quadratic equation. if we compare the equation to standard quadratic equation $a{{x}^{2}}+bx+c$ then
$a=10,b=3$ and $c=-4$
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 split $bx$ to $mx+nx$ then we can factor the equation easily.
In our case
$\begin{align}
  & ac=10\times (-4) \\
 & ac=-40 \\
\end{align}$
And $b=3$
So pair of 2 numbers whose product is -40 are (1,-40) , (2,-20) , (4,-10) , (5,-8), (8,-5),(10,-4),(20,-2) and (40,-1) .
But there is only one pair whose sum is 3 that is (8,-5)
We can spilt $3x$ to $8x-5x$
So $10{{x}^{2}}+3x-4=10{{x}^{2}}+8x-5x-4$
Taking 2x common in the first half of the equation and taking -1 common in the second half of the equation.
$\Rightarrow 10{{x}^{2}}+8x-5x-4=2x\left( 5x+4 \right)-1\left( 5x+4 \right)$
Taking $5x+4$ common
$\Rightarrow 2x\left( 5x+4 \right)-1\left( 5x+4 \right)=\left( 5x+4 \right)\left( 2x-1 \right)$
In this way we can factor a quadratic equation; another method is directly finding the roots of the equation.

Note: While factoring a quadratic equation spilt the $bx$ term in such a way that the product of the coefficient of x is equal to $ac$ (product of the coefficient of ${{x}^{2}}$ and constant term). Sometimes we can’t spilt $bx$ because the roots may come as irrational.in that case we have to first find the roots of equation then we can write $a{{x}^{2}}+bx+c=a\left( x-\alpha \right)\left( x-\beta \right)$ where $\alpha $ and $\beta $ are roots of the equation. Be careful while choosing the sign of the pair 5 and 8. We require the sum to be positive, so 8 must-have positive signs, and 5 must have negative signs.