How do you factor the expression $-10{{x}^{2}}-23x-12$ ?

Hint: The given equation is a quadratic equation. To factorize a quadratic equation $a{{x}^{2}}+bx+c$ , we will write bx = mx + nx such that mn = ac . Then we can take common from $a{{x}^{2}}+mx$ and nx + c and we will find that after taking common the quotient will be the same in both cases. We can take the quotient common from the whole equation.

Complete step by step solution:
The given equation is $-10{{x}^{2}}-23x-12$
If we compare it with $a{{x}^{2}}+bx+c$ we get a = - 10, b = - 23 and c = - 12
The value of ac is 120
So, we have to find 2 numbers such that their sum is - 23 and product is 120. The numbers are - 8 and – 15
So, we can write
 $\Rightarrow -10{{x}^{2}}-23x-12=-10{{x}^{2}}-15x-8x-12$
We can take - 5x common from first 2 terms and - 4 common from last 2 terms
$\Rightarrow -10{{x}^{2}}-23x-12=-5x\left( 2x+3 \right)-4\left( 2x+3 \right)$
Now we can take 2x + 3 common from the whole equation
$\Rightarrow -10{{x}^{2}}-23x-12=\left( -5x-4 \right)\left( 2x+3 \right)$
We can take – 1 out side
$\Rightarrow -10{{x}^{2}}-23x-12=-\left( 5x+4 \right)\left( 2x+3 \right)$
$-\left( 5x+4 \right)\left( 2x+3 \right)$ is the factored form of the equation $-10{{x}^{2}}-23x-12$

Note: The graph of a quadratic equation is always parabolic. In quadratic equation $a{{x}^{2}}+bx+c$ if a is positive, then the parabola will be upward and if a is negative then the parabola will be downward. The parabola will be symmetric with respect to the straight-line x = $-\dfrac{b}{2a}$ . we will get the maximum or minimum value at x = $-\dfrac{b}{2a}$