Solve f(x)=-3x^2-12x-8

$$f ( x ) = - 3 x ^ { 2 } - 12 x - 8$$

Factor

$$-3\left(x-\left(-\frac{2\sqrt{3}}{3}-2\right)\right)\left(x-\left(\frac{2\sqrt{3}}{3}-2\right)\right)$$

Steps Using the Quadratic Formula

Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$-3x^{2}-12x-8=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\left(-3\right)\left(-8\right)}}{2\left(-3\right)}$$

Square -12.

$$x=\frac{-\left(-12\right)±\sqrt{144-4\left(-3\right)\left(-8\right)}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-\left(-12\right)±\sqrt{144+12\left(-8\right)}}{2\left(-3\right)}$$

Multiply 12 times -8.

$$x=\frac{-\left(-12\right)±\sqrt{144-96}}{2\left(-3\right)}$$

Add 144 to -96.

$$x=\frac{-\left(-12\right)±\sqrt{48}}{2\left(-3\right)}$$

Take the square root of 48.

$$x=\frac{-\left(-12\right)±4\sqrt{3}}{2\left(-3\right)}$$

The opposite of -12 is 12.

$$x=\frac{12±4\sqrt{3}}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{12±4\sqrt{3}}{-6}$$

Now solve the equation x=\frac{12±4\sqrt{3}}{-6} when ± is plus. Add 12 to 4\sqrt{3}.

$$x=\frac{4\sqrt{3}+12}{-6}$$

Divide 12+4\sqrt{3} by -6.

$$x=-\frac{2\sqrt{3}}{3}-2$$

Now solve the equation x=\frac{12±4\sqrt{3}}{-6} when ± is minus. Subtract 4\sqrt{3} from 12.

$$x=\frac{12-4\sqrt{3}}{-6}$$

Divide 12-4\sqrt{3} by -6.

$$x=\frac{2\sqrt{3}}{3}-2$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -2-\frac{2\sqrt{3}}{3} for x_{1} and -2+\frac{2\sqrt{3}}{3} for x_{2}.

$$-3x^{2}-12x-8=-3\left(x-\left(-\frac{2\sqrt{3}}{3}-2\right)\right)\left(x-\left(\frac{2\sqrt{3}}{3}-2\right)\right)$$

Steps Using Direct Factoring Method

Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.

$$x ^ 2 +4x +\frac{8}{3} = 0$$

Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C

$$r + s = -4 rs = \frac{8}{3}$$

Two numbers r and s sum up to -4 exactly when the average of the two numbers is \frac{1}{2}*-4 = -2. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = -2 - u s = -2 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{8}{3}

$$(-2 - u) (-2 + u) = \frac{8}{3}$$

Simplify by expanding (a -b) (a + b) = a^2 – b^2

$$4 - u^2 = \frac{8}{3}$$

Simplify the expression by subtracting 4 on both sides

$$-u^2 = \frac{8}{3}-4 = -\frac{4}{3}$$

Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u

$$u^2 = \frac{4}{3} u = \pm\sqrt{\frac{4}{3}} = \pm \frac{2}{\sqrt{3}}$$

The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.

$$r =-2 - \frac{2}{\sqrt{3}} = -3.155 s = -2 + \frac{2}{\sqrt{3}} = -0.845$$

Evaluate

$$-3x^{2}-12x-8$$