Solve -3x^2-12x-12

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

Factor

$$-3\left(x+2\right)^{2}$$

Solution Steps

Factor out 3.

$$3\left(-x^{2}-4x-4\right)$$

Consider -x^{2}-4x-4. Factor the expression by grouping. First, the expression needs to be rewritten as -x^{2}+ax+bx-4. To find a and b, set up a system to be solved.

$$a+b=-4 ab=-\left(-4\right)=4$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 4.

$$-1,-4 -2,-2$$

Calculate the sum for each pair.

$$-1-4=-5 -2-2=-4$$

The solution is the pair that gives sum -4.

$$a=-2 b=-2$$

Rewrite -x^{2}-4x-4 as \left(-x^{2}-2x\right)+\left(-2x-4\right).

$$\left(-x^{2}-2x\right)+\left(-2x-4\right)$$

Factor out -x in the first and -2 in the second group.

$$-x\left(x+2\right)-2\left(x+2\right)$$

Factor out common term x+2 by using distributive property.

$$\left(x+2\right)\left(-x-2\right)$$

Rewrite the complete factored expression.

$$3\left(x+2\right)\left(-x-2\right)$$

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.

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

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.

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

Square -12.

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

Multiply -4 times -3.

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

Multiply 12 times -12.

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

Add 144 to -144.

$$x=\frac{-\left(-12\right)±\sqrt{0}}{2\left(-3\right)}$$

Take the square root of 0.

$$x=\frac{-\left(-12\right)±0}{2\left(-3\right)}$$

The opposite of -12 is 12.

$$x=\frac{12±0}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{12±0}{-6}$$

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

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

Simplify all the expressions of the form p-\left(-q\right) to p+q.

$$-3x^{2}-12x-12=-3\left(x+2\right)\left(x+2\right)$$

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.

$$x ^ 2 +4x +4 = 0$$

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

$$r + s = -4 rs = 4$$

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 = 4

$$(-2 - u) (-2 + u) = 4$$

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

$$4 - u^2 = 4$$

Simplify the expression by subtracting 4 on both sides

$$-u^2 = 4-4 = 0$$

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

$$u^2 = 0 u = 0$$

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

$$r = s = -2$$

Evaluate

$$-3\left(x+2\right)^{2}$$