Solve -4x^2-12x+16

$$- 4 x ^ { 2 } - 12 x + 16$$

Factor

$$4\left(1-x\right)\left(x+4\right)$$

Solution Steps

Factor out 4.

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

Consider -x^{2}-3x+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.

$$a+b=-3 ab=-4=-4$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -4.

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

Calculate the sum for each pair.

$$1-4=-3 2-2=0$$

The solution is the pair that gives sum -3.

$$a=1 b=-4$$

Rewrite -x^{2}-3x+4 as \left(-x^{2}+x\right)+\left(-4x+4\right).

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

Factor out x in the first and 4 in the second group.

$$x\left(-x+1\right)+4\left(-x+1\right)$$

Factor out common term -x+1 by using distributive property.

$$\left(-x+1\right)\left(x+4\right)$$

Rewrite the complete factored expression.

$$4\left(-x+1\right)\left(x+4\right)$$

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.

$$-4x^{2}-12x+16=0$$

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.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\left(-4\right)\times 16}}{2\left(-4\right)}$$

Square -12.

$$x=\frac{-\left(-12\right)±\sqrt{144-4\left(-4\right)\times 16}}{2\left(-4\right)}$$

Multiply -4 times -4.

$$x=\frac{-\left(-12\right)±\sqrt{144+16\times 16}}{2\left(-4\right)}$$

Multiply 16 times 16.

$$x=\frac{-\left(-12\right)±\sqrt{144+256}}{2\left(-4\right)}$$

Add 144 to 256.

$$x=\frac{-\left(-12\right)±\sqrt{400}}{2\left(-4\right)}$$

Take the square root of 400.

$$x=\frac{-\left(-12\right)±20}{2\left(-4\right)}$$

The opposite of -12 is 12.

$$x=\frac{12±20}{2\left(-4\right)}$$

Multiply 2 times -4.

$$x=\frac{12±20}{-8}$$

Now solve the equation x=\frac{12±20}{-8} when ± is plus. Add 12 to 20.

$$x=\frac{32}{-8}$$

Divide 32 by -8.

$$x=-4$$

Now solve the equation x=\frac{12±20}{-8} when ± is minus. Subtract 20 from 12.

$$x=-\frac{8}{-8}$$

Divide -8 by -8.

$$x=1$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -4 for x_{1} and 1 for x_{2}.

$$-4x^{2}-12x+16=-4\left(x-\left(-4\right)\right)\left(x-1\right)$$

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

$$-4x^{2}-12x+16=-4\left(x+4\right)\left(x-1\right)$$

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.

$$x ^ 2 +3x -4 = 0$$

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

$$r + s = -3 rs = -4$$

Two numbers r and s sum up to -3 exactly when the average of the two numbers is \frac{1}{2}*-3 = -\frac{3}{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>

$$r = -\frac{3}{2} - u s = -\frac{3}{2} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -4

$$(-\frac{3}{2} - u) (-\frac{3}{2} + u) = -4$$

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

$$\frac{9}{4} - u^2 = -4$$

Simplify the expression by subtracting \frac{9}{4} on both sides

$$-u^2 = -4-\frac{9}{4} = -\frac{25}{4}$$

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

$$u^2 = \frac{25}{4} u = \pm\sqrt{\frac{25}{4}} = \pm \frac{5}{2}$$

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

$$r =-\frac{3}{2} - \frac{5}{2} = -4 s = -\frac{3}{2} + \frac{5}{2} = 1$$

Evaluate

$$4\left(1-x\right)\left(x+4\right)$$