Solve -8x^2-4x+4

$$- 8 x ^ { 2 } - 4 x + 4$$

Factor

$$4\left(-x-1\right)\left(2x-1\right)$$

Solution Steps

Factor out 4.

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

Consider -2x^{2}-x+1. Factor the expression by grouping. First, the expression needs to be rewritten as -2x^{2}+ax+bx+1. To find a and b, set up a system to be solved.

$$a+b=-1 ab=-2=-2$$

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. The only such pair is the system solution.

$$a=1 b=-2$$

Rewrite -2x^{2}-x+1 as \left(-2x^{2}+x\right)+\left(-2x+1\right).

$$\left(-2x^{2}+x\right)+\left(-2x+1\right)$$

Factor out -x in the first and -1 in the second group.

$$-x\left(2x-1\right)-\left(2x-1\right)$$

Factor out common term 2x-1 by using distributive property.

$$\left(2x-1\right)\left(-x-1\right)$$

Rewrite the complete factored expression.

$$4\left(2x-1\right)\left(-x-1\right)$$

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.

$$-8x^{2}-4x+4=0$$

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.

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

Square -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\left(-8\right)\times 4}}{2\left(-8\right)}$$

Multiply -4 times -8.

$$x=\frac{-\left(-4\right)±\sqrt{16+32\times 4}}{2\left(-8\right)}$$

Multiply 32 times 4.

$$x=\frac{-\left(-4\right)±\sqrt{16+128}}{2\left(-8\right)}$$

Add 16 to 128.

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

Take the square root of 144.

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

The opposite of -4 is 4.

$$x=\frac{4±12}{2\left(-8\right)}$$

Multiply 2 times -8.

$$x=\frac{4±12}{-16}$$

Now solve the equation x=\frac{4±12}{-16} when ± is plus. Add 4 to 12.

$$x=\frac{16}{-16}$$

Divide 16 by -16.

$$x=-1$$

Now solve the equation x=\frac{4±12}{-16} when ± is minus. Subtract 12 from 4.

$$x=-\frac{8}{-16}$$

Reduce the fraction \frac{-8}{-16} to lowest terms by extracting and canceling out 8.

$$x=\frac{1}{2}$$

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

$$-8x^{2}-4x+4=-8\left(x-\left(-1\right)\right)\left(x-\frac{1}{2}\right)$$

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

$$-8x^{2}-4x+4=-8\left(x+1\right)\left(x-\frac{1}{2}\right)$$

Subtract \frac{1}{2} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$-8x^{2}-4x+4=-8\left(x+1\right)\times \frac{-2x+1}{-2}$$

Cancel out 2, the greatest common factor in -8 and 2.

$$-8x^{2}-4x+4=4\left(x+1\right)\left(-2x+1\right)$$

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.

$$x ^ 2 +\frac{1}{2}x -\frac{1}{2} = 0$$

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

$$r + s = -\frac{1}{2} rs = -\frac{1}{2}$$

Two numbers r and s sum up to -\frac{1}{2} exactly when the average of the two numbers is \frac{1}{2}*-\frac{1}{2} = -\frac{1}{4}. 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>

$$r = -\frac{1}{4} - u s = -\frac{1}{4} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{1}{2}

$$(-\frac{1}{4} - u) (-\frac{1}{4} + u) = -\frac{1}{2}$$

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

$$\frac{1}{16} - u^2 = -\frac{1}{2}$$

Simplify the expression by subtracting \frac{1}{16} on both sides

$$-u^2 = -\frac{1}{2}-\frac{1}{16} = -\frac{9}{16}$$

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

$$u^2 = \frac{9}{16} u = \pm\sqrt{\frac{9}{16}} = \pm \frac{3}{4}$$

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

$$r =-\frac{1}{4} - \frac{3}{4} = -1 s = -\frac{1}{4} + \frac{3}{4} = 0.500$$

Evaluate

$$4-4x-8x^{2}$$