Solve 4x-4x^2+8

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

Factor

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

Solution Steps

Factor out 4.

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

Consider x-x^{2}+2. Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

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

Factor the expression by grouping. First, the expression needs to be rewritten as -x^{2}+ax+bx+2. 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 positive, the positive number has greater absolute value than the negative. The only such pair is the system solution.

$$a=2 b=-1$$

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

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

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

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

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

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

Rewrite the complete factored expression.

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

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.

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

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.

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

Square 4.

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

Multiply -4 times -4.

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

Multiply 16 times 8.

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

Add 16 to 128.

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

Take the square root of 144.

$$x=\frac{-4±12}{2\left(-4\right)}$$

Multiply 2 times -4.

$$x=\frac{-4±12}{-8}$$

Now solve the equation x=\frac{-4±12}{-8} when ± is plus. Add -4 to 12.

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

Divide 8 by -8.

$$x=-1$$

Now solve the equation x=\frac{-4±12}{-8} when ± is minus. Subtract 12 from -4.

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

Divide -16 by -8.

$$x=2$$

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 2 for x_{2}.

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

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

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

Evaluate

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