Solve -4x^2+4x

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

Factor

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

Solution Steps

Factor out 4.

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

Consider -x^{2}+x. Factor out x.

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

Rewrite the complete factored expression.

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

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.

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

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.

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

Take the square root of 4^{2}.

$$x=\frac{-4±4}{2\left(-4\right)}$$

Multiply 2 times -4.

$$x=\frac{-4±4}{-8}$$

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

$$x=\frac{0}{-8}$$

Divide 0 by -8.

$$x=0$$

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

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

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 0 for x_{1} and 1 for x_{2}.

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

Evaluate

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