Solve 4x-12x^2

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

Factor

$$4x\left(1-3x\right)$$

Solution Steps

Factor out 4.

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

Consider x-3x^{2}. Factor out x.

$$x\left(1-3x\right)$$

Rewrite the complete factored expression.

$$4x\left(-3x+1\right)$$

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.

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

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.

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

Take the square root of 4^{2}.

$$x=\frac{-4±4}{2\left(-12\right)}$$

Multiply 2 times -12.

$$x=\frac{-4±4}{-24}$$

Now solve the equation x=\frac{-4±4}{-24} when ± is plus. Add -4 to 4.

$$x=\frac{0}{-24}$$

Divide 0 by -24.

$$x=0$$

Now solve the equation x=\frac{-4±4}{-24} when ± is minus. Subtract 4 from -4.

$$x=-\frac{8}{-24}$$

Reduce the fraction \frac{-8}{-24} to lowest terms by extracting and canceling out 8.

$$x=\frac{1}{3}$$

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 \frac{1}{3} for x_{2}.

$$-12x^{2}+4x=-12x\left(x-\frac{1}{3}\right)$$

Subtract \frac{1}{3} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$-12x^{2}+4x=-12x\times \frac{-3x+1}{-3}$$

Cancel out 3, the greatest common factor in -12 and -3.

$$-12x^{2}+4x=4x\left(-3x+1\right)$$

Evaluate

$$4x\left(1-3x\right)$$