Solve f(x)=3x^2-12x

$$f ( x ) = 3 x ^ { 2 } - 12 x$$

Factor

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

Solution Steps

Factor out 3.

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

Consider x^{2}-4x. Factor out x.

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

Rewrite the complete factored expression.

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

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.

$$3x^{2}-12x=0$$

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.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}}}{2\times 3}$$

Take the square root of \left(-12\right)^{2}.

$$x=\frac{-\left(-12\right)±12}{2\times 3}$$

The opposite of -12 is 12.

$$x=\frac{12±12}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{12±12}{6}$$

Now solve the equation x=\frac{12±12}{6} when ± is plus. Add 12 to 12.

$$x=\frac{24}{6}$$

Divide 24 by 6.

$$x=4$$

Now solve the equation x=\frac{12±12}{6} when ± is minus. Subtract 12 from 12.

$$x=\frac{0}{6}$$

Divide 0 by 6.

$$x=0$$

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 0 for x_{2}.

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

Evaluate

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