Solve f(x)=6x^2-18x

$$f ( x ) = 6 x ^ { 2 } - 18 x$$

Factor

$$6x\left(x-3\right)$$

Solution Steps

Factor out 6.

$$6\left(x^{2}-3x\right)$$

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

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

Rewrite the complete factored expression.

$$6x\left(x-3\right)$$

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.

$$6x^{2}-18x=0$$

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.

$$x=\frac{-\left(-18\right)±\sqrt{\left(-18\right)^{2}}}{2\times 6}$$

Take the square root of \left(-18\right)^{2}.

$$x=\frac{-\left(-18\right)±18}{2\times 6}$$

The opposite of -18 is 18.

$$x=\frac{18±18}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{18±18}{12}$$

Now solve the equation x=\frac{18±18}{12} when ± is plus. Add 18 to 18.

$$x=\frac{36}{12}$$

Divide 36 by 12.

$$x=3$$

Now solve the equation x=\frac{18±18}{12} when ± is minus. Subtract 18 from 18.

$$x=\frac{0}{12}$$

Divide 0 by 12.

$$x=0$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 3 for x_{1} and 0 for x_{2}.

$$6x^{2}-18x=6\left(x-3\right)x$$

Evaluate

$$6x\left(x-3\right)$$