Solve 18x-3x^2

$$18x-3 { x }^{ 2 }$$

Factor

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

Solution Steps

Factor out 3.

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

Consider 6x-x^{2}. Factor out x.

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

Rewrite the complete factored expression.

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

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.

$$-3x^{2}+18x=0$$

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.

$$x=\frac{-18±\sqrt{18^{2}}}{2\left(-3\right)}$$

Take the square root of 18^{2}.

$$x=\frac{-18±18}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{-18±18}{-6}$$

Now solve the equation x=\frac{-18±18}{-6} when ± is plus. Add -18 to 18.

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

Divide 0 by -6.

$$x=0$$

Now solve the equation x=\frac{-18±18}{-6} when ± is minus. Subtract 18 from -18.

$$x=-\frac{36}{-6}$$

Divide -36 by -6.

$$x=6$$

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 6 for x_{2}.

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

Evaluate

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