Solve f(x)=-3x^2+9x

$$f ( x ) = - 3 x ^ { 2 } + 9 x$$

Factor

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

Solution Steps

Factor out 3.

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

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

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

Rewrite the complete factored expression.

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

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.

$$-3x^{2}+9x=0$$

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.

$$x=\frac{-9±\sqrt{9^{2}}}{2\left(-3\right)}$$

Take the square root of 9^{2}.

$$x=\frac{-9±9}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{-9±9}{-6}$$

Now solve the equation x=\frac{-9±9}{-6} when ± is plus. Add -9 to 9.

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

Divide 0 by -6.

$$x=0$$

Now solve the equation x=\frac{-9±9}{-6} when ± is minus. Subtract 9 from -9.

$$x=-\frac{18}{-6}$$

Divide -18 by -6.

$$x=3$$

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 3 for x_{2}.

$$-3x^{2}+9x=-3x\left(x-3\right)$$

Evaluate

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