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

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

Factor

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

Solution Steps

Factor out 3. Polynomial x-x^{2}-4 is not factored since it does not have any rational roots.

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

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.

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

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.

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

Square 3.

$$x=\frac{-3±\sqrt{9-4\left(-3\right)\left(-12\right)}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-3±\sqrt{9+12\left(-12\right)}}{2\left(-3\right)}$$

Multiply 12 times -12.

$$x=\frac{-3±\sqrt{9-144}}{2\left(-3\right)}$$

Add 9 to -144.

$$x=\frac{-3±\sqrt{-135}}{2\left(-3\right)}$$

Since the square root of a negative number is not defined in the real field, there are no solutions. Quadratic polynomial cannot be factored.

$$-3x^{2}+3x-12$$

Evaluate

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