Solve F(x)=-3x^2+x

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

Factor

$$x\left(1-3x\right)$$

Solution Steps

Factor out x.

$$x\left(-3x+1\right)$$

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.

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

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.

$$x=\frac{-1±\sqrt{1^{2}}}{2\left(-3\right)}$$

Take the square root of 1^{2}.

$$x=\frac{-1±1}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{-1±1}{-6}$$

Now solve the equation x=\frac{-1±1}{-6} when ± is plus. Add -1 to 1.

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

Divide 0 by -6.

$$x=0$$

Now solve the equation x=\frac{-1±1}{-6} when ± is minus. Subtract 1 from -1.

$$x=-\frac{2}{-6}$$

Reduce the fraction \frac{-2}{-6} to lowest terms by extracting and canceling out 2.

$$x=\frac{1}{3}$$

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 \frac{1}{3} for x_{2}.

$$-3x^{2}+x=-3x\left(x-\frac{1}{3}\right)$$

Subtract \frac{1}{3} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$-3x^{2}+x=-3x\times \frac{-3x+1}{-3}$$

Cancel out 3, the greatest common factor in -3 and -3.

$$-3x^{2}+x=x\left(-3x+1\right)$$

Evaluate

$$x\left(1-3x\right)$$