Solve -4x^2+x

$$-4 { x }^{ 2 } +x$$

Factor

$$x\left(1-4x\right)$$

Solution Steps

Factor out x.

$$x\left(-4x+1\right)$$

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.

$$-4x^{2}+x=0$$

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.

$$x=\frac{-1±\sqrt{1^{2}}}{2\left(-4\right)}$$

Take the square root of 1^{2}.

$$x=\frac{-1±1}{2\left(-4\right)}$$

Multiply 2 times -4.

$$x=\frac{-1±1}{-8}$$

Now solve the equation x=\frac{-1±1}{-8} when ± is plus. Add -1 to 1.

$$x=\frac{0}{-8}$$

Divide 0 by -8.

$$x=0$$

Now solve the equation x=\frac{-1±1}{-8} when ± is minus. Subtract 1 from -1.

$$x=-\frac{2}{-8}$$

Reduce the fraction \frac{-2}{-8} to lowest terms by extracting and canceling out 2.

$$x=\frac{1}{4}$$

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}{4} for x_{2}.

$$-4x^{2}+x=-4x\left(x-\frac{1}{4}\right)$$

Subtract \frac{1}{4} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$-4x^{2}+x=-4x\times \frac{-4x+1}{-4}$$

Cancel out 4, the greatest common factor in -4 and -4.

$$-4x^{2}+x=x\left(-4x+1\right)$$

Evaluate

$$x\left(1-4x\right)$$