Solve =-16t^2+4t

$$= -16 { t }^{ 2 } +4t$$

Factor

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

Solution Steps

Factor out 4.

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

Consider -4t^{2}+t. Factor out t.

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

Rewrite the complete factored expression.

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

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.

$$-16t^{2}+4t=0$$

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.

$$t=\frac{-4±\sqrt{4^{2}}}{2\left(-16\right)}$$

Take the square root of 4^{2}.

$$t=\frac{-4±4}{2\left(-16\right)}$$

Multiply 2 times -16.

$$t=\frac{-4±4}{-32}$$

Now solve the equation t=\frac{-4±4}{-32} when ± is plus. Add -4 to 4.

$$t=\frac{0}{-32}$$

Divide 0 by -32.

$$t=0$$

Now solve the equation t=\frac{-4±4}{-32} when ± is minus. Subtract 4 from -4.

$$t=-\frac{8}{-32}$$

Reduce the fraction \frac{-8}{-32} to lowest terms by extracting and canceling out 8.

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

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}.

$$-16t^{2}+4t=-16t\left(t-\frac{1}{4}\right)$$

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

$$-16t^{2}+4t=-16t\times \frac{-4t+1}{-4}$$

Cancel out 4, the greatest common factor in -16 and -4.

$$-16t^{2}+4t=4t\left(-4t+1\right)$$

Evaluate

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