Solve 4p-20p^2

$$4p-20 { p }^{ 2 }$$

Factor

$$4p\left(1-5p\right)$$

Solution Steps

Factor out 4.

$$4\left(p-5p^{2}\right)$$

Consider p-5p^{2}. Factor out p.

$$p\left(1-5p\right)$$

Rewrite the complete factored expression.

$$4p\left(-5p+1\right)$$

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.

$$-20p^{2}+4p=0$$

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.

$$p=\frac{-4±\sqrt{4^{2}}}{2\left(-20\right)}$$

Take the square root of 4^{2}.

$$p=\frac{-4±4}{2\left(-20\right)}$$

Multiply 2 times -20.

$$p=\frac{-4±4}{-40}$$

Now solve the equation p=\frac{-4±4}{-40} when ± is plus. Add -4 to 4.

$$p=\frac{0}{-40}$$

Divide 0 by -40.

$$p=0$$

Now solve the equation p=\frac{-4±4}{-40} when ± is minus. Subtract 4 from -4.

$$p=-\frac{8}{-40}$$

Reduce the fraction \frac{-8}{-40} to lowest terms by extracting and canceling out 8.

$$p=\frac{1}{5}$$

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}{5} for x_{2}.

$$-20p^{2}+4p=-20p\left(p-\frac{1}{5}\right)$$

Subtract \frac{1}{5} from p by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$-20p^{2}+4p=-20p\times \frac{-5p+1}{-5}$$

Cancel out 5, the greatest common factor in -20 and -5.

$$-20p^{2}+4p=4p\left(-5p+1\right)$$

Evaluate

$$4p\left(1-5p\right)$$