Solve 4p^2+16p

$$4 p ^ { 2 } + 16 p$$

Factor

$$4p\left(p+4\right)$$

Solution Steps

Factor out 4.

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

Consider p^{2}+4p. Factor out p.

$$p\left(p+4\right)$$

Rewrite the complete factored expression.

$$4p\left(p+4\right)$$

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.

$$4p^{2}+16p=0$$

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.

$$p=\frac{-16±\sqrt{16^{2}}}{2\times 4}$$

Take the square root of 16^{2}.

$$p=\frac{-16±16}{2\times 4}$$

Multiply 2 times 4.

$$p=\frac{-16±16}{8}$$

Now solve the equation p=\frac{-16±16}{8} when ± is plus. Add -16 to 16.

$$p=\frac{0}{8}$$

Divide 0 by 8.

$$p=0$$

Now solve the equation p=\frac{-16±16}{8} when ± is minus. Subtract 16 from -16.

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

Divide -32 by 8.

$$p=-4$$

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 -4 for x_{2}.

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

Simplify all the expressions of the form p-\left(-q\right) to p+q.

$$4p^{2}+16p=4p\left(p+4\right)$$

Evaluate

$$4p\left(p+4\right)$$