Solve =p^2+9p

$$= p ^ { 2 } + 9 p$$

Factor

$$p\left(p+9\right)$$

Solution Steps

Factor out p.

$$p\left(p+9\right)$$

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.

$$p^{2}+9p=0$$

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.

$$p=\frac{-9±\sqrt{9^{2}}}{2}$$

Take the square root of 9^{2}.

$$p=\frac{-9±9}{2}$$

Now solve the equation p=\frac{-9±9}{2} when ± is plus. Add -9 to 9.

$$p=\frac{0}{2}$$

Divide 0 by 2.

$$p=0$$

Now solve the equation p=\frac{-9±9}{2} when ± is minus. Subtract 9 from -9.

$$p=-\frac{18}{2}$$

Divide -18 by 2.

$$p=-9$$

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 -9 for x_{2}.

$$p^{2}+9p=p\left(p-\left(-9\right)\right)$$

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

$$p^{2}+9p=p\left(p+9\right)$$

Evaluate

$$p\left(p+9\right)$$