Answer
$p = -9$ or $p = 3$
Work Step by Step
First, we want to rewrite this equation as a quadratic one. The quadratic equation is given as:
$ax^2 + bx + c$, where $a$, $b$, and $c$ are all real numbers.
To do this, we subtract $27$ from each side of the equation:
$p^2 + 6p - 27 = 0$
To factor this equation, we want to find which factors, when multiplied, will give us the product of the $a$ and $c$ terms but when added together will give us the $b$ term.
Let's look at possible factors:
$9$ and $-3$
$27$ and $-1$
It looks like the first combination will work. Let's put the factors together:
$(p + 9)(p - 3) = 0$
According to the zero product property, if the product of two factors equals $0$, then either factor can be $0$; therefore, we can set each of these factors equal to $0$ and solve:
$p + 9 = 0$ or $p - 3 = 0$
Add or subtract to solve:
$p = -9$ or $p = 3$
To check our answer, we substitute each of these values into the original equation to see if the two sides equal one another:
$(-9)^2 + 6(-9) = 27$
Multiply to simplify:
$81 - 54 = 27$
Subtract:
$27 = 27$
The two sides of the equation are equal to one another; therefore, this solution is correct. Let's check the other solution:
$(3)^2 + 6(3) = 27$
Multiply to simplify:
$9 + 18 = 27$
Subtract:
$27 = 27$
The two sides of the equation are equal to one another; therefore, this solution is correct.
