We want to solve the given equation using the Zero Product Property. To do this we need to rewrite the given equation in the factored form. We can start by using factoring.
Factoring
To factor a trinomial with a leading coefficient of 1, think of the process as multiplying two binomials in reverse. Let's start by taking a look at the constant term.
x^2+14x+33=0
In this case, we have 33. This is a positive number, so for the product of the constant terms in the factors to be positive, these constants must have the same sign — both positive or both negative.
Factor Constants
Product of Constants
1 and 33
33
-1 and -33
33
3 and 11
33
-3 and -11
33
Next, let's consider the coefficient of the linear term.
x^2+14x+33=0
For this term we need the sum of the factors that produced the constant term to equal the coefficient of the linear term, 14.
Factors
Sum of Factors
1 and 33
34
-1 and -33
-34
3 and 11
14
-3 and -11
-14
We found the factors whose product is 33 and whose sum is 14.
x^2+14x+33=0
⇕
(x+3)(x+11)=0
Now the equation is written in a factored form.
Zero Product Property
Since the equation is already written in factored form, we can use the Zero Product Property.
We want to solve the quadratic equation by completing the square. To do so we will start by rewriting the equation so all terms with x are on one side of the equation and all constants on the other side.
x^2+14x+33=0 ⇔ x^2+14x=- 33
In a quadratic expression, b is the linear coefficient. For the equation above, we have that b=14. Let's now calculate ( b2 )^2.
The solutions for this equation are x=- 7± 4. Let's separate them into the positive and negative cases.
x=- 7± 4
x_1=- 7+ 4
x_2=- 7- 4
x_1=- 3
x_2=- 11
By completing the square, we found that the solutions are x_1=- 3 and x_2=- 11. Notice that those are the same solutions as obtained previously by factoring and then using Zero Product Property. Both methods are equally good, so they give the same solutions.
