Exercise 71 Page 521

To solve the given equation we will factor it and then apply the Zero Product Property.

Factoring

Let's start by rewriting the equation in such way that all the terms are on the left side of the equality sign. n^2-9n=- 18 ⇕ n^2-9n+18=0 Now, to factor a trinomial with a leading coefficient of 1, think of the process as multiplying two binomials in reverse. Let's take a look at the constant term. n^2-9n+ 18=0In this case, we have 18. 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.)

Next, let's consider the coefficient of the linear term. n^2 -9n+ 18=0 For this term, we need the sum of the factors that produced the constant term to equal the coefficient of the linear term, - 9.

We found the factors whose product is 18 and whose sum is - 9. n^2 -9n+ 18=0 ⇕ (n-3)(n-6)=0

Solving the Equation

Now, as we have already factored the equation, we can apply the Zero Product Property to solve it.

We found that the solutions of the given equation are n=3 and n=6.

Checking the Answer

We can substitute our solutions back into the given equation and simplify to check if our answers are correct. We will start with n=3.

Substituting and simplifying created a true statement, so we know that n=3 is a solution of the equation. Let's move on to n=6.

Again, we created a true statement. n=6 is indeed a solution of the equation.