Answer :
Let's solve the polynomial inequality [tex]\( x^2 - 12x + 27 > 0 \)[/tex] step by step.
### Step 1: Set the polynomial equal to zero
Firstly, we need to find the points where the polynomial [tex]\( x^2 - 12x + 27 \)[/tex] equals zero. These points are called boundary points and can be found by solving the equation:
[tex]\[ x^2 - 12x + 27 = 0 \][/tex]
### Step 2: Solve the quadratic equation
To solve the quadratic equation, we can factorize it:
[tex]\[ x^2 - 12x + 27 = (x - 3)(x - 9) = 0 \][/tex]
Setting each factor to zero gives us the boundary points:
[tex]\[ x - 3 = 0 \implies x = 3 \][/tex]
[tex]\[ x - 9 = 0 \implies x = 9 \][/tex]
So, the boundary points are [tex]\( x = 3 \)[/tex] and [tex]\( x = 9 \)[/tex].
### Step 3: Determine the sign intervals
The boundary points divide the number line into three intervals:
1. [tex]\( (-\infty, 3) \)[/tex]
2. [tex]\( (3, 9) \)[/tex]
3. [tex]\( (9, \infty) \)[/tex]
We need to determine whether [tex]\( x^2 - 12x + 27 \)[/tex] is positive or negative in each of these intervals.
#### Interval [tex]\( (-\infty, 3) \)[/tex]:
Choose a test point less than 3, for example, [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = 0^2 - 12(0) + 27 = 27 \][/tex]
Since 27 is positive, [tex]\( f(x) > 0 \)[/tex] on the interval [tex]\( (-\infty, 3) \)[/tex].
#### Interval [tex]\( (3, 9) \)[/tex]:
Choose a test point between 3 and 9, for example, [tex]\( x = 5 \)[/tex]:
[tex]\[ f(5) = 5^2 - 12(5) + 27 = 25 - 60 + 27 = -8 \][/tex]
Since -8 is negative, [tex]\( f(x) < 0 \)[/tex] on the interval [tex]\( (3, 9) \)[/tex].
#### Interval [tex]\( (9, \infty) \)[/tex]:
Choose a test point greater than 9, for example, [tex]\( x = 10 \)[/tex]:
[tex]\[ f(10) = 10^2 - 12(10) + 27 = 100 - 120 + 27 = 7 \][/tex]
Since 7 is positive, [tex]\( f(x) > 0 \)[/tex] on the interval [tex]\( (9, \infty) \)[/tex].
### Step 4: Write the solution in interval notation
The polynomial [tex]\( x^2 - 12x + 27 > 0 \)[/tex] is true wherever the function is positive. Based on our analysis, these intervals are:
[tex]\[ (-\infty, 3) \cup (9, \infty) \][/tex]
### Step 5: Interpret the solution in the given table format
[tex]\[ \begin{tabular}{|c|c|c|c|} \hline Interval & $(-\infty, 3)$ & $(3, 9)$ & $(9, \infty)$ \\ \hline Sign & positive & negative & positive \\ \hline \end{tabular} \][/tex]
Therefore, the solutions to the polynomial inequality [tex]\( x^2 - 12x + 27 > 0 \)[/tex] can be expressed in interval notation as:
[tex]\[ (-\infty, 3) \cup (9, \infty) \][/tex]
Here you also have the graph representation on the number line:
[tex]\[ \begin{array}{ccc} \text{Negative} & \text{Positive} & \text{Positive} \end{array} \][/tex]
And the intervals for which [tex]\( f(x) > 0 \)[/tex] are:
[tex]\[ x \in (-\infty, 3) \cup (9, \infty) \][/tex]
### Step 1: Set the polynomial equal to zero
Firstly, we need to find the points where the polynomial [tex]\( x^2 - 12x + 27 \)[/tex] equals zero. These points are called boundary points and can be found by solving the equation:
[tex]\[ x^2 - 12x + 27 = 0 \][/tex]
### Step 2: Solve the quadratic equation
To solve the quadratic equation, we can factorize it:
[tex]\[ x^2 - 12x + 27 = (x - 3)(x - 9) = 0 \][/tex]
Setting each factor to zero gives us the boundary points:
[tex]\[ x - 3 = 0 \implies x = 3 \][/tex]
[tex]\[ x - 9 = 0 \implies x = 9 \][/tex]
So, the boundary points are [tex]\( x = 3 \)[/tex] and [tex]\( x = 9 \)[/tex].
### Step 3: Determine the sign intervals
The boundary points divide the number line into three intervals:
1. [tex]\( (-\infty, 3) \)[/tex]
2. [tex]\( (3, 9) \)[/tex]
3. [tex]\( (9, \infty) \)[/tex]
We need to determine whether [tex]\( x^2 - 12x + 27 \)[/tex] is positive or negative in each of these intervals.
#### Interval [tex]\( (-\infty, 3) \)[/tex]:
Choose a test point less than 3, for example, [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = 0^2 - 12(0) + 27 = 27 \][/tex]
Since 27 is positive, [tex]\( f(x) > 0 \)[/tex] on the interval [tex]\( (-\infty, 3) \)[/tex].
#### Interval [tex]\( (3, 9) \)[/tex]:
Choose a test point between 3 and 9, for example, [tex]\( x = 5 \)[/tex]:
[tex]\[ f(5) = 5^2 - 12(5) + 27 = 25 - 60 + 27 = -8 \][/tex]
Since -8 is negative, [tex]\( f(x) < 0 \)[/tex] on the interval [tex]\( (3, 9) \)[/tex].
#### Interval [tex]\( (9, \infty) \)[/tex]:
Choose a test point greater than 9, for example, [tex]\( x = 10 \)[/tex]:
[tex]\[ f(10) = 10^2 - 12(10) + 27 = 100 - 120 + 27 = 7 \][/tex]
Since 7 is positive, [tex]\( f(x) > 0 \)[/tex] on the interval [tex]\( (9, \infty) \)[/tex].
### Step 4: Write the solution in interval notation
The polynomial [tex]\( x^2 - 12x + 27 > 0 \)[/tex] is true wherever the function is positive. Based on our analysis, these intervals are:
[tex]\[ (-\infty, 3) \cup (9, \infty) \][/tex]
### Step 5: Interpret the solution in the given table format
[tex]\[ \begin{tabular}{|c|c|c|c|} \hline Interval & $(-\infty, 3)$ & $(3, 9)$ & $(9, \infty)$ \\ \hline Sign & positive & negative & positive \\ \hline \end{tabular} \][/tex]
Therefore, the solutions to the polynomial inequality [tex]\( x^2 - 12x + 27 > 0 \)[/tex] can be expressed in interval notation as:
[tex]\[ (-\infty, 3) \cup (9, \infty) \][/tex]
Here you also have the graph representation on the number line:
[tex]\[ \begin{array}{ccc} \text{Negative} & \text{Positive} & \text{Positive} \end{array} \][/tex]
And the intervals for which [tex]\( f(x) > 0 \)[/tex] are:
[tex]\[ x \in (-\infty, 3) \cup (9, \infty) \][/tex]
