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Solve the polynomial inequality and graph the solution set on a real number line. Express the solution set in interval notation.

[tex]\[ x^2 - 12x + 27 \ \textgreater \ 0 \][/tex]

Use the inequality in the form [tex]\( f(x) \ \textgreater \ 0 \)[/tex], to write the intervals determined by the boundary points as they appear from left to right on a number line.

[tex]\[
\begin{array}{|c|c|c|c|}
\hline
\text{Interval} & (-\infty, 3) & \square & (9, \infty) \\
\hline
\text{Sign} & \text{positive} & \text{negative} & \text{positive} \\
\hline
\end{array}
\][/tex]

(Type your answers in interval notation. Use ascending order.)

Sagot :

GinnyAnswer
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]
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