Math  /  Algebra

QuestionSolve the polynomial inequality and graph the solution set on a real number line. Express the solution set in interval notation. (x3)(x+9)>0(x-3)(x+9)>0

Studdy Solution

STEP 1

What is this asking? We need to find the values of xx that make the expression (x3)(x+9)(x-3)(x+9) greater than zero, and then show those values on a number line and write them in interval notation! Watch out! Don't forget to consider what happens when (x3)(x+9)(x-3)(x+9) *equals* zero, this helps us find the boundaries of our solution!

STEP 2

1. Find the critical points.
2. Test the intervals.
3. Write the solution.

STEP 3

Let's **find the critical points** by setting each factor equal to zero.
This helps us see where the expression (x3)(x+9)(x-3)(x+9) *could* change its sign (positive or negative).

STEP 4

Setting x3=0x-3=0, we get x=3x=3.
This is our **first critical point**.

STEP 5

Setting x+9=0x+9=0, we get x=9x=-9.
This is our **second critical point**.

STEP 6

Now, we'll **test the intervals** created by these critical points.
We have three intervals to check: x<9x<-9, 9<x<3-9<x<3, and x>3x>3.

STEP 7

Let's pick a **test value** in the interval x<9x<-9, say x=10x=-10.
Plugging it into our expression, we get (103)(10+9)=(13)(1)=13(-10-3)(-10+9) = (-13)(-1) = 13, which is **positive**!

STEP 8

Now, let's pick a **test value** from 9<x<3-9<x<3, say x=0x=0.
Plugging it in, we get (03)(0+9)=(3)(9)=27(0-3)(0+9) = (-3)(9) = -27, which is **negative**!

STEP 9

Finally, let's pick a **test value** from x>3x>3, say x=4x=4.
Plugging it in, we get (43)(4+9)=(1)(13)=13(4-3)(4+9) = (1)(13) = 13, which is **positive**!

STEP 10

We're looking for where (x3)(x+9)(x-3)(x+9) is *greater* than zero, so we want the **positive intervals**.

STEP 11

From our tests, we found that the expression is positive when x<9x<-9 and x>3x>3.

STEP 12

On a number line, we would show open circles at 9-9 and 33 (since we're not including those points) and shade the regions to the left of 9-9 and to the right of 33.

STEP 13

In interval notation, our solution is (,9)(3,)(-\infty, -9) \cup (3, \infty).
The \cup symbol means "union," which combines our two intervals into one solution set!

STEP 14

The solution set is (,9)(3,)(-\infty, -9) \cup (3, \infty).

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