Math / Algebra

QuestionSolve the polynomial inequality and graph the solution set on a real number line. Express the solution set in interval notation. $(x-2)(x+4) \leq 0$

Studdy Solution

STEP 1

What is this asking? We need to find the values of $x$ that make the expression $(x-2)(x+4)$ less than or equal to zero, and then show those values on a number line and write them in interval notation! Watch out! Don't forget to include the endpoints where the expression equals zero!

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 is where our expression could potentially change its sign!

STEP 4

So, $x-2 = 0$ gives us $x = \mathbf{2}$ , and $x+4 = 0$ gives us $x = \mathbf{-4}$ .
These are our critical points!

STEP 5

Now, let's test the intervals created by these critical points.
We have three intervals to check: $x < -4$ , $-4 \leq x \leq 2$ , and $x > 2$ .

STEP 6

For $x < -4$ , let's pick $x = \mathbf{-5}$ .
Plugging this into our expression gives us $(-5-2)(-5+4) = (-7)(-1) = \mathbf{7}$ , which is greater than zero.
So, this interval doesn't satisfy our inequality.

STEP 7

For $-4 \leq x \leq 2$ , let's pick $x = \mathbf{0}$ .
Plugging this in gives us $(0-2)(0+4) = (-2)(4) = \mathbf{-8}$ , which is less than zero.
This interval does satisfy our inequality!

STEP 8

For $x > 2$ , let's pick $x = \mathbf{3}$ .
Plugging this in gives us $(3-2)(3+4) = (1)(7) = \mathbf{7}$ , which is greater than zero.
So, this interval doesn't satisfy our inequality.

STEP 9

The interval that satisfies our inequality is $-4 \leq x \leq 2$ .
Since we're looking for values less than or equal to zero, we include the endpoints!

STEP 10

On the number line, we'll have closed circles at $x = -4$ and $x = 2$ , and we'll shade the region between them.

STEP 11

In interval notation, our solution is $[\mathbf{-4}, \mathbf{2}]$ .
Those square brackets mean we're including the endpoints!

STEP 12

The solution set is $[-4, 2]$ .
This is represented on the number line with closed circles at $-4$ and $2$ , and the region between them shaded.

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QuestionSolve the polynomial inequality and graph the solution set on a real number line. Express the solution set in interval notation. $(x-2)(x+4) \leq 0$

Studdy Solution

STEP 1

What is this asking? We need to find the values of $x$ that make the expression $(x-2)(x+4)$ less than or equal to zero, and then show those values on a number line and write them in interval notation! Watch out! Don't forget to include the endpoints where the expression equals zero!

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 is where our expression could potentially change its sign!

STEP 4

So, $x-2 = 0$ gives us $x = \mathbf{2}$ , and $x+4 = 0$ gives us $x = \mathbf{-4}$ .
These are our critical points!

STEP 5

Now, let's test the intervals created by these critical points.
We have three intervals to check: $x < -4$ , $-4 \leq x \leq 2$ , and $x > 2$ .

STEP 6

For $x < -4$ , let's pick $x = \mathbf{-5}$ .
Plugging this into our expression gives us $(-5-2)(-5+4) = (-7)(-1) = \mathbf{7}$ , which is greater than zero.
So, this interval doesn't satisfy our inequality.

STEP 7

For $-4 \leq x \leq 2$ , let's pick $x = \mathbf{0}$ .
Plugging this in gives us $(0-2)(0+4) = (-2)(4) = \mathbf{-8}$ , which is less than zero.
This interval does satisfy our inequality!

STEP 8

For $x > 2$ , let's pick $x = \mathbf{3}$ .
Plugging this in gives us $(3-2)(3+4) = (1)(7) = \mathbf{7}$ , which is greater than zero.
So, this interval doesn't satisfy our inequality.

STEP 9

The interval that satisfies our inequality is $-4 \leq x \leq 2$ .
Since we're looking for values less than or equal to zero, we include the endpoints!

STEP 10

On the number line, we'll have closed circles at $x = -4$ and $x = 2$ , and we'll shade the region between them.

STEP 11

In interval notation, our solution is $[\mathbf{-4}, \mathbf{2}]$ .
Those square brackets mean we're including the endpoints!

STEP 12

The solution set is $[-4, 2]$ .
This is represented on the number line with closed circles at $-4$ and $2$ , and the region between them shaded.

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QuestionSolve the polynomial inequality and graph the solution set on a real number line. Express the solution set in interval notation. (x−2)(x+4)≤0(x-2)(x+4) \leq 0(x−2)(x+4)≤0

Studdy Solution

STEP 1

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 is where our expression could potentially change its sign!

STEP 4

So, x−2=0x-2 = 0x−2=0 gives us x=2x = \mathbf{2}x=2, and x+4=0x+4 = 0x+4=0 gives us x=−4x = \mathbf{-4}x=−4.These are our **critical points**!

STEP 5

Now, let's **test the intervals** created by these critical points.We have three intervals to check: x<−4x < -4x<−4, −4≤x≤2-4 \leq x \leq 2−4≤x≤2, and x>2x > 2x>2.

STEP 6

For x<−4x < -4x<−4, let's pick x=−5x = \mathbf{-5}x=−5.Plugging this into our expression gives us (−5−2)(−5+4)=(−7)(−1)=7(-5-2)(-5+4) = (-7)(-1) = \mathbf{7}(−5−2)(−5+4)=(−7)(−1)=7, which is *greater than zero*.So, this interval *doesn't* satisfy our inequality.

STEP 7

For −4≤x≤2-4 \leq x \leq 2−4≤x≤2, let's pick x=0x = \mathbf{0}x=0.Plugging this in gives us (0−2)(0+4)=(−2)(4)=−8(0-2)(0+4) = (-2)(4) = \mathbf{-8}(0−2)(0+4)=(−2)(4)=−8, which is *less than zero*.This interval *does* satisfy our inequality!

STEP 8

For x>2x > 2x>2, let's pick x=3x = \mathbf{3}x=3.Plugging this in gives us (3−2)(3+4)=(1)(7)=7(3-2)(3+4) = (1)(7) = \mathbf{7}(3−2)(3+4)=(1)(7)=7, which is *greater than zero*.So, this interval *doesn't* satisfy our inequality.

STEP 9

The interval that satisfies our inequality is −4≤x≤2-4 \leq x \leq 2−4≤x≤2.Since we're looking for values *less than or equal to* zero, we **include the endpoints**!

STEP 10

On the number line, we'll have **closed circles** at x=−4x = -4x=−4 and x=2x = 2x=2, and we'll shade the region between them.

STEP 11

In interval notation, our solution is [−4,2][\mathbf{-4}, \mathbf{2}][−4,2].Those square brackets mean we're including the endpoints!

STEP 12

The solution set is [−4,2][-4, 2][−4,2].This is represented on the number line with closed circles at −4-4−4 and 222, and the region between them shaded.

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QuestionSolve the polynomial inequality and graph the solution set on a real number line. Express the solution set in interval notation. $(x-2)(x+4) \leq 0$

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

Let's find the critical points by setting each factor equal to zero.
This is where our expression could potentially change its sign!

So, $x-2 = 0$ gives us $x = \mathbf{2}$ , and $x+4 = 0$ gives us $x = \mathbf{-4}$ .
These are our critical points!

Now, let's test the intervals created by these critical points.
We have three intervals to check: $x < -4$ , $-4 \leq x \leq 2$ , and $x > 2$ .

For $x < -4$ , let's pick $x = \mathbf{-5}$ .
Plugging this into our expression gives us $(-5-2)(-5+4) = (-7)(-1) = \mathbf{7}$ , which is greater than zero.
So, this interval doesn't satisfy our inequality.

For $-4 \leq x \leq 2$ , let's pick $x = \mathbf{0}$ .
Plugging this in gives us $(0-2)(0+4) = (-2)(4) = \mathbf{-8}$ , which is less than zero.
This interval does satisfy our inequality!

For $x > 2$ , let's pick $x = \mathbf{3}$ .
Plugging this in gives us $(3-2)(3+4) = (1)(7) = \mathbf{7}$ , which is greater than zero.
So, this interval doesn't satisfy our inequality.

The interval that satisfies our inequality is $-4 \leq x \leq 2$ .
Since we're looking for values less than or equal to zero, we include the endpoints!

On the number line, we'll have closed circles at $x = -4$ and $x = 2$ , and we'll shade the region between them.

In interval notation, our solution is $[\mathbf{-4}, \mathbf{2}]$ .
Those square brackets mean we're including the endpoints!

The solution set is $[-4, 2]$ .
This is represented on the number line with closed circles at $-4$ and $2$ , and the region between them shaded.