Solve (3x+1)(1-3x): Finding Values Where Function is Negative

Q: Why can't I just expand the expression first?

While you could expand $ (3x+1)(1-3x) $ to get $ 1-9x^2 $, the factored form makes sign analysis much easier! You can see exactly where each factor changes sign.

Q: How do I remember which intervals are negative?

Use the "sign chart" method: Draw a number line with your roots $ -\frac{1}{3} $ and $ \frac{1}{3} $, then test one point in each of the three intervals to determine the sign pattern.

Q: What if I get confused about which factor is positive or negative?

For $ 3x+1 $: it's negative when x and positive when x > -1/3. For $ 1-3x $: it's positive when x and negative when x > 1/3. Practice with simple test values!

Q: Why don't we include the boundary points in our answer?

The question asks for $ f(x) (strictly less than zero). At the boundary points \( x = -\frac{1}{3} $ and $ x = \frac{1}{3} $, the function equals zero, not less than zero.

Q: How can I double-check my intervals are correct?

Pick one test point from your final answer regions and substitute it back into the original expression. For example, try $ x = 1 $: \( (3(1)+1)(1-3(1)) = (4)(-2) = -8 ✓

Quadratic Inequalities with Sign Analysis

Look at the following function:

$y=\left(3x+1\right)\left(1-3x\right)$

Determine for which values of $x$ the following is true:

$f\left(x\right) < 0$

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Step-by-step written solution

Follow each step carefully to understand the complete solution

Understand the problem

Look at the following function:

$y=\left(3x+1\right)\left(1-3x\right)$

Determine for which values of $x$ the following is true:

$f\left(x\right) < 0$

Step-by-step solution

To determine for which values of $x$ the expression $y = (3x+1)(1-3x)$ is less than zero, we follow these steps:

Find the roots of each factor:

The factor $3x+1 = 0$ gives the root $x = -\frac{1}{3}$ .
The factor $1-3x = 0$ gives the root $x = \frac{1}{3}$ .

Determine the intervals created by these roots:

Interval 1: $x < -\frac{1}{3}$
Interval 2: $-\frac{1}{3} < x < \frac{1}{3}$
Interval 3: $x > \frac{1}{3}$

Test the sign of the product within each interval:

For $x < -\frac{1}{3}$ , choose $x = -1$ :

$3x+1$ is negative, and $1-3x$ is positive. Product = negative.

For $-\frac{1}{3} < x < \frac{1}{3}$ , choose $x = 0$ :

$3x+1$ is positive, and $1-3x$ is positive. Product = positive.

For $x > \frac{1}{3}$ , choose $x = 1$ :

$3x+1$ is positive, and $1-3x$ is negative. Product = negative.

Conclusion: The product $(3x+1)(1-3x)$ is less than zero for:

$x > \frac{1}{3}$ or $x < -\frac{1}{3}$

Final Answer

$x > \frac{1}{3}$ or $x < -\frac{1}{3}$

Key Points to Remember

Essential concepts to master this topic

Zero Finding: Set each factor equal to zero to find boundary points
Interval Testing: Choose test points like x = -1, 0, 1 to determine signs
Solution Check: Verify negative regions match your interval notation answer ✓

Common Mistakes

Avoid these frequent errors

Testing signs incorrectly at boundary points
Don't substitute the actual roots like x = -1/3 into the factors = always gives zero! This tells you nothing about the sign in each interval. Always choose test points that are clearly inside each interval, not on the boundaries.

Practice Quiz

Test your knowledge with interactive questions

The graph of the function below intersects the X-axis at points A and B.

The vertex of the parabola is marked at point C.

Find all values of $$ x $$ where $f\left(x\right) > 0$ .

FAQ

Everything you need to know about this question

Why can't I just expand the expression first?

While you could expand $(3x+1)(1-3x)$ to get $1-9x^2$ , the factored form makes sign analysis much easier! You can see exactly where each factor changes sign.

How do I remember which intervals are negative?

Use the "sign chart" method: Draw a number line with your roots $-\frac{1}{3}$ and $\frac{1}{3}$ , then test one point in each of the three intervals to determine the sign pattern.

What if I get confused about which factor is positive or negative?

For $3x+1$ : it's negative when x < -1/3 and positive when x > -1/3. For $1-3x$ : it's positive when x < 1/3 and negative when x > 1/3. Practice with simple test values!

Why don't we include the boundary points in our answer?

The question asks for $f(x) < 0$ (strictly less than zero). At the boundary points $x = -\frac{1}{3}$ and $x = \frac{1}{3}$ , the function equals zero, not less than zero.

How can I double-check my intervals are correct?

Pick one test point from your final answer regions and substitute it back into the original expression. For example, try $x = 1$ : $(3(1)+1)(1-3(1)) = (4)(-2) = -8 < 0$ ✓

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Solve (3x+1)(1-3x): Finding Values Where Function is Negative

Quadratic Inequalities with Sign Analysis

❤️ Continue Your Math Journey!

Step-by-step written solution

Understand the problem

Step-by-step solution

Final Answer

Key Points to Remember

Common Mistakes

Practice Quiz

FAQ

Why can't I just expand the expression first?

How do I remember which intervals are negative?

What if I get confused about which factor is positive or negative?

Why don't we include the boundary points in our answer?

How can I double-check my intervals are correct?

🌟 Unlock Your Math Potential

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