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

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}