Factorizing the Quadratic Expression

Oct 13, 2025 by magicalceylontours.com 36 views

Introduction

In this article, we will solve the given inequality $\left(x-\frac{1}{2}\right)\left(2x^2-2x-12\right) \geq 0$ and find the intervals where this inequality holds true. We will use algebraic methods to factorize the quadratic expression and then analyze the sign of the expression in different intervals.

Factorizing the Quadratic Expression

To factorize the quadratic expression $2x^2-2x-12$ , we need to find two numbers whose product is $-24$ and whose sum is $-2$ . These numbers are $-6$ and $4$ . Therefore, we can write the quadratic expression as:

2x^2-2x-12 = 2(x^2-x-6) = 2(x-3)(x+2)

Analyzing the Sign of the Expression

Now, we need to analyze the sign of the expression $\left(x-\frac{1}{2}\right)\left(2(x-3)(x+2)\right)$ in different intervals. To do this, we need to find the critical points of the expression, which are the values of $x$ that make the expression equal to zero.

The critical points of the expression are:

$x = \frac{1}{2}$ , which makes the first factor equal to zero
$x = 3$ , which makes the second factor equal to zero
$x = -2$ , which makes the third factor equal to zero

Finding the Intervals

Now, we need to find the intervals where the expression is non-negative. To do this, we can use a sign chart to analyze the sign of the expression in different intervals.

Interval	Sign of $x-\frac{1}{2}$	Sign of $2(x-3)(x+2)$	Sign of the expression
$(-\infty, -2)$	-	-	+
$(-2, \frac{1}{2})$	-	+	-
$(\frac{1}{2}, 3)$	+	+	+
$(3, \infty)$	+	-	-

Conclusion

From the sign chart, we can see that the expression is non-negative in the intervals $(-\infty, -2)$ and $(\frac{1}{2}, 3)$ . Therefore, the solution to the inequality is:

x \leq -2 \text{ or } \frac{1}{2} \leq x \leq 3

Discussion

In this article, we solved the inequality $\left(x-\frac{1}{2}\right)\left(2x^2-2x-12\right) \geq 0$ and found the intervals where this inequality holds true. We used algebraic methods to factorize the quadratic expression and then analyzed the sign of the expression in different intervals. The solution to the inequality is $x \leq -2 \text{ or } \frac{1}{2} \leq x \leq 3$ .

Final Answer

The final answer is $\boxed{x \leq -2 \text{ or } \frac{1}{2} \leq x \leq 3}$ .

Introduction

In our previous article, we solved the inequality $\left(x-\frac{1}{2}\right)\left(2x^2-2x-12\right) \geq 0$ and found the intervals where this inequality holds true. In this article, we will answer some frequently asked questions related to solving this inequality.

Q: What is the first step in solving the inequality?

A: The first step in solving the inequality is to factorize the quadratic expression $2x^2-2x-12$ . We need to find two numbers whose product is $-24$ and whose sum is $-2$ . These numbers are $-6$ and $4$ . Therefore, we can write the quadratic expression as:

2x^2-2x-12 = 2(x^2-x-6) = 2(x-3)(x+2)

Q: How do we find the critical points of the expression?

A: To find the critical points of the expression, we need to find the values of $x$ that make the expression equal to zero. In this case, the critical points are:

$x = \frac{1}{2}$ , which makes the first factor equal to zero
$x = 3$ , which makes the second factor equal to zero
$x = -2$ , which makes the third factor equal to zero

Q: How do we use a sign chart to analyze the sign of the expression in different intervals?

A: To use a sign chart, we need to analyze the sign of each factor in different intervals. We can do this by choosing a test point in each interval and evaluating the sign of each factor at that point. For example, in the interval $(-\infty, -2)$ , we can choose the test point $x = -3$ . Evaluating the sign of each factor at this point, we get:

$x-\frac{1}{2} = -3 - \frac{1}{2} = -\frac{7}{2} < 0$
$2(x-3)(x+2) = 2(-3-3)(-3+2) = 2(-6)(-1) = 12 > 0$
The expression is positive in this interval.

Q: What is the solution to the inequality?

A: From the sign chart, we can see that the expression is non-negative in the intervals $(-\infty, -2)$ and $(\frac{1}{2}, 3)$ . Therefore, the solution to the inequality is:

x \leq -2 \text{ or } \frac{1}{2} \leq x \leq 3

Q: How do we check our solution?

A: To check our solution, we can plug in a test point in each interval and verify that the expression is non-negative. For example, we can choose the test point $x = -3$ in the interval $(-\infty, -2)$ and verify that the expression is non-negative:

\left(-3-\frac{1}{2}\right)\left(2(-3-3)(-3+2)\right) = \left(-\frac{7}{2}\right)(12) = -42 \geq 0

Q: What are some common mistakes to avoid when solving inequalities?

A: Some common mistakes to avoid when solving inequalities include:

Not factoring the quadratic expression correctly
Not finding all the critical points of the expression
Not using a sign chart to analyze the sign of the expression in different intervals
Not checking the solution by plugging in test points

Conclusion

In this article, we answered some frequently asked questions related to solving the inequality $\left(x-\frac{1}{2}\right)\left(2x^2-2x-12\right) \geq 0$ . We hope that this article has been helpful in clarifying any doubts you may have had about solving this inequality.