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Quadratic and Other Polynomial Inequalities Solve. For \(g(x)=(x-2)(x-3)(x+1),\) find all \(x\) -values for which \(g(x)>0\).

Short Answer

Expert verified
The solution is \( -1 < x < 2 \) and \( x > 3 \).

Step by step solution

01

Identify the inequality

We need to solve the inequality \((x-2)(x-3)(x+1) > 0\). This means we are looking for values of \(x\) that make the product of these three factors positive.
02

Find the roots

Identify the roots of the polynomial by setting each factor equal to zero: \(x-2=0\), \(x-3=0\), \(x+1=0\). Thus, the roots are \(x=2\), \(x=3\), and \(x=-1\).
03

Determine the intervals

The roots divide the real number line into four intervals: \((-\infty, -1)\), \((-1, 2)\), \((2, 3)\), and \((3, \infty)\). We will test values from each interval to see where the product \((x-2)(x-3)(x+1)\) is positive.
04

Test intervals

Choose a test point in each interval and evaluate the sign of the polynomial at these points:- For \(x < -1\), choose \(x = -2\). The sign is negative since \((-2-2)(-2-3)(-2+1) = (-4)(-5)(-1) = -20\).- For \(-1 < x < 2\), choose \(x = 0\). The sign is positive since \((0-2)(0-3)(0+1) = (-2)(-3)(1) = 6\).- For \(2 < x < 3\), choose \(x = 2.5\). The sign is negative since \((2.5-2)(2.5-3)(2.5+1) = (0.5)(-0.5)(3.5) = -0.875\).- For \(x > 3\), choose \(x = 4\). The sign is positive since \((4-2)(4-3)(4+1) = (2)(1)(5) = 10\).
05

Compile the solution

The polynomial is positive in the intervals where the test points yielded a positive result, which are \((-1, 2)\) and \((3, \infty)\). Therefore, the solution to \( (x-2)(x-3)(x+1) > 0 \) is \( -1 < x < 2 \) and \( x > 3 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

polynomial roots
To solve polynomial inequalities, understanding the concept of polynomial roots is essential. Roots are the values of x for which the polynomial equals zero. For instance, given the polynomial function \( g(x) = (x-2)(x-3)(x+1) \), we find its roots by setting each factor to zero. This means solving \( x-2=0 \), \( x-3=0 \), and \( x+1=0 \), which gives us the roots \( x = 2 \), \( x = 3 \), and \( x = -1 \). These roots are the points where the polynomial changes its sign. Hence, we use these roots to organize the number line into different intervals for further testing. By finding and understanding the polynomial roots, you create a framework for solving and testing the inequality.
testing intervals
After identifying the polynomial roots, the next step involves testing the intervals created by these roots. These intervals help us determine where the polynomial is positive or negative. In our example, the roots \( x = 2 \), \( x = 3 \), and \( x = -1 \) split the number line into four intervals:
  • \( (-\infty, -1) \)
  • \( (-1, 2) \)
  • \( (2, 3) \)
  • \( (3, \infty) \)
For each interval, pick a test point. Substitute this point into the polynomial to check the sign of the product.
For instance:
  • Pick \( x = -2 \) in \( (-\infty, -1) \): \( ((-2)-2)((-2)-3)((-2)+1)=(-4)(-5)(-1)=-20 \)
  • Pick \( x = 0 \) in \( (-1, 2) \): \( (0-2)(0-3)(0+1) = (-2)(-3)(1)=6 \)
  • Pick \( x = 2.5 \) in \( (2, 3) \): \( (2.5-2)(2.5-3)(2.5+1) = (0.5)(-0.5)(3.5)=-0.875 \)
  • Pick \( x = 4 \) in \( (3, \infty) \): \( (4-2)(4-3)(4+1) = (2)(1)(5)=10 \)
By identifying whether the polynomial is positive or negative in these intervals, we can compile the final solution.
inequality solution
The final step in solving a polynomial inequality involves compiling the results from the interval tests to form the solution to the inequality. In this case, we want \( (x-2)(x-3)(x+1) > 0 \). From our interval tests, we found out that the polynomial is positive in the intervals \( (-1, 2) \) and \( (3, \infty) \).
This means the solution to the inequality \( (x-2)(x-3)(x+1) > 0 \) includes all \( x \)-values in these intervals. Therefore, the final solution is:
  • \( -1 < x < 2 \)
  • \( x > 3 \)
Checking and combining the intervals makes sure we cover all possible values that satisfy the inequality and present the most comprehensive solution to the problem.

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