Solve the polynomial inequality x^2-3x-10> 0

As x² - 3x - 10 factors, we can use the zero product property and pull it apart into factors. It factors because 2 and 5 have a difference of 3 and product of 10

x² - 3x - 10 > 0

(x - 5)(x + 2) > 0

To find where the inequality is true, we use a sign test. When x-5 > 0, it means x >5, and 5 is one of our sign test values. Similarly, -2 is the other. We can then make three intervals - A (less than -2), B (between -2 and 5), and C (more than 5).

< -2 5 >

A B C

To do a sign test, we choose values to see what happens. First, test x = -5.

(-5 - 5)(-5 + 2) = (-10)(-3) = 30. Because 30 > 0, the inequality is true for x < -2. So interval A is true.

Next, test x = 0.

(0 - 5) (0 + 2) = (-5)(2) = -10. Because -10 < 0 is false, the inequality is false for x > -2 and x < 5. So interval B is false.

Last, test x = 10

(10 - 5) (10 + 2) = (5) (8) = 40. Because 40 > 0 is true, the inequality is true for x > 5. So interval C is true.

We put the true intervals together.

(-∞, -2) ∪ (5, ∞) is where this inequality is true,

{x : x < -2 or x > 5} is also how it's said in set notation.

the answer is b i just did that

step-by-step explanation: