x+2 > 10 solves to x > 8 after we subtract 2 from both sides
So set A is the set of real numbers that are larger than 8. The value 8 itself is not in set A. The same can be said about 5 as well.
Set B is the set of values that are larger than 5 since 2x > 10 turns into x > 5 after dividing both sides by 2. The value x = 5 is not in set B since x > 5 would turn into 5 > 5 which is false. The values x = 6, x = 8, and x = 9 are in set B.
----------------
Summarizing everything, we can say...
5 is not in set A. True
5 is in set B. False
6 is in set A. False
6 is not in set B. False
8 is not in set A. True
8 is in set B. True
9 is in set A. True
9 is not in set B. False
If they intersect eachother then the answer is true
Recall that zeroes can be transformed into factors by subtracting them from x. This gives us the following factors:
(x - 1)(x + 3)(x - 4)
Now, if you multiply the first two factors together, you get the following:
(x² + 2x - 3)
Multiply that by the last factor, (x - 4), and you get this:
(x³ + 2x² - 3x - 4x² - 8x + 12)
This can be simplified:
(x³ - 2x² - 11x + 12)
And there's your final answer. Hope this helped!
