Question #6aada

1 Answer
Dec 8, 2016

First, factor the polynomial: #(x-3)(x+1)# and solve each factor for 0.


On a number line, the numbers -1 and 3 would be the "zeros" of the function. Those specific values would give you a value EQUAL to zero, and that is not what you want.

So, test points (numbers) that are outside of those two numbers, and in between those numbers to look for POSITIVE values. (truth)

Test -2: #(-2-3)(-2+1) = (-5)(-1) = 5# TRUE!

Test 0: #(0-3)(0+1) = (-3)(1) = -3# FALSE!

Test 4: #(4-3)(4+1) = (1)(5) = 5# TRUE!

So, it seems as if values less than -1 are true, and values greater than 3 are true. Interval notation would look like this:
#( - oo ,-1)uu(3, oo )#

Notice that the parentheses are used to NOT include the values at the ends of the intervals!