# 008A Sample Final A, Question 4

Question: Solve. Provide your solution in interval notation. ${\displaystyle (x-4)(2x+1)(x-1)<0}$

Foundations:
1) What are the zeros of the left hand side?
2) Can the function be both positive and negative between consecutive zeros?
1) The zeros are ${\displaystyle -{\frac {1}{2}}}$, 1, and 4.
2) No. If the function is positive between 1 and 4 it must be positive for any value of x between 1 and 4.

Solution:

Step 1:
The zeros of the left hand side are ${\displaystyle -{\frac {1}{2}}}$, 1, and 4
Step 2:
The zeros split the real number line into 4 intervals: ${\displaystyle (-\infty ,-{\frac {1}{2}}),(-{\frac {1}{2}},1),(1,4),}$ and ${\displaystyle (4,\infty )}$.
We now pick one number from each interval: -1, 0, 2, and 5. We will use these numbers to determine if the left hand side function is positive or negative in each interval.
x = -1: (-1 -4)(2(-1) + 1)(-1 - 1) = (-5)(-1)(-2) = -10 < 0
x = 0: (-4)(1)(-1) = 4 > 0
x = 2: (2-4)(2(2) + 1)(2 - 1) = (-2)(5)(1) = -10 < 0
x = 5: (5 - 4)(2(5) + 1)(5 - 1) = (1)(11)(4) = 44 > 0
 ${\displaystyle x:}$ ${\displaystyle x=-1}$ ${\displaystyle x=0}$ ${\displaystyle x=2}$ ${\displaystyle x=5}$ ${\displaystyle f(x):}$ ${\displaystyle (-)}$ ${\displaystyle (+)}$ ${\displaystyle (-)}$ ${\displaystyle (+)}$
Step 3:
We take the intervals for which our test point led the function being negative, (${\displaystyle -\infty ,-{\frac {1}{2}}}$), and (1, 4).
Since we we are solving a strict inequality we do not need to change the parenthesis to square brackets, and the final answer is ${\displaystyle (-\infty ,-{\frac {1}{2}})\cup (1,4)}$
${\displaystyle (-\infty ,-{\frac {1}{2}})\cup (1,4)}$