Solve using test intervals |x 1|> or equal to 4

Solve x5 + 3x4 – 23x3 – 51x2 + 94x + 120 > 0. First, I factor to find the zeroes:x5 + 3x4 – 23x3 – 51x2 + 94x + 120= (x + 5)(x + 3)(x + 1)(x – 2)(x – 4) = 0...so x = –5, –3, –1, 2, and 4 are the zeroes of this polynomial. (Review how to solve polynomials, if you're not sure how to get this solution.)To solve by the Test-Point Method, I would pick a sample point in each interval, the intervals being (negative infinity, –5), (–5, –3), (–3, –1), (–1, 2), (2, 4), and (4, positive infinity). As you can see, if your polynomial or rational function has many factors, the Test-Point Method can become quite time-consuming.To solve by the Factor Method, I would solve each factor for its positivity: x + 5 > 0 for x > –5;x + 3 > 0 for x > –3; x + 1 > 0 for x > –1; x – 2 > 0 for x > 2; and x – 4 > 0 for x > 4. Then I draw the grid:...and fill it in:...and solve:Then the solution (remembering to include the endpoints, because this is an "or equal to" inequality) is the set of x-values in the intervals [–5, –3], [–1, 2], and [4, positive infinity]. 

As you can see, if your polynomial or rational function has many factors, the Factor Method can be much faster.