STEPS: x4 + 3x3 – 12x2 + 12x < 0 x(x – 2)2 (x + 3) < 0 Graph the polynomial function and then find the intervals that apply to the inequality. STEPS: Find and graph zeros. Find intervals. Cross vs. Bump Find direction of endpoints: + up, – down (on right) and turning points. x4 + 3x3 – 12x2 + 12x < 0 In factored form. x(x – 2)2 (x + 3) < 0

STEP 1: Find and graph zeros. Find intervals. x(x – 2)2(x + 3) < 0 x = 0, 2, –3 Intervals: (-, -3) (-3, 0) (0, 2) (2, )

STEP 2: Cross vs. Bump x(x – 2)2(x + 3) < 0 Even exponents are always positive so they indicate a bump of the x-axis. Odd exponents can be negative so they cross the x-axis. BUMP

STEP 3: Find endpoint directions: + up, – down (on right) and turning points to sketch the graph. x(x – 2)2(x + 3) < 0 If not given the Leading Coefficient would be: x·x2·x = x4 POSITIVE leading coefficient means up on right and an EVEN exponent means same direction on left. x4 indicates 3 turning points. 3 turning points

There’s no need for exact points for peaks/valleys because we’re looking for specific intervals below the x-axis (< 0). x(x – 2)2(x + 3) < 0 (1, 4) Zeros: 0, 2, –3 Intervals: (-, -3) (-3, 0) (0, 2) (2, ) (-3, 0) NOTE: If it were < the interval would be [-3, 0] and {2}.

Since we’re just looking for intervals, polynomial inequalities can be graphed on a number line. x(x – 2)2(x + 3) < 0 Zeros: 0, 2, –3 Intervals: (-, -3) (-3, 0) (0, 2) (2, ) (-3, 0) or -3 < x < 0