Solution for Problem 1.6.2

Solve for x:    x3 + 3x2  – 4x – 12 = 0 

Solve

Step

 x3 + 3x2  – 4x – 12 = 0 

Use the grouping method

x2 (x + 3)  – 4x – 12 = 0

 Look for a pattern.  Factor a x2 form the first 2 terms. 

x2 (x + 3)  – 4(x + 3) = 0

 I see a x + 3 pattern.  Factor out a -4 from the last 2 terms. 
(x + 3)(x2  – 4) = 0 Now factor out a (x + 3) from each part. 

(x + 3)(x – 2)(x + 2) = 0

 Finish factoring!  Difference of two squares. 

x + 3 = 0 or x – 2 or 
x + 2 = 0 

 Set each factor to zero

x = -3 , x = 2 or x = -2

 Done

Check 

If x  = -3, then
(-3)3 + 3(-3)2 – 4(-3) – 12 = 0
    -27 + 27 + 12 – 12 = 0
0 = 0

If x  = 2, then
(2)3 + 3(2)2 – 4(2) – 12 = 0
    8 + 12 – 8 – 12 = 0
0 = 0

If x  = -2, then
(-2)3 + 3(-2)2 – 4(-2) – 12 = 0
    -8 + 12 + 8 – 12 = 0
0 = 0