Solve the equation in the real number system: 9x^3+9x^2-4x-4=0... (MULTIPLE CHOICE)

Question

Solve the equation in the real number system: 9x³+9x²-4x-4=0

 

a. -3, -1, 3

b. -5, -1, 5

c. -1, -2/3, 2/3

d. -2/3, 2/3, 1

 

Choose one thanks

Answer 1

Note the pattern in the coefficients here:  2 9s, followed by 2 -4s.

This leads one to realize that we can factor the first 2 terms 9x³ + 9x² as 9x²(x+1) and

the second 2 terms -4x - 4 as -4(x+1)

Thus,  9x³ + 9x² - 4x - 4 = 9x²(x+1)  -4(x+1) = (9x² - 4) (x+1)

As 9x² - 4 = (3x+2)(3x-2),

(9x² - 4) (x+1) = (3x+2)(3x-2)(x+1)

Thus, the solutions are -2/3, 2/3, -1

The solution is c.

Answer 2

9x³ + 9x² - 4x - 4 = 0

 

Factoring by grouping, we have:

 

9x²(x + 1) - 4(x + 1) = 0

 

(x + 1)(9x² - 4) = 0

 

(x + 1)(3x - 2)(3x + 2) = 0

 

Setting each factor equal to zero, we get x = -1, 2/3, -2/3.