Question

using remainder theorem completely factorize
3x³+2x²-19x+6
very urgent

Answer 1

3x³+2x²-19x+6

when x=2

Since the remainder is 0, x = 2 is an actual zero. This means that x - 2 is one of the factors. The numbers on the bottom represent the coefficients of the other factor. Notice that the exponents are reduced by 1. 
3x³ + 2x² - 19x + 6 = (x - 2)(3x² + 8x - 3) 

You can factor 3x² + 8x - 3 further. 
3x³ + 2x² - 19x + 6 = (x - 2)(3x² + 8x - 3) = (x - 2)(3x - 1)(x + 3) 

Set each factor to 0 and solve. 
x - 2 = 0 
x = 2 

3x - 1 = 0 
3x = 1 
3x / 3 = 1 / 3 
x = 1 / 3 

x + 3 = 0 
x = -3 

ANSWER: {-3, 1 / 3, 2} 

Answer 2

Answer:

3x³+2x²-19x+6

Step-by-step explanation:

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