2a)
To verify, set the factors to zero.
(x - 2) = 0, x = 2
(x + 3) = 0, x = -3
Insert zero's in the equation:
when x = 2
3(2)³ + 2(2)² - 19(2) + 6 = 0
when x = -3
3(-3)³ + 2(-3)² - 19(-3) + 6 = 0
Hence verified the factors.
2b)
[tex]= \sf \dfrac{\left(3x^3+2x^2-19x+6\right)}{\left(x-2\right)\left(x+3\right)}[/tex]
[tex]\sf = \dfrac{\left(3x-1\right)\left(x-2\right)\left(x+3\right)}{\left(x-2\right)\left(x+3\right)}[/tex]
[tex]\sf = 3x -1[/tex]
The remaining factor is (3x - 1)
2c)
Set the factors to zero to find real zeros of f
2d)
Complete factorization: (3x - 1)(x - 2)(x + 3)
