Determine whether (3x-2) is a factor of the polynomial 3x^3-14x^2+26x-12. Explain. ?

x = a is a root of a polynomial P(x) if P(a) = 0

The factor theorem states that if x = a is a root, then P(x) has a factor of (x - a)

(3x - 2) corresponds to the root x = 2/3

So if 3x³ - 14x² + 26x - 12 = 0 at x = 2/3 then (3x - 2) must be a factor:

3(2/3)³ - 14(2/3)² + 26(2/3) - 12 = 3(8/27) - 14(4/9) + 52/3 - 12
= 8/9 - 56/9 + 156/9 - 108/9
= 0

So (3x - 2) is a factor.

Put f(x) = 3x^3 - 14x^2 + 26x -12. f(2/3) = 3*(2/3)^3 -14(2/3)^2 + 26(2/3) - 12
= 8/9 - 56/9 + 156/9 -108/9 =(8+156)/9 - (56+ 108)/9 = 0. Therefore 3x-2
is a factor of f(x).

3x - 2 = 3 ( x - 2/3 )
Divide by x - 2/3 by using synthetic division :-

2/3 | 3______- 14_______26______- 12
___|_________2_______- 8_______ 12
___| 3______- 12________18_______0

Answer is 3x² - 12 x + 18

Thus 3x - 2 is a factor

(ax-b) is a linear factor of P(x) if and only if P(b/a)=0

Assume function is:
(3x-2)(x+a)(x+b) = 0
(3x-2) = 0
x = 2/3

Now if (3x-2) is a factor f(2/3) = 0
3(2/3)^3-14(2/3)^2+26(2/3)-12 = 0

Yeep it's a factor.

3x^3 - 14x^2 + 26x -12
If you divide 3x^3 - 14x^2 + 26x - 12 by 3x - 2 you get x² - 4x + 6 with no remainder.
Answer:
Yes

You can assume it is a factor by using long division.