To determine which polynomials are divisible by [tex]\( (x-1) \)[/tex], we need to find out if the remainder is zero when each polynomial is divided by [tex]\( (x-1) \)[/tex]. This can be checked using the Remainder Theorem, which states that for a polynomial [tex]\( P(x) \)[/tex], the remainder of the division of [tex]\( P(x) \)[/tex] by [tex]\( (x - a) \)[/tex] is [tex]\( P(a) \)[/tex].
In this case, we will check the value of each polynomial at [tex]\( x = 1 \)[/tex]. If [tex]\( P(1) = 0 \)[/tex], then the polynomial [tex]\( P(x) \)[/tex] is divisible by [tex]\( (x-1) \)[/tex].
Let's evaluate each polynomial at [tex]\( x = 1 \)[/tex]:
1. For [tex]\( A(x) = 3x^3 + 2x^2 - x \)[/tex]:
[tex]\[
A(1) = 3(1)^3 + 2(1)^2 - (1) = 3 + 2 - 1 = 4 \neq 0
\][/tex]
Hence, [tex]\( A(x) \)[/tex] is not divisible by [tex]\( (x-1) \)[/tex].
2. For [tex]\( B(x) = 5x^3 - 4x^2 - x \)[/tex]:
[tex]\[
B(1) = 5(1)^3 - 4(1)^2 - (1) = 5 - 4 - 1 = 0
\][/tex]
Hence, [tex]\( B(x) \)[/tex] is divisible by [tex]\( (x-1) \)[/tex].
3. For [tex]\( C(x) = 2x^3 - 3x^2 + 2x - 1 \)[/tex]:
[tex]\[
C(1) = 2(1)^3 - 3(1)^2 + 2(1) - 1 = 2 - 3 + 2 - 1 = 0
\][/tex]
Hence, [tex]\( C(x) \)[/tex] is divisible by [tex]\( (x-1) \)[/tex].
4. For [tex]\( D(x) = x^3 + 2x^2 + 3x + 2 \)[/tex]:
[tex]\[
D(1) = (1)^3 + 2(1)^2 + 3(1) + 2 = 1 + 2 + 3 + 2 = 8 \neq 0
\][/tex]
Hence, [tex]\( D(x) \)[/tex] is not divisible by [tex]\( (x-1) \)[/tex].
Therefore, the polynomials that are divisible by [tex]\( (x-1) \)[/tex] are:
[tex]\[
\boxed{\text{B and C}}
\][/tex]
