Answer:
Step-by-step explanation:
For us to be able to determine the polynomials that are divisible by (x-1), this means that x-1 must be a factor for the functon to be able to divide any of the polynimial.
Since x-1 is a factor, we can get the value of x
x-1 = 0
x =0+1
x = 1
Next is for to substitute x - 1 into the polynomial and see the ones that will give us zero
For A(x)=3x^3+2x^2-x
A(1) = 3(1)^3+2(1)^2-(1)
A(1) = 3+2-(1)
A(1) = 5-1
A(1) = 4
Since A(1) ≠ 0, then x-1 is not divisible by the polynomial function.
<u>For B(x)=5x^3-4x^2-x</u>
B(1)=5(1)^3-4(1)^2-1
B(1)=5-4-1
B(1)=1-1 = 0
Since B(1) = 0, hence x-1 is divisible by 5x^3-4x^2-x
For the polynomial C(x)= 2x^3-3x^2+2x-1
C(1)=2(1)^3-3(1)^2+2(1)-1
C(1)=2-3+2-1
C(1)= -1+1
C(1)= 0
Since C(1) = 0, hence x-1 is divisible by<u> the </u>
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<u>F</u>or the polynomial D(x)=x^3+2x^2+3x+2
D(1)=1^3+2(1)^2+3(1)+2
D(1)=1+2+3+2
D(x) = 8
Hence the polynomial D(x) is not divisible by x-1
Hence the correct options are B(x)=5x^3-4x^2-x and 2x^3-3x^2+2x-1
) of the square of the number j (j²)