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Daniel [21]
3 years ago
5

Select all polynomials that are divisible by (x-1)(x−1)left parenthesis, x, minus, 1, right parenthesis. Choose all answers that

apply: Choose all answers that apply: (Choice A) A A(x)=3x^3+2x^2-xA(x)=3x 3 +2x 2 −xA, left parenthesis, x, right parenthesis, equals, 3, x, cubed, plus, 2, x, squared, minus, x (Choice B) B B(x)=5x^3-4x^2-xB(x)=5x 3 −4x 2 −xB, left parenthesis, x, right parenthesis, equals, 5, x, cubed, minus, 4, x, squared, minus, x (Choice C) C C(x)=2x^3-3x^2+2x-1C(x)=2x 3 −3x 2 +2x−1C, left parenthesis, x, right parenthesis, equals, 2, x, cubed, minus, 3, x, squared, plus, 2, x, minus, 1 (Choice D) D D(x)=x^3+2x^2+3x+2D(x)=x 3 +2x 2 +3x+2
Mathematics
1 answer:
alexdok [17]3 years ago
8 0

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>

<u />

<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

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11b-7-3n+8-5b+9n what is the answer
Sergio [31]

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6 n + 6 b + 1

Step-by-step explanation:

Simplify the following:

11 b - 7 - 3 n + 8 - 5 b + 9 n

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