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Answer :
Sure, let's determine which of these polynomials have [tex]\((x-2)\)[/tex] as a factor. To do this, we'll apply the Factor Theorem, which states that a polynomial [tex]\(f(x)\)[/tex] has a factor of [tex]\((x-a)\)[/tex] if and only if [tex]\(f(a) = 0\)[/tex].
Here, we need to check if substituting [tex]\(x = 2\)[/tex] into each polynomial results in zero.
Let's check each polynomial:
(A) [tex]\(A(x) = 6x^2 - 7x - 5\)[/tex]
[tex]\[ A(2) = 6(2)^2 - 7(2) - 5 \][/tex]
[tex]\[ = 6(4) - 14 - 5 \][/tex]
[tex]\[ = 24 - 14 - 5 \][/tex]
[tex]\[ = 5 \][/tex]
Since [tex]\(A(2) \neq 0\)[/tex], [tex]\(A(x)\)[/tex] does not have [tex]\((x-2)\)[/tex] as a factor.
(B) [tex]\(B(x) = 3x^2 + 15x - 42\)[/tex]
[tex]\[ B(2) = 3(2)^2 + 15(2) - 42 \][/tex]
[tex]\[ = 3(4) + 30 - 42 \][/tex]
[tex]\[ = 12 + 30 - 42 \][/tex]
[tex]\[ = 0 \][/tex]
Since [tex]\(B(2) = 0\)[/tex], [tex]\(B(x)\)[/tex] has [tex]\((x-2)\)[/tex] as a factor.
(C) [tex]\(C(x) = 2x^3 + 13x^2 + 16x + 5\)[/tex]
[tex]\[ C(2) = 2(2)^3 + 13(2)^2 + 16(2) + 5 \][/tex]
[tex]\[ = 2(8) + 13(4) + 32 + 5 \][/tex]
[tex]\[ = 16 + 52 + 32 + 5 \][/tex]
[tex]\[ = 105 \][/tex]
Since [tex]\(C(2) \neq 0\)[/tex], [tex]\(C(x)\)[/tex] does not have [tex]\((x-2)\)[/tex] as a factor.
(D) [tex]\(D(x) = 3x^3 - 2x^2 - 15x + 14\)[/tex]
[tex]\[ D(2) = 3(2)^3 - 2(2)^2 - 15(2) + 14 \][/tex]
[tex]\[ = 3(8) - 2(4) - 30 + 14 \][/tex]
[tex]\[ = 24 - 8 - 30 + 14 \][/tex]
[tex]\[ = 0 \][/tex]
Since [tex]\(D(2) = 0\)[/tex], [tex]\(D(x)\)[/tex] has [tex]\((x-2)\)[/tex] as a factor.
(E) [tex]\(E(x) = 8x^4 - 41x^3 - 18x^2 + 101x + 70\)[/tex]
[tex]\[ E(2) = 8(2)^4 - 41(2)^3 - 18(2)^2 + 101(2) + 70 \][/tex]
[tex]\[ = 8(16) - 41(8) - 18(4) + 202 + 70 \][/tex]
[tex]\[ = 128 - 328 - 72 + 202 + 70 \][/tex]
[tex]\[ = 0 \][/tex]
Since [tex]\(E(2) = 0\)[/tex], [tex]\(E(x)\)[/tex] has [tex]\((x-2)\)[/tex] as a factor.
(F) [tex]\(F(x) = x^4 + 5x^3 - 27x^2 - 101x - 70\)[/tex]
[tex]\[ F(2) = (2)^4 + 5(2)^3 - 27(2)^2 - 101(2) - 70 \][/tex]
[tex]\[ = 16 + 40 - 108 - 202 - 70 \][/tex]
[tex]\[ = -16 \][/tex]
Since [tex]\(F(2) \neq 0\)[/tex], [tex]\(F(x)\)[/tex] does not have [tex]\((x-2)\)[/tex] as a factor.
The polynomials that have [tex]\((x-2)\)[/tex] as a factor are [tex]\(B(x)\)[/tex], [tex]\(D(x)\)[/tex], and [tex]\(E(x)\)[/tex].
Here, we need to check if substituting [tex]\(x = 2\)[/tex] into each polynomial results in zero.
Let's check each polynomial:
(A) [tex]\(A(x) = 6x^2 - 7x - 5\)[/tex]
[tex]\[ A(2) = 6(2)^2 - 7(2) - 5 \][/tex]
[tex]\[ = 6(4) - 14 - 5 \][/tex]
[tex]\[ = 24 - 14 - 5 \][/tex]
[tex]\[ = 5 \][/tex]
Since [tex]\(A(2) \neq 0\)[/tex], [tex]\(A(x)\)[/tex] does not have [tex]\((x-2)\)[/tex] as a factor.
(B) [tex]\(B(x) = 3x^2 + 15x - 42\)[/tex]
[tex]\[ B(2) = 3(2)^2 + 15(2) - 42 \][/tex]
[tex]\[ = 3(4) + 30 - 42 \][/tex]
[tex]\[ = 12 + 30 - 42 \][/tex]
[tex]\[ = 0 \][/tex]
Since [tex]\(B(2) = 0\)[/tex], [tex]\(B(x)\)[/tex] has [tex]\((x-2)\)[/tex] as a factor.
(C) [tex]\(C(x) = 2x^3 + 13x^2 + 16x + 5\)[/tex]
[tex]\[ C(2) = 2(2)^3 + 13(2)^2 + 16(2) + 5 \][/tex]
[tex]\[ = 2(8) + 13(4) + 32 + 5 \][/tex]
[tex]\[ = 16 + 52 + 32 + 5 \][/tex]
[tex]\[ = 105 \][/tex]
Since [tex]\(C(2) \neq 0\)[/tex], [tex]\(C(x)\)[/tex] does not have [tex]\((x-2)\)[/tex] as a factor.
(D) [tex]\(D(x) = 3x^3 - 2x^2 - 15x + 14\)[/tex]
[tex]\[ D(2) = 3(2)^3 - 2(2)^2 - 15(2) + 14 \][/tex]
[tex]\[ = 3(8) - 2(4) - 30 + 14 \][/tex]
[tex]\[ = 24 - 8 - 30 + 14 \][/tex]
[tex]\[ = 0 \][/tex]
Since [tex]\(D(2) = 0\)[/tex], [tex]\(D(x)\)[/tex] has [tex]\((x-2)\)[/tex] as a factor.
(E) [tex]\(E(x) = 8x^4 - 41x^3 - 18x^2 + 101x + 70\)[/tex]
[tex]\[ E(2) = 8(2)^4 - 41(2)^3 - 18(2)^2 + 101(2) + 70 \][/tex]
[tex]\[ = 8(16) - 41(8) - 18(4) + 202 + 70 \][/tex]
[tex]\[ = 128 - 328 - 72 + 202 + 70 \][/tex]
[tex]\[ = 0 \][/tex]
Since [tex]\(E(2) = 0\)[/tex], [tex]\(E(x)\)[/tex] has [tex]\((x-2)\)[/tex] as a factor.
(F) [tex]\(F(x) = x^4 + 5x^3 - 27x^2 - 101x - 70\)[/tex]
[tex]\[ F(2) = (2)^4 + 5(2)^3 - 27(2)^2 - 101(2) - 70 \][/tex]
[tex]\[ = 16 + 40 - 108 - 202 - 70 \][/tex]
[tex]\[ = -16 \][/tex]
Since [tex]\(F(2) \neq 0\)[/tex], [tex]\(F(x)\)[/tex] does not have [tex]\((x-2)\)[/tex] as a factor.
The polynomials that have [tex]\((x-2)\)[/tex] as a factor are [tex]\(B(x)\)[/tex], [tex]\(D(x)\)[/tex], and [tex]\(E(x)\)[/tex].
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