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Answer :
To determine which of these polynomials have [tex]\((x-2)\)[/tex] as a factor, we can use the fact that if [tex]\((x-2)\)[/tex] is a factor of a polynomial, then substituting [tex]\(x = 2\)[/tex] into the polynomial should result in a value of zero. Let's go through each option to check this:
1. Polynomial [tex]\( A(x) = 6x^2 - 7x - 5 \)[/tex]
Substitute [tex]\(x = 2\)[/tex]:
[tex]\( A(2) = 6(2)^2 - 7(2) - 5 = 24 - 14 - 5 = 5 \)[/tex]
Since [tex]\(A(2) \neq 0\)[/tex], [tex]\((x-2)\)[/tex] is not a factor of [tex]\(A(x)\)[/tex].
2. Polynomial [tex]\( B(x) = 3x^2 + 15x - 42 \)[/tex]
Substitute [tex]\(x = 2\)[/tex]:
[tex]\( B(2) = 3(2)^2 + 15(2) - 42 = 12 + 30 - 42 = 0 \)[/tex]
Since [tex]\(B(2) = 0\)[/tex], [tex]\((x-2)\)[/tex] is a factor of [tex]\(B(x)\)[/tex].
3. Polynomial [tex]\( C(x) = 2x^3 + 13x^2 + 16x + 5 \)[/tex]
Substitute [tex]\(x = 2\)[/tex]:
[tex]\( C(2) = 2(2)^3 + 13(2)^2 + 16(2) + 5 = 16 + 52 + 32 + 5 = 105 \)[/tex]
Since [tex]\(C(2) \neq 0\)[/tex], [tex]\((x-2)\)[/tex] is not a factor of [tex]\(C(x)\)[/tex].
4. Polynomial [tex]\( D(x) = 3x^3 - 2x^2 - 15x + 14 \)[/tex]
Substitute [tex]\(x = 2\)[/tex]:
[tex]\( D(2) = 3(2)^3 - 2(2)^2 - 15(2) + 14 = 24 - 8 - 30 + 14 = 0 \)[/tex]
Since [tex]\(D(2) = 0\)[/tex], [tex]\((x-2)\)[/tex] is a factor of [tex]\(D(x)\)[/tex].
5. Polynomial [tex]\( E(x) = 8x^4 - 41x^3 - 18x^2 + 101x + 70 \)[/tex]
Substitute [tex]\(x = 2\)[/tex]:
Calculate [tex]\(E(2)\)[/tex] and it equals 0, confirming [tex]\((x-2)\)[/tex] is a factor of [tex]\(E(x)\)[/tex].
6. Polynomial [tex]\( F(x) = x^4 + 5x^3 - 27x^2 - 101x - 70 \)[/tex]
Substitute [tex]\(x = 2\)[/tex]:
Calculate [tex]\(F(2)\)[/tex] and it does not equal 0, confirming [tex]\((x-2)\)[/tex] is not a factor of [tex]\(F(x)\)[/tex].
Therefore, the polynomials that could have [tex]\((x-2)\)[/tex] as a factor are [tex]\( B(x) \)[/tex], [tex]\( D(x) \)[/tex], and [tex]\( E(x) \)[/tex].
1. Polynomial [tex]\( A(x) = 6x^2 - 7x - 5 \)[/tex]
Substitute [tex]\(x = 2\)[/tex]:
[tex]\( A(2) = 6(2)^2 - 7(2) - 5 = 24 - 14 - 5 = 5 \)[/tex]
Since [tex]\(A(2) \neq 0\)[/tex], [tex]\((x-2)\)[/tex] is not a factor of [tex]\(A(x)\)[/tex].
2. Polynomial [tex]\( B(x) = 3x^2 + 15x - 42 \)[/tex]
Substitute [tex]\(x = 2\)[/tex]:
[tex]\( B(2) = 3(2)^2 + 15(2) - 42 = 12 + 30 - 42 = 0 \)[/tex]
Since [tex]\(B(2) = 0\)[/tex], [tex]\((x-2)\)[/tex] is a factor of [tex]\(B(x)\)[/tex].
3. Polynomial [tex]\( C(x) = 2x^3 + 13x^2 + 16x + 5 \)[/tex]
Substitute [tex]\(x = 2\)[/tex]:
[tex]\( C(2) = 2(2)^3 + 13(2)^2 + 16(2) + 5 = 16 + 52 + 32 + 5 = 105 \)[/tex]
Since [tex]\(C(2) \neq 0\)[/tex], [tex]\((x-2)\)[/tex] is not a factor of [tex]\(C(x)\)[/tex].
4. Polynomial [tex]\( D(x) = 3x^3 - 2x^2 - 15x + 14 \)[/tex]
Substitute [tex]\(x = 2\)[/tex]:
[tex]\( D(2) = 3(2)^3 - 2(2)^2 - 15(2) + 14 = 24 - 8 - 30 + 14 = 0 \)[/tex]
Since [tex]\(D(2) = 0\)[/tex], [tex]\((x-2)\)[/tex] is a factor of [tex]\(D(x)\)[/tex].
5. Polynomial [tex]\( E(x) = 8x^4 - 41x^3 - 18x^2 + 101x + 70 \)[/tex]
Substitute [tex]\(x = 2\)[/tex]:
Calculate [tex]\(E(2)\)[/tex] and it equals 0, confirming [tex]\((x-2)\)[/tex] is a factor of [tex]\(E(x)\)[/tex].
6. Polynomial [tex]\( F(x) = x^4 + 5x^3 - 27x^2 - 101x - 70 \)[/tex]
Substitute [tex]\(x = 2\)[/tex]:
Calculate [tex]\(F(2)\)[/tex] and it does not equal 0, confirming [tex]\((x-2)\)[/tex] is not a factor of [tex]\(F(x)\)[/tex].
Therefore, the polynomials that could 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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