Answer :
To determine which of the given polynomials have [tex]\((x-2)\)[/tex] as a factor, we can apply the Remainder Theorem. The Remainder Theorem 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]. This means that if [tex]\(P(a) = 0\)[/tex], then [tex]\((x-a)\)[/tex] is a factor of [tex]\(P(x)\)[/tex].
In this case, we want to check if [tex]\(x-2\)[/tex] is a factor, which requires us to evaluate the polynomial at [tex]\(x = 2\)[/tex]. If the result is zero, then [tex]\((x-2)\)[/tex] is a factor. Let's evaluate each polynomial:
1. For [tex]\(A(x) = 6x^2 - 7x - 5\)[/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. For [tex]\(B(x) = 3x^2 + 15x - 42\)[/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. For [tex]\(C(x) = 2x^3 + 13x^2 + 16x + 5\)[/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. For [tex]\(D(x) = 3x^3 - 2x^2 - 15x + 14\)[/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. For [tex]\(E(x) = 8x^4 - 41x^3 - 18x^2 + 101x + 70\)[/tex]:
[tex]\[
E(2) = 8(2)^4 - 41(2)^3 - 18(2)^2 + 101(2) + 70 = 128 - 328 - 72 + 202 + 70 = 0
\][/tex]
Since [tex]\(E(2) = 0\)[/tex], [tex]\((x-2)\)[/tex] is a factor of [tex]\(E(x)\)[/tex].
6. For [tex]\(F(x) = x^4 + 5x^3 - 27x^2 - 101x - 70\)[/tex]:
[tex]\[
F(2) = (2)^4 + 5(2)^3 - 27(2)^2 - 101(2) - 70 = 16 + 40 - 108 - 202 - 70 = -324
\][/tex]
Since [tex]\(F(2) \neq 0\)[/tex], [tex]\((x-2)\)[/tex] is not a factor of [tex]\(F(x)\)[/tex].
Thus, 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].
In this case, we want to check if [tex]\(x-2\)[/tex] is a factor, which requires us to evaluate the polynomial at [tex]\(x = 2\)[/tex]. If the result is zero, then [tex]\((x-2)\)[/tex] is a factor. Let's evaluate each polynomial:
1. For [tex]\(A(x) = 6x^2 - 7x - 5\)[/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. For [tex]\(B(x) = 3x^2 + 15x - 42\)[/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. For [tex]\(C(x) = 2x^3 + 13x^2 + 16x + 5\)[/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. For [tex]\(D(x) = 3x^3 - 2x^2 - 15x + 14\)[/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. For [tex]\(E(x) = 8x^4 - 41x^3 - 18x^2 + 101x + 70\)[/tex]:
[tex]\[
E(2) = 8(2)^4 - 41(2)^3 - 18(2)^2 + 101(2) + 70 = 128 - 328 - 72 + 202 + 70 = 0
\][/tex]
Since [tex]\(E(2) = 0\)[/tex], [tex]\((x-2)\)[/tex] is a factor of [tex]\(E(x)\)[/tex].
6. For [tex]\(F(x) = x^4 + 5x^3 - 27x^2 - 101x - 70\)[/tex]:
[tex]\[
F(2) = (2)^4 + 5(2)^3 - 27(2)^2 - 101(2) - 70 = 16 + 40 - 108 - 202 - 70 = -324
\][/tex]
Since [tex]\(F(2) \neq 0\)[/tex], [tex]\((x-2)\)[/tex] is not a factor of [tex]\(F(x)\)[/tex].
Thus, 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].
