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1. Which of these polynomials could have [tex](x-2)[/tex] as a factor?

A. [tex]A(x)=6x^2-7x-5[/tex]

B. [tex]B(x)=3x^2+15x-42[/tex]

C. [tex]C(x)=2x^3+13x^2+16x+5[/tex]

D. [tex]D(x)=3x^3-2x^2-15x+14[/tex]

E. [tex]E(x)=8x^4-41x^3-18x^2+101x+70[/tex]

F. [tex]F(x)=x^4+5x^3-27x^2-101x-70[/tex]

Answer :

To determine which of these polynomials could have [tex]\((x-2)\)[/tex] as a factor, we need to evaluate each polynomial at [tex]\(x=2\)[/tex]. When [tex]\((x-2)\)[/tex] is a factor of a polynomial, substituting [tex]\(x=2\)[/tex] into the polynomial will result in a value of 0.

Let's evaluate each polynomial:

1. Polynomial A: [tex]\(A(x) = 6x^2 - 7x - 5\)[/tex]
- Plug in [tex]\(x=2\)[/tex]:
[tex]\[
A(2) = 6(2)^2 - 7(2) - 5
= 24 - 14 - 5
= 5
\][/tex]
- The result is 5, so [tex]\((x-2)\)[/tex] is not a factor of [tex]\(A(x)\)[/tex].

2. Polynomial B: [tex]\(B(x) = 3x^2 + 15x - 42\)[/tex]
- Plug in [tex]\(x=2\)[/tex]:
[tex]\[
B(2) = 3(2)^2 + 15(2) - 42
= 12 + 30 - 42
= 0
\][/tex]
- The result is 0, so [tex]\((x-2)\)[/tex] is a factor of [tex]\(B(x)\)[/tex].

3. Polynomial C: [tex]\(C(x) = 2x^3 + 13x^2 + 16x + 5\)[/tex]
- Plug in [tex]\(x=2\)[/tex]:
[tex]\[
C(2) = 2(2)^3 + 13(2)^2 + 16(2) + 5
= 16 + 52 + 32 + 5
= 105
\][/tex]
- The result is 105, so [tex]\((x-2)\)[/tex] is not a factor of [tex]\(C(x)\)[/tex].

4. Polynomial D: [tex]\(D(x) = 3x^3 - 2x^2 - 15x + 14\)[/tex]
- Plug in [tex]\(x=2\)[/tex]:
[tex]\[
D(2) = 3(2)^3 - 2(2)^2 - 15(2) + 14
= 24 - 8 - 30 + 14
= 0
\][/tex]
- The result is 0, so [tex]\((x-2)\)[/tex] is a factor of [tex]\(D(x)\)[/tex].

5. Polynomial E: [tex]\(E(x) = 8x^4 - 41x^3 - 18x^2 + 101x + 70\)[/tex]
- Plug in [tex]\(x=2\)[/tex]:
[tex]\[
E(2) = 8(2)^4 - 41(2)^3 - 18(2)^2 + 101(2) + 70
= 128 - 328 - 72 + 202 + 70
= 0
\][/tex]
- The result is 0, so [tex]\((x-2)\)[/tex] is a factor of [tex]\(E(x)\)[/tex].

6. Polynomial F: [tex]\(F(x) = x^4 + 5x^3 - 27x^2 - 101x - 70\)[/tex]
- Plug in [tex]\(x=2\)[/tex]:
[tex]\[
F(2) = (2)^4 + 5(2)^3 - 27(2)^2 - 101(2) - 70
= 16 + 40 - 108 - 202 - 70
= -324
\][/tex]
- The result is -324, so [tex]\((x-2)\)[/tex] is not a factor of [tex]\(F(x)\)[/tex].

Based on these evaluations, the polynomials that could have [tex]\((x-2)\)[/tex] as a factor are:
- [tex]\(B(x) = 3x^2 + 15x - 42\)[/tex]
- [tex]\(D(x) = 3x^3 - 2x^2 - 15x + 14\)[/tex]
- [tex]\(E(x) = 8x^4 - 41x^3 - 18x^2 + 101x + 70\)[/tex]

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