Answer :
To determine which polynomials have [tex]\( (x-2) \)[/tex] as a factor, we can use the Factor Theorem. The Factor Theorem states that a polynomial has [tex]\( (x-c) \)[/tex] as a factor if and only if substituting [tex]\( x=c \)[/tex] into the polynomial results in [tex]\( 0 \)[/tex].
Let's check each polynomial by substituting [tex]\( x=2 \)[/tex] into them and see if the result is zero.
1. Polynomial A(x) = [tex]\(6x^2 - 7x - 5\)[/tex]:
Substitute [tex]\( x = 2 \)[/tex]:
[tex]\[
6(2)^2 - 7(2) - 5 = 6(4) - 14 - 5 = 24 - 14 - 5 = 5
\][/tex]
Since 5 is not equal to 0, [tex]\( (x-2) \)[/tex] is not a factor of A(x).
2. Polynomial B(x) = [tex]\(3x^2 + 15x - 42\)[/tex]:
Substitute [tex]\( x = 2 \)[/tex]:
[tex]\[
3(2)^2 + 15(2) - 42 = 3(4) + 30 - 42 = 12 + 30 - 42 = 0
\][/tex]
Since the result is 0, [tex]\( (x-2) \)[/tex] is a factor of B(x).
3. Polynomial C(x) = [tex]\(2x^3 + 13x^2 + 16x + 5\)[/tex]:
Substitute [tex]\( x = 2 \)[/tex]:
[tex]\[
2(2)^3 + 13(2)^2 + 16(2) + 5 = 2(8) + 13(4) + 32 + 5 = 16 + 52 + 32 + 5 = 105
\][/tex]
Since 105 is not equal to 0, [tex]\( (x-2) \)[/tex] is not a factor of C(x).
4. Polynomial D(x) = [tex]\(3x^3 - 2x^2 - 15x + 14\)[/tex]:
Substitute [tex]\( x = 2 \)[/tex]:
[tex]\[
3(2)^3 - 2(2)^2 - 15(2) + 14 = 3(8) - 2(4) - 30 + 14 = 24 - 8 - 30 + 14 = 0
\][/tex]
Since the result is 0, [tex]\( (x-2) \)[/tex] is a factor of D(x).
5. Polynomial E(x) = [tex]\(8x^4 - 41x^3 - 18x^2 + 101x + 70\)[/tex]:
Substitute [tex]\( x = 2 \)[/tex]:
[tex]\[
8(2)^4 - 41(2)^3 - 18(2)^2 + 101(2) + 70 = 8(16) - 41(8) - 18(4) + 202 + 70 = 0
\][/tex]
Since the result is 0, [tex]\( (x-2) \)[/tex] is a factor of E(x).
6. Polynomial F(x) = [tex]\(x^4 + 5x^3 - 27x^2 - 101x - 70\)[/tex]:
Substitute [tex]\( x = 2 \)[/tex]:
[tex]\[
(2)^4 + 5(2)^3 - 27(2)^2 - 101(2) - 70 = 16 + 40 - 108 - 202 - 70 = -324
\][/tex]
Since -324 is not equal to 0, [tex]\( (x-2) \)[/tex] is not a factor of F(x).
In conclusion, the polynomials that have [tex]\( (x-2) \)[/tex] as a factor are:
- Polynomial B(x): [tex]\(3x^2 + 15x - 42\)[/tex]
- Polynomial D(x): [tex]\(3x^3 - 2x^2 - 15x + 14\)[/tex]
- Polynomial E(x): [tex]\(8x^4 - 41x^3 - 18x^2 + 101x + 70\)[/tex]
Let's check each polynomial by substituting [tex]\( x=2 \)[/tex] into them and see if the result is zero.
1. Polynomial A(x) = [tex]\(6x^2 - 7x - 5\)[/tex]:
Substitute [tex]\( x = 2 \)[/tex]:
[tex]\[
6(2)^2 - 7(2) - 5 = 6(4) - 14 - 5 = 24 - 14 - 5 = 5
\][/tex]
Since 5 is not equal to 0, [tex]\( (x-2) \)[/tex] is not a factor of A(x).
2. Polynomial B(x) = [tex]\(3x^2 + 15x - 42\)[/tex]:
Substitute [tex]\( x = 2 \)[/tex]:
[tex]\[
3(2)^2 + 15(2) - 42 = 3(4) + 30 - 42 = 12 + 30 - 42 = 0
\][/tex]
Since the result is 0, [tex]\( (x-2) \)[/tex] is a factor of B(x).
3. Polynomial C(x) = [tex]\(2x^3 + 13x^2 + 16x + 5\)[/tex]:
Substitute [tex]\( x = 2 \)[/tex]:
[tex]\[
2(2)^3 + 13(2)^2 + 16(2) + 5 = 2(8) + 13(4) + 32 + 5 = 16 + 52 + 32 + 5 = 105
\][/tex]
Since 105 is not equal to 0, [tex]\( (x-2) \)[/tex] is not a factor of C(x).
4. Polynomial D(x) = [tex]\(3x^3 - 2x^2 - 15x + 14\)[/tex]:
Substitute [tex]\( x = 2 \)[/tex]:
[tex]\[
3(2)^3 - 2(2)^2 - 15(2) + 14 = 3(8) - 2(4) - 30 + 14 = 24 - 8 - 30 + 14 = 0
\][/tex]
Since the result is 0, [tex]\( (x-2) \)[/tex] is a factor of D(x).
5. Polynomial E(x) = [tex]\(8x^4 - 41x^3 - 18x^2 + 101x + 70\)[/tex]:
Substitute [tex]\( x = 2 \)[/tex]:
[tex]\[
8(2)^4 - 41(2)^3 - 18(2)^2 + 101(2) + 70 = 8(16) - 41(8) - 18(4) + 202 + 70 = 0
\][/tex]
Since the result is 0, [tex]\( (x-2) \)[/tex] is a factor of E(x).
6. Polynomial F(x) = [tex]\(x^4 + 5x^3 - 27x^2 - 101x - 70\)[/tex]:
Substitute [tex]\( x = 2 \)[/tex]:
[tex]\[
(2)^4 + 5(2)^3 - 27(2)^2 - 101(2) - 70 = 16 + 40 - 108 - 202 - 70 = -324
\][/tex]
Since -324 is not equal to 0, [tex]\( (x-2) \)[/tex] is not a factor of F(x).
In conclusion, the polynomials that have [tex]\( (x-2) \)[/tex] as a factor are:
- Polynomial B(x): [tex]\(3x^2 + 15x - 42\)[/tex]
- Polynomial D(x): [tex]\(3x^3 - 2x^2 - 15x + 14\)[/tex]
- Polynomial E(x): [tex]\(8x^4 - 41x^3 - 18x^2 + 101x + 70\)[/tex]
