Answer :
To determine which of the given polynomials could have [tex]\((x-2)\)[/tex] as a factor, we'll use the Remainder Theorem. This theorem states that if a polynomial [tex]\( f(x) \)[/tex] has [tex]\((x-a)\)[/tex] as a factor, then [tex]\( f(a) = 0 \)[/tex].
In this problem, we want to check if [tex]\((x-2)\)[/tex] is a factor, so we will substitute [tex]\( x = 2 \)[/tex] into each polynomial. If the result is 0, then [tex]\((x-2)\)[/tex] is a factor of that polynomial.
Let's evaluate each polynomial:
a. [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.
b. [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.
c. [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.
d. [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.
e. [tex]\( E(x) = 8x^4 - 41x^3 - 18x^2 + 101x + 70 \)[/tex]
- Substitute [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]
- Since [tex]\( E(2) = 0 \)[/tex], [tex]\((x-2)\)[/tex] is a factor.
f. [tex]\( F(x) = x^4 + 5x^3 - 27x^2 - 101x - 70 \)[/tex]
- Substitute [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]
- Since [tex]\( F(2) \neq 0 \)[/tex], [tex]\((x-2)\)[/tex] is not a factor.
Conclusion:
The polynomials that 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]
In this problem, we want to check if [tex]\((x-2)\)[/tex] is a factor, so we will substitute [tex]\( x = 2 \)[/tex] into each polynomial. If the result is 0, then [tex]\((x-2)\)[/tex] is a factor of that polynomial.
Let's evaluate each polynomial:
a. [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.
b. [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.
c. [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.
d. [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.
e. [tex]\( E(x) = 8x^4 - 41x^3 - 18x^2 + 101x + 70 \)[/tex]
- Substitute [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]
- Since [tex]\( E(2) = 0 \)[/tex], [tex]\((x-2)\)[/tex] is a factor.
f. [tex]\( F(x) = x^4 + 5x^3 - 27x^2 - 101x - 70 \)[/tex]
- Substitute [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]
- Since [tex]\( F(2) \neq 0 \)[/tex], [tex]\((x-2)\)[/tex] is not a factor.
Conclusion:
The polynomials that 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]
