Question #14

Given the following polynomial function, which answer choice is NOT a factor?

[tex]\[ f(x) = x^4 - x^3 - 7x^2 + x + 6 \][/tex]

A. [tex]\((x + 2)\)[/tex]

B. [tex]\((x + 1)\)[/tex]

C. [tex]\((x - 1)\)[/tex]

D. [tex]\((x + 3)\)[/tex]



Answer :

To determine which of the given answers is not a factor of the polynomial [tex]\( f(x) = x^4 - x^3 - 7x^2 + x + 6 \)[/tex], we will use the Remainder Theorem. The Remainder Theorem states that if a polynomial [tex]\( f(x) \)[/tex] is divided by [tex]\( x - c \)[/tex], the remainder of this division is [tex]\( f(c) \)[/tex]. Therefore, if [tex]\( f(c) = 0 \)[/tex], then [tex]\( x - c \)[/tex] is a factor of [tex]\( f(x) \)[/tex].

Let's check each of the given factors one by one:

1. Check [tex]\( x + 2 \)[/tex] (which means [tex]\( c = -2 \)[/tex]):
[tex]\[ f(-2) = (-2)^4 - (-2)^3 - 7(-2)^2 + (-2) + 6 \][/tex]
[tex]\[ = 16 + 8 - 28 - 2 + 6 \][/tex]
[tex]\[ = 0 \][/tex]
Since [tex]\( f(-2) = 0 \)[/tex], [tex]\( x + 2 \)[/tex] is a factor of [tex]\( f(x) \)[/tex].

2. Check [tex]\( x + 1 \)[/tex] (which means [tex]\( c = -1 \)[/tex]):
[tex]\[ f(-1) = (-1)^4 - (-1)^3 - 7(-1)^2 + (-1) + 6 \][/tex]
[tex]\[ = 1 + 1 - 7 - 1 + 6 \][/tex]
[tex]\[ = 0 \][/tex]
Since [tex]\( f(-1) = 0 \)[/tex], [tex]\( x + 1 \)[/tex] is a factor of [tex]\( f(x) \)[/tex].

3. Check [tex]\( x - 1 \)[/tex] (which means [tex]\( c = 1 \)[/tex]):
[tex]\[ f(1) = (1)^4 - (1)^3 - 7(1)^2 + (1) + 6 \][/tex]
[tex]\[ = 1 - 1 - 7 + 1 + 6 \][/tex]
[tex]\[ = 0 \][/tex]
Since [tex]\( f(1) = 0 \)[/tex], [tex]\( x - 1 \)[/tex] is a factor of [tex]\( f(x) \)[/tex].

4. Check [tex]\( x + 3 \)[/tex] (which means [tex]\( c = -3 \)[/tex]):
[tex]\[ f(-3) = (-3)^4 - (-3)^3 - 7(-3)^2 + (-3) + 6 \][/tex]
[tex]\[ = 81 + 27 - 63 - 3 + 6 \][/tex]
[tex]\[ = 48 \][/tex]
Since [tex]\( f(-3) \neq 0 \)[/tex] (it equals 48), [tex]\( x + 3 \)[/tex] is not a factor of [tex]\( f(x) \)[/tex].

Therefore, the factor that is not a factor of the polynomial [tex]\( f(x) = x^4 - x^3 - 7x^2 + x + 6 \)[/tex] is [tex]\((x + 3)\)[/tex].