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Use the Factor Theorem to determine whether [tex]$x-1$[/tex] is a factor of [tex]$P(x) = -x^4 + x^3 + 6x^2 - 9$[/tex].

Specifically, evaluate [tex][tex]$P$[/tex][/tex] at the proper value, and then determine whether [tex]$x-1$[/tex] is a factor.

[tex]P(1) = \square[/tex]

A. [tex]x-1[/tex] is a factor of [tex]P(x)[/tex].
B. [tex]x-1[/tex] is not a factor of [tex]P(x)[/tex].


Sagot :

To determine whether [tex]\( x - 1 \)[/tex] is a factor of [tex]\( P(x) = -x^4 + x^3 + 6x^2 - 9 \)[/tex], we can use the Factor Theorem. The Factor Theorem states that [tex]\( x - c \)[/tex] is a factor of the polynomial [tex]\( P(x) \)[/tex] if and only if [tex]\( P(c) = 0 \)[/tex].

In this case, we need to evaluate [tex]\( P(1) \)[/tex] and check if it equals zero.

First, we evaluate the polynomial at [tex]\( x = 1 \)[/tex]:

[tex]\[ P(1) = - (1)^4 + (1)^3 + 6(1)^2 - 9 \][/tex]

Calculate each term step-by-step:

[tex]\[ (1)^4 = 1 \implies - (1)^4 = -1 \][/tex]

[tex]\[ (1)^3 = 1 \][/tex]

[tex]\[ 6(1)^2 = 6 \][/tex]

[tex]\[ -9 \text{ (constant term)} \][/tex]

Combine all the terms:

[tex]\[ P(1) = -1 + 1 + 6 - 9 \][/tex]

Simplify the expression:

[tex]\[ P(1) = 7 - 9 = -3 \][/tex]

So, [tex]\( P(1) = -3 \)[/tex].

Since [tex]\( P(1) \neq 0 \)[/tex], by the Factor Theorem, [tex]\( x - 1 \)[/tex] is not a factor of [tex]\( P(x) \)[/tex].

Thus, the detailed answer is:
[tex]\[ P(1) = -3 \][/tex]

And

[tex]\[ x-1 \text{ is not a factor of } P(x). \][/tex]