Let g(x)=(x^(2)-x-12)/(x-4) when x!=4. g is continuous for all real numbers. Find g(4). Choose 1 answer: (A) -4 (B) 7 (c) -3 (D) 4

Factorize numerator: Simplify the expression for g(x) by factoring the numerator. The numerator x^2 - x - 12 can be factored as (x - 4)(x + 3). So, g(x) = [(x - 4)(x + 3)] / (x - 4). Cancel common factor: Cancel out the common factor (x - 4) from the numerator and the denominator. This gives g(x) = x + 3 for x ≠ 4. Find g(4): To find g(4), substitute x = 4 into the simplified expression g(x) = x + 3. So, g(4) = 4 + 3 = 7.

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Let
g(x)=(x^(2)-x-12)/(x-4) when
x!=4.

g is continuous for all real numbers.
Find
g(4).
Choose 1 answer:
(A) -4
(B) 7
(c) -3
(D) 4

Let $g(x)=\frac{x^{2}-x-12}{x-4}$ when $x\neq 4$ . $\newline$ $g$ is continuous for all real numbers. $\newline$ Find $g(4)$ . $\newline$ Choose $1$ answer: $\newline$ (A) $-4$ $\newline$ (B) $7$ $\newline$ (C) $-3$ $\newline$ (D) $4$

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Algebra 2

Conjugate root theorems

Full solution

Q. Let

g(x)=\frac{x^{2}-x-12}{x-4}

when

x\neq 4

\newline

g

is continuous for all real numbers.

\newline

Find

g(4)

\newline

Choose

1

answer:

\newline

(A)

-4

\newline

(B)

7

\newline

(C)

-3

\newline

(D)

4

Factorize numerator: Simplify the expression for $g(x)$ by factoring the numerator. The numerator $x^2 - x - 12$ can be factored as $(x - 4)(x + 3)$ . So, $g(x) = \frac{(x - 4)(x + 3)}{(x - 4)}$ .
Cancel common factor: Cancel out the common factor $(x - 4)$ from the numerator and the denominator. This gives $g(x) = x + 3$ for $x \neq 4$ .
Find $g(4)$ : To find $g(4)$ , substitute $x = 4$ into the simplified expression $g(x) = x + 3$ . So, $g(4) = 4 + 3 = 7$ .