Question

f(x) = (x^2 + 4) g(x) = x + 3 h(x) = x - 1
find f(g(x)) • h(x)
a. $x^2 + x + 12$
b. $x^3 + 5x^2 + 7x - 13$
c. $x^3 - x^2 + 13x - 13$
d. $x^4 - 2x^3 + x^2 + 8x - 12$

Explanation:

Step1: Find \( f(g(x)) \)

Substitute \( g(x) = x + 3 \) into \( f(x) \). So \( f(g(x)) = f(x + 3) = ( (x + 3)^2 + 4 ) \). Expand \( (x + 3)^2 \): \( (x + 3)^2 = x^2 + 6x + 9 \). Then \( f(g(x)) = x^2 + 6x + 9 + 4 = x^2 + 6x + 13 \).

Step2: Multiply \( f(g(x)) \) by \( h(x) \)

\( h(x) = x - 1 \), so we multiply \( (x^2 + 6x + 13) \) by \( (x - 1) \). Using the distributive property (FOIL method for polynomials):
\[

$$\begin{align*} &(x^2 + 6x + 13)(x - 1)\\ =& x^2 \cdot x + x^2 \cdot (-1) + 6x \cdot x + 6x \cdot (-1) + 13 \cdot x + 13 \cdot (-1)\\ =& x^3 - x^2 + 6x^2 - 6x + 13x - 13\\ =& x^3 + 5x^2 + 7x - 13 \end{align*}$$

\]
Wait, but let's check again. Wait, original \( f(x) \) is \( (x^2 + 4) \)? Wait, maybe I misread. Wait, the problem says \( f(x) = (x^2 + 4) \)? Wait, no, maybe it's \( f(x) = (x^2 + 4) \) or is it \( f(x) = (x^3 + 4) \)? Wait, the first line: \( f(x) = (x^2 + 4) \)? Wait, the user's image: \( f(x) = (x^2 + 4) \), \( g(x) = x + 3 \), \( h(x) = x - 1 \). Wait, no, maybe I made a mistake. Wait, let's re-express:

Wait, if \( f(x) = x^2 + 4 \), then \( f(g(x)) = f(x + 3) = (x + 3)^2 + 4 = x^2 + 6x + 9 + 4 = x^2 + 6x + 13 \). Then multiply by \( h(x) = x - 1 \):

\( (x^2 + 6x + 13)(x - 1) = x^3 - x^2 + 6x^2 - 6x + 13x - 13 = x^3 + 5x^2 + 7x - 13 \), which is option b. But wait, let's check the options again. Wait, the options are:

a. \( x^2 + x + 12 \)

b. \( x^3 + 5x^2 + 7x - 13 \)

c. \( x^3 - x^2 + 13x - 13 \)

d. \( x^4 - 2x^3 + x^2 + 8x - 12 \)

Wait, maybe I misread \( f(x) \). Wait, maybe \( f(x) = x^3 + 4 \)? Let's check. If \( f(x) = x^3 + 4 \), then \( f(g(x)) = (x + 3)^3 + 4 \). Expand \( (x + 3)^3 = x^3 + 9x^2 + 27x + 27 \), so \( f(g(x)) = x^3 + 9x^2 + 27x + 27 + 4 = x^3 + 9x^2 + 27x + 31 \). Then multiply by \( h(x) = x - 1 \): \( (x^3 + 9x^2 + 27x + 31)(x - 1) = x^4 - x^3 + 9x^3 - 9x^2 + 27x^2 - 27x + 31x - 31 = x^4 + 8x^3 + 18x^2 + 4x - 31 \), which is not any option. So maybe \( f(x) = x^2 + 4 \) is correct. Then the multiplication:

\( (x^2 + 6x + 13)(x - 1) = x^3 - x^2 + 6x^2 - 6x + 13x - 13 = x^3 + 5x^2 + 7x - 13 \), which is option b.

Answer:

b. \( x^3 + 5x^2 + 7x - 13 \)