Question

f(x) = x⁴ + 4x² + 3; g(x) = 3x⁴ + 12x³ + 18x² + 12x + 3

Explanation:

Since the problem is not clearly stated, I assume we might be asked to analyze the functions \( f(x) \) and \( g(x) \), perhaps factor them or describe their transformations. Let's start with factoring \( f(x) \).

Step 1: Factor \( f(x) = x^4 + 4x^2 + 3 \)

Let \( u = x^2 \), then \( f(x) = u^2 + 4u + 3 \). Factoring the quadratic in \( u \): \( u^2 + 4u + 3=(u + 1)(u + 3) \). Substituting back \( u = x^2 \), we get \( f(x)=(x^2 + 1)(x^2 + 3) \).

Step 2: Factor \( g(x) = 3x^4 + 12x^3 + 18x^2 + 12x + 3 \)

First, factor out a 3: \( g(x)=3(x^4 + 4x^3 + 6x^2 + 4x + 1) \). Notice that \( x^4 + 4x^3 + 6x^2 + 4x + 1 \) is a perfect square trinomial in the form of \( (a + b)^4 \)? Wait, actually, \( (x + 1)^4=x^4 + 4x^3 + 6x^2 + 4x + 1 \). Let's check: \( (x + 1)^2=x^2 + 2x + 1 \), then \( (x + 1)^4=(x^2 + 2x + 1)^2=x^4 + 4x^3 + 6x^2 + 4x + 1 \). So \( g(x)=3(x + 1)^4 \).

If the question was about the relationship between the graphs, \( f(x) \) is a even function (symmetric about y-axis) with vertex at (0, 3), and \( g(x) \) is a transformed version, shifted left by 1 unit (since \( (x + 1)^4 \)) and vertically stretched by 3, with vertex at (-1, 0).

Answer:

For \( f(x) \), factored form is \( \boldsymbol{(x^2 + 1)(x^2 + 3)} \); for \( g(x) \), factored form is \( \boldsymbol{3(x + 1)^4} \). If analyzing the graph, \( f(x) \) is symmetric about y - axis, \( g(x) \) is symmetric about \( x=-1 \), \( g(x) \) is a vertical stretch and horizontal shift of a related function to \( f(x) \). (The answer depends on the specific question, but factoring is shown here.)