Q2(ii):

Use the Factor Theorem to determine whether $g(x)$ is a factor of $p(x)$ in each of the following cases: (ii) $p(x) = x^3 + 3x^2 + 3x + 1, g(x) = x + 2$

Given Variables & Initial Setup

We are tasked with determining whether the linear polynomial $g(x)$ is a factor of the cubic polynomial $p(x)$. The given functions are:

• Dividend Polynomial: $p(x) = x^3 + 3x^2 + 3x + 1$
• Divisor Polynomial: $g(x) = x + 2$

Theoretical Foundation [The Factor Theorem]: The Factor Theorem states that for any polynomial $p(x)$ of degree $n \ge 1$ and any real number $c$, the linear binomial $(x - c)$ is a factor of $p(x)$ if and only if $p(c) = 0$. In other words, the remainder upon dividing $p(x)$ by $(x - c)$ must be exactly zero.

Step 1: Determining the Root of the Divisor $g(x)$

To apply the Factor Theorem, we must first find the zero (or root) of the divisor $g(x)$. We do this by setting $g(x)$ equal to zero and solving for $x$:

$g(x) = 0$
$x + 2 = 0$
$x = -2$

Here, our test value is $c = -2$. According to the Factor Theorem, $g(x)$ will be a factor of $p(x)$ if and only if $p(-2) = 0$.

Step 2: Evaluating the Polynomial $p(x)$ at $x = -2$

We substitute $x = -2$ into the polynomial $p(x)$ to calculate the remainder.

$p(-2) = (-2)^3 + 3(-2)^2 + 3(-2) + 1$

Now, we resolve each term sequentially following the order of operations (exponents, then multiplication):

• Cubic term: $(-2)^3 = -8$
• Quadratic term: $3(-2)^2 = 3(4) = 12$
• Linear term: $3(-2) = -6$
• Constant term: $1$

Substituting these evaluated terms back into the equation yields:

$p(-2) = -8 + 12 - 6 + 1$

Combining the terms from left to right:

$p(-2) = 4 - 6 + 1$
$p(-2) = -2 + 1$
$p(-2) = -1$

Step 3: Analytical Insight & Verification (Algebraic Identity)

To demonstrate rigorous mathematical fluency, we can verify this result by recognizing the structure of $p(x)$. The polynomial $x^3 + 3x^2 + 3x + 1$ is the standard binomial expansion of $(x + 1)^3$ [Per the algebraic identity $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$].

Rewriting $p(x)$:
$p(x) = (x + 1)^3$

Evaluating at $x = -2$:
$p(-2) = (-2 + 1)^3 = (-1)^3 = -1$

This confirms our previous arithmetic calculation with absolute certainty.

Step 4: Graphical Representation of the Remainder

The graph below illustrates the function $p(x) = (x+1)^3$. The Factor Theorem dictates that for $x+2$ to be a factor, the graph must cross the x-axis at $x = -2$. As shown, at $x = -2$, the function evaluates to $-1$, visually confirming a non-zero remainder.

Step 5: Conclusion

Because the evaluation of the polynomial at $x = -2$ yields a remainder of $-1$ rather than $0$ (i.e., $p(-2) \neq 0$), the condition required by the Factor Theorem is not satisfied.

Final Solution: Since $p(-2) = -1 \neq 0$, by the Factor Theorem, $g(x) = x + 2$ is NOT a factor of $p(x) = x^3 + 3x^2 + 3x + 1$.