Q1(iii):

Determine which of the following polynomials has $(x + 1)$ a factor : (iii) $x^4 + 3x^3 + 3x^2 + x + 1$

Given Variables & Initial Setup

We are tasked with determining whether the linear binomial $D(x) = x + 1$ is a factor of the given quartic polynomial:

$P(x) = x^4 + 3x^3 + 3x^2 + x + 1$

To solve this, we utilize the Factor Theorem. [Per the Factor Theorem of Polynomials, a linear polynomial $(x - c)$ is a factor of a polynomial $P(x)$ if and only if $P(c) = 0$].

Step 1: Determining the Root of the Divisor

First, we must find the zero (or root) of the divisor $D(x) = x + 1$. We do this by equating the divisor to zero and solving for $x$:

$x + 1 = 0$

$x = -1$

[By the properties of equality, subtracting 1 from both sides isolates $x$, yielding the test value $c = -1$].

Step 2: Applying the Factor Theorem

We now substitute the root $x = -1$ into the original polynomial $P(x)$ to evaluate the remainder. [According to the Remainder Theorem, evaluating $P(-1)$ yields the exact remainder of the division $\frac{P(x)}{x+1}$].

$P(-1) = (-1)^4 + 3(-1)^3 + 3(-1)^2 + (-1) + 1$

Step 3: Algebraic Evaluation

We systematically evaluate each term of the polynomial, adhering strictly to the order of operations (exponents before multiplication):

• First term: $(-1)^4 = 1$ [An even power of a negative number results in a positive value].
• Second term: $3(-1)^3 = 3(-1) = -3$ [An odd power of a negative number remains negative].
• Third term: $3(-1)^2 = 3(1) = 3$ [An even power yields a positive, multiplied by 3].
• Fourth term: $(-1) = -1$
• Fifth term: $1 = 1$ [Constant term remains unchanged].

Substituting these evaluated terms back into the polynomial expression:

$P(-1) = 1 - 3 + 3 - 1 + 1$

Step 4: Final Arithmetic Simplification

We now sum the terms sequentially from left to right:

$P(-1) = (1 - 3) + 3 - 1 + 1$

$P(-1) = -2 + 3 - 1 + 1$

$P(-1) = 1 - 1 + 1$

$P(-1) = 0 + 1$

$P(-1) = 1$

Step 5: Final Analysis & Conclusion

The calculated remainder is $1$. Because the remainder is non-zero ($P(-1) \neq 0$), the polynomial $P(x)$ is not perfectly divisible by $(x + 1)$. [By the logical contrapositive of the Factor Theorem, if $P(c) \neq 0$, then $(x - c)$ cannot be a factor].

Final Solution: Since $P(-1) = 1 \neq 0$, the binomial $(x + 1)$ is NOT a factor of the polynomial $x^4 + 3x^3 + 3x^2 + x + 1$.