Given Variables & Theoretical Foundation

We are given the following polynomial and a linear divisor:

• Dividend Polynomial: $p(x) = x^3 - 4x^2 + x + 6$
• Divisor Polynomial: $g(x) = x - 3$

Theoretical Justification [The Factor Theorem]: The Factor Theorem states that a polynomial $p(x)$ has a factor $(x - c)$ if and only if $p(c) = 0$. Therefore, to determine if $g(x)$ is a factor of $p(x)$, we must evaluate the polynomial at the root of $g(x)$ and check if the remainder is zero.

Step 1: Determining the Root of the Divisor

To find the value of $c$ to substitute into $p(x)$, we set the divisor $g(x)$ equal to zero and solve for $x$:

$g(x) = 0$
$x - 3 = 0$
$x = 3$

Here, our test value is $c = 3$.

Step 2: Evaluating the Polynomial $p(x)$ at $x = 3$

We substitute $x = 3$ into the original polynomial $p(x)$ to find the remainder [Per the Remainder Theorem]:

$p(3) = (3)^3 - 4(3)^2 + (3) + 6$

Now, we perform the arithmetic operations step-by-step:

• Calculate the cube of 3: $(3)^3 = 27$
• Calculate the square of 3, then multiply by -4: $-4(3)^2 = -4(9) = -36$
• Substitute these values back into the expression:

$p(3) = 27 - 36 + 3 + 6$

Group the positive and negative terms to simplify the addition:

$p(3) = (27 + 3 + 6) - 36$
$p(3) = 36 - 36$
$p(3) = 0$

Step 3: Graphical Verification of the Roots

Because $p(3) = 0$, $x = 3$ is an $x$-intercept (or root) of the polynomial curve $y = x^3 - 4x^2 + x + 6$. The graph below plots the exact coordinates of the polynomial, visually confirming that the curve intersects the $x$-axis exactly at $x = 3$.

Step 4: Logical Conclusion

Since the evaluation of the polynomial at $x = 3$ yields a remainder of exactly $0$ ($p(3) = 0$), the condition of the Factor Theorem is perfectly satisfied.

Final Solution: By the Factor Theorem, since $p(3) = 0$, $g(x) = x - 3$ is indeed a factor of the polynomial $p(x) = x^3 - 4x^2 + x + 6$.

Initial Setup & Polynomial Identification

We are tasked with factorising the quadratic polynomial $P(x) = 12x^2 - 7x + 1$. [Per the Fundamental Theorem of Algebra and the properties of quadratic expressions, a polynomial of degree 2 can generally be factored into the product of two linear binomials]. We will utilize the method of splitting the middle term (also known as the AC method).

Step 1: Determine the Coefficients for the AC Method

A standard quadratic polynomial is expressed in the form $ax^2 + bx + c$. By comparing our given polynomial to the standard form, we identify the coefficients:

• $a = 12$ (Coefficient of $x^2$)
• $b = -7$ (Coefficient of $x$)
• $c = 1$ (Constant term)

Step 2: Find the Splitting Factors

To split the middle term, we must find two integers, let us define them as $p$ and $q$, that satisfy two specific conditions simultaneously:

1. Product Condition: $p \times q = a \times c = 12 \times 1 = 12$
2. Sum Condition: $p + q = b = -7$

[This algebraic condition ensures that the middle term is partitioned in a way that maintains the polynomial's equivalence while allowing for factorisation by grouping].

We list the factor pairs of $12$: $(1, 12)$, $(2, 6)$, and $(3, 4)$. Because the product ($+12$) is positive and the sum ($-7$) is negative, both factors $p$ and $q$ must be negative integers. Let us test the negative pairs:

• $(-1) + (-12) = -13$ (Incorrect)
• $(-2) + (-6) = -8$ (Incorrect)
• $(-3) + (-4) = -7$ (Correct)

Thus, the required splitting factors are $p = -4$ and $q = -3$.

Step 3: Rewrite the Polynomial by Splitting the Middle Term

We substitute the middle term $-7x$ with the equivalent expression $-4x - 3x$:

$P(x) = 12x^2 - 4x - 3x + 1$

Step 4: Factorise by Grouping

We now group the four terms into two distinct pairs to extract the Greatest Common Factor (GCF) from each pair:

$(12x^2 - 4x) - (3x - 1)$

Analyzing the first group: $(12x^2 - 4x)$
The highest common numerical factor of $12$ and $4$ is $4$. The highest common variable factor of $x^2$ and $x$ is $x$. Therefore, the GCF is $4x$.
Factoring out $4x$ yields: $4x(3x - 1)$

Analyzing the second group: $-(3x - 1)$
To ensure the binomial inside the parentheses matches the first group, we factor out $-1$.
Factoring out $-1$ yields: $-1(3x - 1)$

[By the Distributive Property of Multiplication over Addition, $ab + ac = a(b+c)$. We apply this in reverse to factor out the GCF].

Step 5: Extract the Common Binomial Factor

Substitute the factored groups back into the main expression:

$4x(3x - 1) - 1(3x - 1)$

Observe that the binomial $(3x - 1)$ is now a common factor to both major terms. We factor out $(3x - 1)$ from the entire expression:

$(3x - 1)(4x - 1)$

Final Solution: The factorised form of the polynomial $12x^2 - 7x + 1$ is $(4x - 1)(3x - 1)$.

Initial Setup & Theoretical Foundation

We are given the quadratic polynomial:

$p(x) = x^2 + x + k$

We are also given the condition that the linear binomial $(x - 1)$ is a factor of $p(x)$. Our objective is to determine the precise value of the unknown constant $k$ that satisfies this condition.

Step 1: Applying the Factor Theorem

To solve for $k$, we utilize the Factor Theorem. [Per the Factor Theorem of Polynomials, a linear polynomial $(x - a)$ is a factor of a polynomial $p(x)$ if and only if the polynomial evaluated at $x = a$ is equal to zero, i.e., $p(a) = 0$].

To find the root of the divisor, we set the linear factor to zero:

$x - 1 = 0 \implies x = 1$

Therefore, for $(x - 1)$ to be a factor of $p(x)$, the condition $p(1) = 0$ must hold true.

Step 2: Evaluating the Polynomial at the Root

We substitute $x = 1$ into the given polynomial $p(x)$:

$p(1) = (1)^2 + (1) + k$

Simplifying the numerical terms [By applying standard exponentiation and addition]:

$p(1) = 1 + 1 + k$

$p(1) = 2 + k$

Step 3: Solving for the Unknown Constant $k$

According to the condition established in Step 1, we equate $p(1)$ to $0$:

$2 + k = 0$

Isolating $k$ by subtracting $2$ from both sides of the equation [Per the Subtraction Property of Equality]:

$k = -2$

Graphical Verification of the Polynomial

By substituting $k = -2$ back into the original expression, we obtain the complete polynomial: $p(x) = x^2 + x - 2$. Geometrically, the roots of this polynomial correspond to the x-intercepts of its parabolic graph. As shown below, the parabola crosses the x-axis exactly at $x = 1$, visually confirming that $(x - 1)$ is a factor.

Final Solution: The value of $k$ for which $(x - 1)$ is a factor of $p(x) = x^2 + x + k$ is $k = -2$.

Initial Setup & Given Polynomial

We are tasked with factorising the quadratic polynomial:

$P(x) = 6x^2 + 5x - 6$

This expression is in the standard quadratic form $ax^2 + bx + c$. To factorise it, we will employ the Splitting the Middle Term technique, also known as the AC Method [Per the fundamental principles of polynomial factorisation over integers].

Step 1: Coefficient Analysis & The AC Method

First, we identify the coefficients of the quadratic polynomial:

• Leading coefficient ($a$) = $6$
• Middle coefficient ($b$) = $5$
• Constant term ($c$) = $-6$

According to the AC Method, we must find the product of the leading coefficient and the constant term ($a \times c$):

$a \times c = 6 \times (-6) = -36$

Step 2: Splitting the Middle Term

We must now find two integers, let's call them $p$ and $q$, that satisfy two conditions simultaneously:

1. Their product must equal $ac$: $p \times q = -36$
2. Their sum must equal $b$: $p + q = 5$

Since the product ($-36$) is negative, the two numbers must have opposite signs. Since their sum ($5$) is positive, the number with the larger absolute value must be positive. Let us systematically evaluate the factor pairs of $36$:

The correct integers are $9$ and $-4$. We will use these to split the middle term ($5x$).

Step 3: Algebraic Substitution and Grouping

Substitute $5x$ with $9x - 4x$ in the original polynomial:

$6x^2 + 9x - 4x - 6$

Next, we group the terms into pairs to factor out the Greatest Common Divisor (GCD) from each pair [Applying the Distributive Property $ab + ac = a(b+c)$]:

$= (6x^2 + 9x) - (4x + 6)$

Extract the GCD from the first group ($6x^2 + 9x$). The GCD of $6$ and $9$ is $3$, and the GCD of $x^2$ and $x$ is $x$. Thus, we factor out $3x$:

$= 3x(2x + 3) - (4x + 6)$

Extract the GCD from the second group ($4x + 6$). The GCD of $4$ and $6$ is $2$. To ensure the binomial inside the parentheses matches the first group, we factor out $-2$:

$= 3x(2x + 3) - 2(2x + 3)$

Step 4: Factoring out the Common Binomial

Notice that the binomial $(2x + 3)$ is now a common factor in both terms. We factor it out [By the reverse distributive property]:

$= (2x + 3)(3x - 2)$

Geometric Verification: Area Model

To rigorously verify our algebraic manipulation, we can map the grouped terms to a geometric area model. The total area of the rectangle represents the polynomial $6x^2 + 5x - 6$, while the side lengths represent its factors $(2x + 3)$ and $(3x - 2)$.

Final Solution: The factorised form of the polynomial $6x^2 + 5x - 6$ is $(2x + 3)(3x - 2)$.

Initial Setup & Theoretical Foundation

We are tasked with factorising the quadratic polynomial:

$P(x) = 2x^2 + 7x + 3$

To factorise a quadratic polynomial of the standard form $ax^2 + bx + c$, we employ the "Splitting the Middle Term" method (also known as the AC method). This technique relies on finding two integers, $p$ and $q$, that satisfy two specific conditions simultaneously:

• Sum Condition: $p + q = b$ [The coefficient of the middle term]
• Product Condition: $p \times q = a \times c$ [The product of the leading coefficient and the constant term]

Step 1: Identifying Coefficients for the AC Method

By comparing the given polynomial $2x^2 + 7x + 3$ with the standard quadratic form $ax^2 + bx + c$, we extract the following coefficients:

• Leading coefficient ($a$) = $2$
• Middle coefficient ($b$) = $7$
• Constant term ($c$) = $3$

Next, we calculate the target product ($ac$):

$a \times c = 2 \times 3 = 6$

Step 2: Determining the Splitting Factors

We must find two integers $p$ and $q$ such that:

$p + q = 7$

$p \times q = 6$

Let us systematically evaluate the factor pairs of $6$:

The correct integers are $6$ and $1$.

Step 3: Rewriting the Polynomial

We substitute the middle term $7x$ with the sum of $6x$ and $1x$ [Per the distributive property, $7x = (6 + 1)x = 6x + x$]:

$2x^2 + 6x + x + 3$

Step 4: Factorisation by Grouping

We now group the four terms into two pairs to extract the Greatest Common Factor (GCF) from each pair:

$= (2x^2 + 6x) + (x + 3)$

Extracting the GCF from the first group: The terms $2x^2$ and $6x$ share a common factor of $2x$.

$2x(x + 3)$

Extracting the GCF from the second group: The terms $x$ and $3$ share no common factors other than $1$.

$+ 1(x + 3)$

Reassembling the expression yields:

$= 2x(x + 3) + 1(x + 3)$

Notice that the binomial $(x + 3)$ is now a common factor to both major terms. We factor out $(x + 3)$ [By the Distributive Property of Multiplication over Addition, $AB + CB = (A + C)B$]:

$= (x + 3)(2x + 1)$

Visual Representation: Area Model of Factorisation

The area model geometrically validates our algebraic factorisation. The total area of the rectangle represents the polynomial $2x^2 + 7x + 3$, while the side lengths represent the factors $(2x + 1)$ and $(x + 3)$.

Final Solution: The factorised form of the polynomial $2x^2 + 7x + 3$ is $(x + 3)(2x + 1)$.

Given Variables & Initial Setup

We are given a quadratic polynomial $p(x)$ with an unknown coefficient $k$, and a linear binomial $g(x)$ which is stated to be a factor of $p(x)$.

• The Polynomial: $p(x) = 2x^2 + kx + \sqrt{2}$
• The Linear Divisor: $g(x) = x - 1$

Step 1: Determining the Zero of the Divisor

To apply polynomial remainder and factor theorems, we must first find the root (or zero) of the linear divisor $g(x)$. [Per the Fundamental Theorem of Algebra, a linear polynomial of the form $ax + b$ has exactly one real root].

We set the divisor equal to zero:

$x - 1 = 0$

$x = 1$

The zero of the divisor is $x = 1$.

Step 2: Application of the Factor Theorem

The Factor Theorem states that a polynomial $p(x)$ has a factor $(x - c)$ if and only if $p(c) = 0$. [This is a direct corollary of the Polynomial Remainder Theorem, which states that the remainder of $p(x)$ divided by $(x - c)$ is $p(c)$].

Since the problem explicitly states that $(x - 1)$ is a factor of $p(x)$, we can definitively establish the following equation:

$p(1) = 0$

Step 3: Algebraic Substitution and Evaluation

We now substitute $x = 1$ into the original polynomial expression $p(x) = 2x^2 + kx + \sqrt{2}$ and equate the entire expression to zero.

$p(1) = 2(1)^2 + k(1) + \sqrt{2} = 0$

Evaluate the exponential and multiplicative terms [By the identity property of multiplication, $1^2 = 1$ and $k \cdot 1 = k$]:

$2(1) + k + \sqrt{2} = 0$

$2 + k + \sqrt{2} = 0$

Step 4: Isolating the Unknown Variable $k$

To find the value of $k$, we must isolate it on one side of the equation. We achieve this by subtracting $2$ and $\sqrt{2}$ from both sides of the equation [Per the Subtraction Property of Equality].

$k = -2 - \sqrt{2}$

For mathematical elegance and standard algebraic formatting, we can factor out the negative sign from the binomial on the right side:

$k = -(2 + \sqrt{2})$

Final Solution: The value of $k$ is $-(2 + \sqrt{2})$.

Initial Setup & Given Polynomial

We are tasked with factorising the cubic polynomial. Let the given polynomial be defined as:

$P(x) = x^3 - 3x^2 - 9x - 5$

Step 1: Applying the Rational Root Theorem & Factor Theorem

To find the first linear factor of the cubic polynomial, we utilize the Rational Root Theorem. The theorem states that any rational root of the polynomial must be a divisor of the constant term ($-5$) divided by a divisor of the leading coefficient ($1$).

The possible rational roots are the divisors of $-5$, which are: $\pm 1, \pm 5$.

We evaluate $P(x)$ at these test values to find a root (where $P(x) = 0$):

• Testing $x = 1$:
  $P(1) = (1)^3 - 3(1)^2 - 9(1) - 5 = 1 - 3 - 9 - 5 = -16$
  Since $P(1) \neq 0$, $x = 1$ is not a root.


• Testing $x = -1$:
  $P(-1) = (-1)^3 - 3(-1)^2 - 9(-1) - 5$
  $P(-1) = -1 - 3(1) + 9 - 5$
  $P(-1) = -1 - 3 + 9 - 5 = 0$

Since $P(-1) = 0$, we conclude [By the Factor Theorem] that $(x + 1)$ is a factor of the polynomial $P(x)$.

Step 2: Algebraic Manipulation (Factorisation by Grouping)

Knowing that $(x + 1)$ is a factor, we can rewrite the terms of the original polynomial $P(x)$ such that $(x + 1)$ can be factored out from grouped terms. This method is mathematically more elegant than polynomial long division.

We systematically split the middle terms to force the appearance of $(x + 1)$:

$P(x) = x^3 - 3x^2 - 9x - 5$

[Split $-3x^2$ into $+x^2 - 4x^2$ to pair with $x^3$]
$P(x) = x^3 + x^2 - 4x^2 - 9x - 5$

[Split $-9x$ into $-4x - 5x$ to pair with $-4x^2$]
$P(x) = x^3 + x^2 - 4x^2 - 4x - 5x - 5$

[Group the terms into pairs]
$P(x) = (x^3 + x^2) - (4x^2 + 4x) - (5x + 5)$

[Factor out the greatest common monomial from each group]
$P(x) = x^2(x + 1) - 4x(x + 1) - 5(x + 1)$

[Factor out the common binomial $(x + 1)$]
$P(x) = (x + 1)(x^2 - 4x - 5)$

Step 3: Factorising the Quadratic Quotient

We are now left with a linear factor and a quadratic factor: $Q(x) = x^2 - 4x - 5$. We must factorise this quadratic expression by splitting the middle term.

We seek two numbers that multiply to the constant term ($-5$) and add up to the coefficient of the middle term ($-4$). These numbers are $-5$ and $1$.

$x^2 - 4x - 5$
$= x^2 - 5x + 1x - 5$
$= x(x - 5) + 1(x - 5)$
$= (x - 5)(x + 1)$

Step 4: Final Synthesis of Factors

Substitute the fully factorised quadratic back into the equation from Step 2:

$P(x) = (x + 1) \cdot [(x - 5)(x + 1)]$
$P(x) = (x + 1)(x + 1)(x - 5)$
$P(x) = (x + 1)^2(x - 5)$

Visual Verification: Graph of the Polynomial

The graph below illustrates the function $y = x^3 - 3x^2 - 9x - 5$. Notice that the curve is tangent to the x-axis at $x = -1$ (indicating a root with a multiplicity of 2, hence $(x+1)^2$) and intersects the x-axis at $x = 5$ (indicating a root with a multiplicity of 1, hence $(x-5)$).

Final Solution: The completely factorised form of the polynomial $x^3 - 3x^2 - 9x - 5$ is $(x + 1)^2(x - 5)$.

Initial Setup & Theoretical Foundation

We are given the polynomial function $p(x)$ and a linear binomial $g(x)$ which is stated to be a factor of $p(x)$:

• Dividend Polynomial: $p(x) = kx^2 - 3x + k$
• Linear Factor: $g(x) = x - 1$

[Per the Factor Theorem, a polynomial $p(x)$ has a factor $(x - c)$ if and only if the polynomial evaluates to zero at $x = c$, meaning $p(c) = 0$. This is a direct corollary of the Remainder Theorem, where a remainder of zero indicates perfect divisibility.]

Step 1: Determining the Root of the Linear Factor

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

$x - 1 = 0$
$x = 1$

[This establishes that $c = 1$. Therefore, if $x - 1$ is a factor of $p(x)$, evaluating $p(x)$ at $x = 1$ must yield exactly $0$.]

Step 2: Applying the Factor Theorem

We substitute $x = 1$ into the polynomial $p(x)$ and set the entire expression equal to zero:

$p(1) = k(1)^2 - 3(1) + k = 0$

Step 3: Algebraic Evaluation for $k$

Now, we simplify the equation to isolate and solve for the unknown constant $k$:

• Evaluate the exponent and multiplication: $k(1) - 3 + k = 0$
• Combine like terms ($k + k$): $2k - 3 = 0$
• Add $3$ to both sides of the equation: $2k = 3$
• Divide by $2$: $k = \frac{3}{2}$

Graphical Verification (Visualizing the Polynomial)

By substituting $k = \frac{3}{2}$ back into the original equation, the polynomial becomes $p(x) = \frac{3}{2}x^2 - 3x + \frac{3}{2}$. Factoring out $\frac{3}{2}$ yields $p(x) = \frac{3}{2}(x^2 - 2x + 1) = \frac{3}{2}(x - 1)^2$. This reveals that $x = 1$ is a repeated root, meaning the parabola's vertex rests exactly on the x-axis at $x = 1$.

Final Solution: The value of $k$ for which $x - 1$ is a factor of $p(x) = kx^2 - 3x + k$ is $k = \frac{3}{2}$.

Initial Setup & Given Polynomial

We are tasked with completely factorising the cubic polynomial:

$P(x) = x^3 + 13x^2 + 32x + 20$

Since the highest degree of the polynomial is $3$, it will have at most three linear factors. We will determine these factors systematically using the Factor Theorem and algebraic manipulation.

Step 1: Applying the Rational Root Theorem & Factor Theorem

To find the first linear factor, we look for an integer root by testing the factors of the constant term. The constant term is $20$.

The possible integer roots are the divisors of $20$: $\pm 1, \pm 2, \pm 4, \pm 5, \pm 10, \pm 20$.

[Logical Deduction: Notice that all coefficients in $P(x)$ are positive. Therefore, substituting any positive value for $x$ will result in a positive sum, meaning $P(x)$ cannot equal zero. Thus, we only need to test the negative divisors.]

Let us test $x = -1$:

$P(-1) = (-1)^3 + 13(-1)^2 + 32(-1) + 20$

$P(-1) = -1 + 13(1) - 32 + 20$

$P(-1) = -1 + 13 - 32 + 20$

$P(-1) = 12 - 32 + 20 = -20 + 20 = 0$

Since $P(-1) = 0$, we conclude that $(x - (-1))$ or $(x + 1)$ is a factor of $P(x)$ [Per the Factor Theorem].

Step 2: Factoring out $(x + 1)$ via Algebraic Splitting

Now that we know $(x + 1)$ is a factor, we can rewrite the terms of the original polynomial to factor out $(x + 1)$ by grouping. We split the $x^2$ and $x$ terms to match the coefficients required to pull out $(x + 1)$.

$P(x) = x^3 + 13x^2 + 32x + 20$

• Split $13x^2$ into $x^2 + 12x^2$
• Split $32x$ into $12x + 20x$

Substituting these into the polynomial:

$P(x) = x^3 + x^2 + 12x^2 + 12x + 20x + 20$

Now, group the terms in pairs:

$P(x) = (x^3 + x^2) + (12x^2 + 12x) + (20x + 20)$

Factor out the greatest common monomial from each pair:

$P(x) = x^2(x + 1) + 12x(x + 1) + 20(x + 1)$

Factor out the common binomial $(x + 1)$:

$P(x) = (x + 1)(x^2 + 12x + 20)$

Step 3: Factorising the Quadratic Quotient

We are left with a quadratic polynomial: $Q(x) = x^2 + 12x + 20$. We will factorise this by splitting the middle term.

We need to find two numbers, $a$ and $b$, such that:

• Sum: $a + b = 12$ (the coefficient of $x$)
• Product: $a \cdot b = 20$ (the constant term)

The factors of $20$ are $(1, 20)$, $(2, 10)$, and $(4, 5)$. The pair that adds up to $12$ is $10$ and $2$.

Rewrite the middle term $12x$ as $10x + 2x$:

$Q(x) = x^2 + 10x + 2x + 20$

Group the terms and factorise:

$Q(x) = x(x + 10) + 2(x + 10)$

Factor out the common binomial $(x + 10)$:

$Q(x) = (x + 2)(x + 10)$

Step 4: Final Synthesis & Geometric Interpretation

Combining the linear factor from Step 1 with the two linear factors from Step 3, we get the complete factorisation of the cubic polynomial:

$P(x) = (x + 1)(x + 2)(x + 10)$

[Geometric Justification: The roots of the polynomial are the $x$-intercepts of its graph. Setting $P(x) = 0$ yields $x = -1$, $x = -2$, and $x = -10$. The graph below accurately plots $y = x^3 + 13x^2 + 32x + 20$, demonstrating the curve crossing the $x$-axis exactly at these three calculated roots.]

Final Solution: The complete factorisation of the polynomial is $(x + 1)(x + 2)(x + 10)$.

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 $p(x) = x^3 + 3x^2 + 3x + 1$.

Initial Setup & Theoretical Foundation

We are given two polynomials:

• The dividend polynomial: $p(x) = 2x^3 + x^2 - 2x - 1$
• The divisor polynomial: $g(x) = x + 1$

To determine whether $g(x)$ is a factor of $p(x)$, we utilize the Factor Theorem. [Per the Factor Theorem, a linear polynomial $x - c$ is a factor of a polynomial $p(x)$ if and only if the polynomial evaluated at $c$ equals zero, i.e., $p(c) = 0$.]

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

First, we must find the root (or zero) of the linear divisor $g(x)$. We do this by setting the polynomial equal to zero and solving for $x$:

$g(x) = 0$
$x + 1 = 0$
$x = -1$

Thus, the value to be substituted into the dividend polynomial $p(x)$ is $c = -1$.

Step 2: Evaluating the Polynomial $p(x)$ at $x = -1$

We now substitute $x = -1$ into the polynomial $p(x)$ to find the remainder. [By the Remainder Theorem, evaluating $p(-1)$ yields the exact remainder of the division of $p(x)$ by $x + 1$.]

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

Evaluating each term sequentially based on the order of operations (exponentiation first):

• $(-1)^3 = -1 \implies 2(-1)^3 = 2(-1) = -2$
• $(-1)^2 = 1$
• $-2(-1) = 2$
• The constant term remains $-1$

Substituting these evaluated terms back into the equation:

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

Grouping the positive and negative terms to simplify the arithmetic:

$p(-1) = (-2 + 2) + (1 - 1)$
$p(-1) = 0 + 0$
$p(-1) = 0$

Step 3: Applying the Factor Theorem

The evaluation yields $p(-1) = 0$. Because the remainder is exactly zero, the condition of the Factor Theorem is perfectly satisfied. This proves that there is no remainder when $2x^3 + x^2 - 2x - 1$ is divided by $x + 1$.

Final Solution: Since $p(-1) = 0$, by the Factor Theorem, $g(x) = x + 1$ is a factor of the polynomial $p(x) = 2x^3 + x^2 - 2x - 1$.

Initial Setup & Given Polynomial

We are tasked with finding the complete linear factorization of the given cubic polynomial. Let the polynomial be defined as:

$P(x) = x^3 - 2x^2 - x + 2$

Because the degree of the polynomial is $3$ (a cubic polynomial), the Fundamental Theorem of Algebra dictates that it will have exactly $3$ roots in the complex plane, which corresponds to up to $3$ linear factors of the form $(x - r_i)$, where $r_i$ are the roots.

Step 1: Method of Grouping Terms

For polynomials with four terms, the most efficient initial strategy is factorisation by grouping. We partition the polynomial into two distinct binomial groups to identify common algebraic structures.

Group the first two terms and the last two terms:

$P(x) = (x^3 - 2x^2) - (x - 2)$

[Note: A negative sign is factored out from the third and fourth terms, which reverses the sign of the constant term inside the parenthesis, ensuring mathematical equivalence to the original expression.]

Step 2: Extracting the Greatest Common Factor (GCF)

Next, we extract the Greatest Common Factor from each individual binomial group.

• From the first group $(x^3 - 2x^2)$, the GCF is $x^2$. Factoring this out yields: $x^2(x - 2)$.
• From the second group $(x - 2)$, the GCF is simply $1$. Factoring this out yields: $1(x - 2)$.

Substituting these back into the polynomial expression:

$P(x) = x^2(x - 2) - 1(x - 2)$

Step 3: Factoring Out the Common Binomial

Observe that the binomial $(x - 2)$ is now a common factor to both terms in the expression. [Per the Distributive Property of Multiplication over Addition, $ab - cb = (a - c)b$]. We factor out $(x - 2)$:

$P(x) = (x - 2)(x^2 - 1)$

Step 4: Applying Algebraic Identities

The expression is now a product of a linear binomial and a quadratic binomial. The quadratic factor $(x^2 - 1)$ is a classic "Difference of Two Squares".

[Per the standard algebraic identity: $a^2 - b^2 = (a - b)(a + b)$]

By setting $a = x$ and $b = 1$, we can expand the quadratic term:

$x^2 - 1 = (x - 1)(x + 1)$

Substituting this complete factorization back into our equation for $P(x)$:

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

Alternative Verification: The Factor Theorem

To ensure absolute rigor, we can verify this result using the Rational Root Theorem and the Factor Theorem.

The Rational Root Theorem states that any rational root $\frac{p}{q}$ must have $p$ as a factor of the constant term ($2$) and $q$ as a factor of the leading coefficient ($1$). The possible rational roots are $\pm 1, \pm 2$.

Testing $x = 1$:

$P(1) = (1)^3 - 2(1)^2 - (1) + 2 = 1 - 2 - 1 + 2 = 0$

[Per the Factor Theorem, since $P(1) = 0$, $(x - 1)$ is a guaranteed factor of $P(x)$].

Performing polynomial long division or synthetic division of $(x^3 - 2x^2 - x + 2)$ by $(x - 1)$ yields the quotient $(x^2 - x - 2)$. Factoring this resulting quadratic equation yields $(x - 2)(x + 1)$, perfectly corroborating our grouping method.

Visual Representation: Factorization Tree

The following diagram illustrates the hierarchical breakdown of the polynomial into its constituent linear factors.

Final Solution: The complete linear factorization of the polynomial $x^3 - 2x^2 - x + 2$ is $(x - 2)(x - 1)(x + 1)$.

Initial Setup & Polynomial Identification

We are tasked with factorising the given quadratic polynomial:

$P(x) = 3x^2 - x - 4$

[Per the Fundamental Theorem of Algebra and the structure of polynomials], a quadratic polynomial in a single variable is expressed in the standard form $ax^2 + bx + c$, where $a$, $b$, and $c$ are real constants, and $a \neq 0$. By comparing our given polynomial to the standard form, we can identify the coefficients:

• Leading coefficient ($a$) = $3$
• Linear coefficient ($b$) = $-1$
• Constant term ($c$) = $-4$

Step 1: The AC Method & Coefficient Analysis

To factorise a quadratic polynomial of the form $ax^2 + bx + c$ over the integers, we employ the method of splitting the middle term (also known as the AC Method). This requires us to find two integers, let us call them $p$ and $q$, that satisfy two simultaneous conditions:

1. The product of the two integers must equal the product of the leading coefficient and the constant term: $p \times q = a \times c$.
2. The sum of the two integers must equal the linear coefficient: $p + q = b$.

Calculating the required product ($a \times c$):

$a \times c = (3) \times (-4) = -12$

The required sum ($b$) is $-1$.

Step 2: Determining the Factors

We must systematically evaluate the integer factor pairs of $-12$ to find the specific pair that sums to $-1$.

The integers that satisfy both conditions are $3$ and $-4$.

Step 3: Splitting the Middle Term

We now rewrite the original linear term ($-x$) as the sum of the two terms derived from our factors: $3x$ and $-4x$.

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

Step 4: Factorisation by Grouping

[By the Distributive Property of Multiplication over Addition], we can group the polynomial into two binomial pairs and extract the greatest common factor (GCF) from each pair.

Group the terms:

$(3x^2 + 3x) - (4x + 4)$

Extract the GCF from the first group ($3x^2 + 3x$). The GCF is $3x$:

$3x(x + 1)$

Extract the GCF from the second group ($-4x - 4$). To ensure the binomial inside the parentheses matches the first group, we extract $-4$:

$-4(x + 1)$

Substitute these back into the expression:

$3x(x + 1) - 4(x + 1)$

Notice that $(x + 1)$ is now a common binomial factor. We factor out $(x + 1)$:

$(x + 1)(3x - 4)$

Visual Representation: Area Model of Factorisation

The algebraic grouping can be geometrically verified using an area model. The total area of the rectangle represents the quadratic polynomial $3x^2 - x - 4$, while the dimensions (length and width) represent its linear factors $(3x - 4)$ and $(x + 1)$.

Final Verification

To ensure absolute rigor, we expand the factored form to verify it yields the original polynomial:

$(x + 1)(3x - 4) = x(3x - 4) + 1(3x - 4)$

$= 3x^2 - 4x + 3x - 4$

$= 3x^2 - x - 4$

The expansion perfectly matches the initial polynomial, confirming the accuracy of the factorisation.

Final Solution: The factorised form of the polynomial $3x^2 - x - 4$ is $(x + 1)(3x - 4)$.

Initial Setup & Given Polynomial

We are tasked with factorising the following cubic polynomial:

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

Because this is a polynomial of degree 3, it can have up to three linear factors. We will approach this factorization using the Method of Grouping, which is highly efficient for four-term polynomials where a proportional relationship exists between the coefficients.

Step 1: Grouping the Terms

We begin by dividing the four terms of the polynomial into two distinct pairs. This allows us to extract the Greatest Common Factor (GCF) from each pair independently.

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

[Per the Associative Property of Addition, grouping terms does not alter the value of the expression. Note that factoring out a negative sign from the last two terms changes $-2y - 1$ to $-(2y + 1)$].

Step 2: Extracting the Greatest Common Factor (GCF)

Next, we factor out the GCF from each binomial group.

• From the first group $(2y^3 + y^2)$, the GCF is $y^2$. Factoring this out yields: $y^2(2y + 1)$.
• From the second group $-(2y + 1)$, the GCF is $1$. Factoring this out yields: $-1(2y + 1)$.

Substituting these back into the expression, we obtain:

$P(y) = y^2(2y + 1) - 1(2y + 1)$

Step 3: Factoring out the Common Binomial

Observe that the binomial $(2y + 1)$ is now a common factor in both terms. We can factor this binomial out of the entire expression.

$P(y) = (2y + 1)(y^2 - 1)$

[Per the Distributive Property of Multiplication over Addition, $ab + ac = a(b+c)$, where $a = (2y + 1)$].

Step 4: Applying the Difference of Squares Identity

The expression is now a product of a linear binomial and a quadratic binomial. The quadratic factor, $(y^2 - 1)$, is a classic "Difference of Squares."

[Per the fundamental algebraic identity: $a^2 - b^2 = (a - b)(a + b)$]

Letting $a = y$ and $b = 1$, we expand the quadratic term:

$y^2 - 1 = (y - 1)(y + 1)$

Substituting this back into our main equation yields the complete prime factorization of the cubic polynomial:

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

Verification via the Factor Theorem (Analytical Check)

To ensure absolute mathematical rigor, we can verify our factors using the Factor Theorem, which states that a polynomial $P(y)$ has a factor $(y - c)$ if and only if $P(c) = 0$.

Let us test the root $y = 1$ (derived from the factor $y - 1$):

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

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

$P(1) = 2 + 1 - 2 - 1 = 0$

Because $P(1) = 0$, $(y - 1)$ is definitively a valid factor. Similar checks for $y = -1$ and $y = -1/2$ will also yield $0$, confirming that all three linear binomials are the correct and complete factors of the given cubic polynomial.

Final Solution: The complete factorization of the polynomial $2y^3 + y^2 - 2y - 1$ is $(2y + 1)(y - 1)(y + 1)$.

Initial Setup & Theoretical Foundation

We are tasked with determining whether the linear polynomial $D(x) = x + 1$ is a factor of the given cubic polynomial $P(x) = x^3 + x^2 + x + 1$.

[Per the Factor Theorem of Polynomials], for any polynomial $P(x)$ of degree $n \ge 1$ and any real number $a$, the linear binomial $(x - a)$ is a factor of $P(x)$ if and only if $P(a) = 0$. This theorem is a direct corollary of the Remainder Theorem, which states that dividing $P(x)$ by $(x - a)$ yields a remainder equal to $P(a)$. If the remainder is zero, the divisor is a perfect factor.

Step 1: Determining the Zero of the Divisor

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

$x + 1 = 0$
$x = -1$

Here, our value for $a$ is $-1$. We must now evaluate $P(-1)$.

Step 2: Evaluating the Polynomial at $x = -1$

We substitute $x = -1$ into the original polynomial $P(x) = x^3 + x^2 + x + 1$.

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

To ensure absolute precision, we break down the exponentiation of the negative integer:

Substituting these evaluated terms back into the polynomial equation:

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

$P(-1) = 0$

Step 3: Graphical Verification

Graphically, the real roots of a polynomial correspond to the $x$-intercepts of its graph. Because $P(-1) = 0$, the graph of $y = x^3 + x^2 + x + 1$ must intersect the $x$-axis exactly at the coordinate $(-1, 0)$.

Step 4: Analytical Conclusion

Because the evaluation of the polynomial at $x = -1$ yields exactly $0$, the remainder of the division $\frac{x^3 + x^2 + x + 1}{x + 1}$ is $0$. Therefore, the polynomial divides perfectly without any fractional remainder.

Final Solution: Since $P(-1) = 0$, by the Factor Theorem, $(x + 1)$ is indeed a factor of the polynomial $x^3 + x^2 + x + 1$.

Initial Setup & Theoretical Foundation

We are given the quadratic polynomial:

$p(x) = kx^2 - \sqrt{2}x + 1$

We are also given a linear binomial, which acts as a divisor:

$g(x) = x - 1$

The objective is to determine the exact value of the unknown coefficient $k$ under the strict condition that $g(x)$ is a factor of $p(x)$.

Step 1: Application of the Factor Theorem

[Per the Factor Theorem of Polynomials], a linear polynomial of the form $x - c$ is a factor of a polynomial $p(x)$ if and only if the polynomial evaluates to zero at $x = c$. Mathematically, this is expressed as:

$p(c) = 0$

By comparing our given divisor $g(x) = x - 1$ to the standard form $x - c$, we identify the root of the divisor as:

$x - 1 = 0 \implies x = 1$

Therefore, for $x - 1$ to be a factor of $p(x)$, the polynomial must satisfy the condition:

$p(1) = 0$

Step 2: Substitution and Algebraic Manipulation

We substitute $x = 1$ into the original polynomial $p(x)$ to establish our equation:

• $p(1) = k(1)^2 - \sqrt{2}(1) + 1$
• $p(1) = k(1) - \sqrt{2} + 1$
• $p(1) = k - \sqrt{2} + 1$

Equating this expression to zero [as mandated by the Factor Theorem]:

$k - \sqrt{2} + 1 = 0$

Step 3: Solving for the Unknown Constant $k$

To isolate $k$, we transpose the constant terms to the right side of the equation. When moving terms across the equals sign, their signs invert [by the properties of equality]:

• Add $\sqrt{2}$ to both sides: $k + 1 = \sqrt{2}$
• Subtract $1$ from both sides: $k = \sqrt{2} - 1$

Graphical Verification & Geometric Interpretation

Substituting $k = \sqrt{2} - 1$ back into the original equation yields the specific polynomial:

$p(x) = (\sqrt{2} - 1)x^2 - \sqrt{2}x + 1$

Geometrically, the fact that $x - 1$ is a factor means that the parabola represented by $y = p(x)$ must intersect the x-axis exactly at the coordinate $(1, 0)$. The graph below plots this precise quadratic function, visually confirming the root at $x = 1$.

Final Solution: The value of $k$ that makes $x - 1$ a factor of the polynomial $p(x) = kx^2 - \sqrt{2}x + 1$ is $k = \sqrt{2} - 1$.

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 $x^4 + 3x^3 + 3x^2 + x + 1$.

Initial Setup & Theoretical Foundation

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

$P(x) = x^3 - x^2 - (2 + \sqrt{2})x + \sqrt{2}$

[Per the Factor Theorem], a polynomial $D(x) = x - c$ is a factor of a polynomial $P(x)$ if and only if the polynomial evaluated at $c$ yields a remainder of zero, i.e., $P(c) = 0$. This is a direct corollary of the Remainder Theorem, which states that dividing $P(x)$ by $(x - c)$ leaves a remainder equal to $P(c)$.

Step 1: Determining the Root of the Divisor

To apply the Factor Theorem, we must first find the zero (or root) of the divisor $D(x) = x + 1$. We do this by setting the divisor equal to zero:

$x + 1 = 0$

$x = -1$

Thus, our test value is $c = -1$. We must evaluate $P(-1)$.

Step 2: Applying the Factor Theorem

Substitute $x = -1$ into the original polynomial $P(x)$:

$P(-1) = (-1)^3 - (-1)^2 - (2 + \sqrt{2})(-1) + \sqrt{2}$

Step 3: Algebraic Evaluation & Simplification

We will now expand and simplify each term of the expression systematically:

• First term: $(-1)^3 = -1$ [Since an odd power of a negative number remains negative]
• Second term: $-(-1)^2 = -(1) = -1$ [Since an even power of a negative number is positive, but the term is preceded by a negative sign]
• Third term: $-(2 + \sqrt{2})(-1) = +(2 + \sqrt{2}) = 2 + \sqrt{2}$ [The product of two negative signs yields a positive sign]
• Fourth term: $+\sqrt{2}$ [Remains unchanged]

Substituting these simplified terms back into the equation for $P(-1)$:

$P(-1) = -1 - 1 + 2 + \sqrt{2} + \sqrt{2}$

Combine the rational (integer) terms and the irrational terms separately:

$P(-1) = (-1 - 1 + 2) + (\sqrt{2} + \sqrt{2})$

$P(-1) = (-2 + 2) + 2\sqrt{2}$

$P(-1) = 0 + 2\sqrt{2}$

$P(-1) = 2\sqrt{2}$

Step 4: Final Analysis & Conclusion

According to the Factor Theorem, $(x + 1)$ is a factor of $P(x)$ if and only if $P(-1) = 0$. Our algebraic evaluation demonstrates that:

$P(-1) = 2\sqrt{2} \neq 0$

Because the remainder is non-zero, the polynomial $P(x)$ is not perfectly divisible by $(x + 1)$.

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

Initial Setup & Theoretical Foundation

We are given the polynomial expression:

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

We must determine whether the linear binomial $D(x) = x + 1$ is a factor of $P(x)$.

[Per the Factor Theorem, a polynomial $P(x)$ has a factor $(x - c)$ if and only if the polynomial evaluated at $c$ equals zero, i.e., $P(c) = 0$. This is a direct corollary of the Remainder Theorem, which states that the remainder of $P(x)$ divided by $(x - c)$ is $P(c)$.]

Step 1: Determining the Zero of the Divisor

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

$x + 1 = 0$

$x = -1$

Thus, our test value is $c = -1$.

Step 2: Applying the Factor Theorem (Substitution)

We substitute $x = -1$ into the original polynomial $P(x)$ to find the remainder:

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

[Theoretical Justification: When a negative number is raised to an even power, the result is positive. When raised to an odd power, the result is negative. Specifically, $(-1)^{2k} = 1$ and $(-1)^{2k+1} = -1$ for any integer $k$.]

• Fourth degree term: $(-1)^4 = 1$
• Cubic term: $(-1)^3 = -1$
• Quadratic term: $(-1)^2 = 1$
• Linear term: $(-1)^1 = -1$
• Constant term: $1$

Step 3: Evaluating the Polynomial

Now, we sum the evaluated terms sequentially:

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

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

$P(-1) = 0 + 0 + 1$

$P(-1) = 1$

Step 4: Logical Conclusion

The remainder of the division is $1$. Because $P(-1) \neq 0$, the condition required by the Factor Theorem is not satisfied. Therefore, the polynomial does not divide evenly by $(x + 1)$.

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