Initial Setup & Theoretical Foundation

We are given the linear polynomial:

$p(x) = 3x - 2$

[Per the definition of the zero of a polynomial, a real number $a$ is a zero of a polynomial $p(x)$ if and only if $p(a) = 0$. Geometrically, this corresponds to the x-coordinate of the point where the graph of the polynomial intersects the x-axis.]

Step 1: Equating the Polynomial to Zero

To find the zero of the given polynomial, we must set the polynomial expression equal to zero. This transforms our polynomial into a linear equation.

$p(x) = 0$

$3x - 2 = 0$

Step 2: Algebraic Isolation of the Variable

We now solve for $x$ using standard algebraic manipulation [By the properties of equality].

• Addition Property of Equality: Add $2$ to both sides of the equation to isolate the term containing $x$.
  $3x - 2 + 2 = 0 + 2$
  $3x = 2$
• Division Property of Equality: Divide both sides by $3$ to solve for $x$.
  $\frac{3x}{3} = \frac{2}{3}$
  $x = \frac{2}{3}$

Step 3: Verification of the Zero

To ensure absolute mathematical rigor, we substitute $x = \frac{2}{3}$ back into the original polynomial to verify that it evaluates to zero.

$p\left(\frac{2}{3}\right) = 3\left(\frac{2}{3}\right) - 2$

$p\left(\frac{2}{3}\right) = 2 - 2$

$p\left(\frac{2}{3}\right) = 0$

[Since the evaluation yields exactly zero, the calculated root is verified as correct.]

Graphical Representation of the Zero

The graph of the linear polynomial $y = 3x - 2$ is a straight line. The zero of the polynomial is the exact point where this line crosses the x-axis (where $y = 0$). As calculated, this intersection occurs at the coordinate $\left(\frac{2}{3}, 0\right)$.

Final Solution: The zero of the polynomial $p(x) = 3x - 2$ is $x = \frac{2}{3}$.

Initial Setup & Given Expression

We are given the following algebraic expression:

$P(r) = r^2$

Our objective is to classify this polynomial as linear, quadratic, or cubic based on its mathematical properties.

Step 1: Theoretical Foundation of Polynomial Classification

In algebra, polynomials are classified according to their degree. The degree of a polynomial in one variable is defined as the highest power (exponent) of the variable present in the expression with a non-zero coefficient. [Per the Fundamental Theorem of Algebra and standard polynomial definitions].

The standard classifications based on degree are as follows:

• Linear Polynomial: A polynomial of degree $1$. General form: $ax + b$ (where $a \neq 0$).
• Quadratic Polynomial: A polynomial of degree $2$. General form: $ax^2 + bx + c$ (where $a \neq 0$).
• Cubic Polynomial: A polynomial of degree $3$. General form: $ax^3 + bx^2 + cx + d$ (where $a \neq 0$).

Step 2: Analyzing the Degree of the Given Polynomial

Let us examine the given polynomial:

$P(r) = r^2$

This expression is a monomial (a polynomial consisting of a single term). We analyze the components of this term:

• The variable is $r$.
• The coefficient is $1$ (since $r^2$ is equivalent to $1 \cdot r^2$).
• The exponent of the variable $r$ is $2$.

Because there are no other terms in the polynomial, the highest power of the variable $r$ is exactly $2$. Therefore, the degree of the polynomial $P(r) = r^2$ is $2$.

Step 3: Visualizing the Quadratic Nature

To rigorously confirm the nature of this polynomial, we can observe its graphical representation. A polynomial of degree $2$ forms a parabola when plotted on a Cartesian coordinate system. The precise geometric mapping of $P(r) = r^2$ demonstrates a non-linear, symmetric curve with a single vertex at the origin $(0,0)$.

Step 4: Final Classification

Because the highest exponent of the variable $r$ is $2$, the polynomial satisfies the strict definition of a quadratic polynomial. It does not possess a degree of $1$ (which would make it linear) nor a degree of $3$ (which would make it cubic).

Final Solution: The polynomial $r^2$ is a quadratic polynomial.

Initial Setup & Algebraic Analysis

We are tasked with factorising the following binomial expression:

$27y^3 + 125z^3$

To factorise this polynomial, we must first analyze the coefficients and the degrees of the variables to identify any underlying algebraic structures. We observe that both terms are perfect cubes.

Step 1: Expressing Terms as Perfect Cubes

We determine the cube roots of the numerical coefficients and the variables:

• For the first term: The prime factorization of $27$ is $3 \times 3 \times 3 = 3^3$. Therefore, $27y^3$ can be rewritten as $(3y)^3$ [Applying the power of a product rule: $x^n y^n = (xy)^n$].
• For the second term: The prime factorization of $125$ is $5 \times 5 \times 5 = 5^3$. Therefore, $125z^3$ can be rewritten as $(5z)^3$.

Rewriting the original expression, we get:

$(3y)^3 + (5z)^3$

Step 2: Stating the Relevant Algebraic Identity

The expression is now explicitly in the form of the sum of two cubes, $a^3 + b^3$. [Per the fundamental algebraic identity for the sum of cubes, derived from the expansion of $(a+b)^3 - 3ab(a+b)$], we know that:

$a^3 + b^3 = (a + b)(a^2 - ab + b^2)$

Step 3: Substitution and Expansion

By mapping our terms to the identity, we set $a = 3y$ and $b = 5z$. Substituting these into the right-hand side of the identity yields:

$(3y)^3 + (5z)^3 = (3y + 5z) \left[ (3y)^2 - (3y)(5z) + (5z)^2 \right]$

Step 4: Simplification of the Quadratic Factor

We must now rigorously simplify each term inside the second set of parentheses (the quadratic trinomial factor):

• Square of the first term: $(3y)^2 = 3^2 \cdot y^2 = 9y^2$
• Product of the two terms: $(3y)(5z) = (3 \cdot 5)(y \cdot z) = 15yz$
• Square of the second term: $(5z)^2 = 5^2 \cdot z^2 = 25z^2$

Replacing the unsimplified terms in our expanded equation with these calculated values, we obtain the final factorised expression:

$(3y + 5z)(9y^2 - 15yz + 25z^2)$

Final Solution: The completely factorised form of $27y^3 + 125z^3$ is $(3y + 5z)(9y^2 - 15yz + 25z^2)$.

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 given the linear polynomial:

$p(x) = ax$

with the strict mathematical constraint that $a \neq 0$.

Step 1: Formulating the Equation for the Zero

The "zero" (or root) of a polynomial is defined as the specific value of the independent variable $x$ that evaluates the polynomial to zero. [By the Fundamental Theorem of Algebra and the definition of polynomial roots].

To find this value, we must set the polynomial expression equal to zero:

$p(x) = 0$

$ax = 0$

Step 2: Algebraic Isolation of the Variable

To isolate $x$, we must divide both sides of the equation by the coefficient $a$. We are mathematically permitted to perform this operation strictly because the problem explicitly states the condition $a \neq 0$. [Per the Division Property of Equality, division by any non-zero real number is defined and preserves the equality].

$x = \frac{0}{a}$

Since zero divided by any non-zero number is strictly zero [Per the Zero Property of Division]:

$x = 0$

Step 3: Verification of the Root

To ensure absolute analytical accuracy, we substitute our derived root, $x = 0$, back into the original polynomial function:

$p(0) = a(0)$

$p(0) = 0$

Because the polynomial evaluates to $0$, the value $x = 0$ is rigorously verified as the correct zero of the polynomial.

Graphical Representation & Geometric Interpretation

Geometrically, the zero of a polynomial represents the exact x-coordinate where the graph of the function intersects the x-axis (where $p(x) = 0$). For any non-zero real value of $a$, the linear equation $p(x) = ax$ represents a straight line passing directly through the origin $(0,0)$. The steepness and direction of the line depend on $a$, but the x-intercept remains invariant at the origin.

Final Solution: The zero of the polynomial $p(x) = ax$ (where $a \neq 0$) is $x = 0$.

Initial Setup & Objective

We are tasked with verifying the fundamental algebraic identity for the difference of two cubes. The identity is given as:

$x^3 - y^3 = (x - y)(x^2 + xy + y^2)$

To verify this rigorously, we will expand the Right-Hand Side (RHS) and demonstrate that it simplifies exactly to the Left-Hand Side (LHS). We will also provide a secondary derivation using the binomial cube expansion to establish deep theoretical consistency.

Method 1: Direct Algebraic Expansion

Step 1: Applying the Distributive Property

We begin with the expression on the Right-Hand Side (RHS):

$RHS = (x - y)(x^2 + xy + y^2)$

[Per the Distributive Law of Multiplication over Addition, $A(B + C + D) = AB + AC + AD$]. We must distribute both terms of the binomial $(x)$ and $(-y)$ across the entire trinomial:

$RHS = x(x^2 + xy + y^2) - y(x^2 + xy + y^2)$

Step 2: Term-by-Term Multiplication

Next, we distribute $x$ into the first trinomial and $-y$ into the second trinomial. [By the laws of exponents, $x^a \cdot x^b = x^{a+b}$]:

• Multiplying by $x$:
  $x \cdot x^2 = x^3$
  $x \cdot xy = x^2y$
  $x \cdot y^2 = xy^2$
• Multiplying by $-y$:
  $-y \cdot x^2 = -x^2y$ [Note: Commutativity allows us to write $-yx^2$ as $-x^2y$]
  $-y \cdot xy = -xy^2$
  $-y \cdot y^2 = -y^3$

Substituting these expanded terms back into our equation yields:

$RHS = (x^3 + x^2y + xy^2) + (-x^2y - xy^2 - y^3)$

Step 3: Aggregation and Simplification

We now drop the parentheses and group the like terms together to identify additive inverses (terms that sum to zero):

$RHS = x^3 + x^2y - x^2y + xy^2 - xy^2 - y^3$

Rearranging the terms for absolute clarity:

$RHS = x^3 + (x^2y - x^2y) + (xy^2 - xy^2) - y^3$

The terms $x^2y$ and $-x^2y$ cancel each other out. Similarly, $xy^2$ and $-xy^2$ cancel each other out:

$RHS = x^3 + 0 + 0 - y^3$

$RHS = x^3 - y^3$

Since the simplified RHS perfectly matches the LHS, the identity is verified.

Visualizing the Expansion (Area Model Matrix)

The following matrix visually maps the distribution of $(x - y)$ across $(x^2 + xy + y^2)$. The color-coded cells demonstrate how the intermediate terms cancel each other out, leaving only the cubic terms.

Method 2: Derivation from the Binomial Cube Identity

To provide exhaustive proof, we can also derive this factorization directly from the standard identity for the cube of a binomial:

$ (x - y)^3 = x^3 - y^3 - 3xy(x - y) $

Step 1: Isolate the Difference of Cubes

We want to solve for $x^3 - y^3$. By adding $3xy(x - y)$ to both sides of the equation, we get:

$ x^3 - y^3 = (x - y)^3 + 3xy(x - y) $

Step 2: Factor out the Common Binomial

Observe that both terms on the right side share a common factor of $(x - y)$. We extract this factor:

$ x^3 - y^3 = (x - y) \left[ (x - y)^2 + 3xy \right] $

Step 3: Expand and Simplify the Inner Bracket

We expand the squared binomial $(x - y)^2 = x^2 - 2xy + y^2$ inside the brackets:

$ x^3 - y^3 = (x - y) \left[ (x^2 - 2xy + y^2) + 3xy \right] $

Combine the like terms ($-2xy$ and $+3xy$):

$ -2xy + 3xy = +xy $

Substituting this back into the bracket yields the final verified identity:

$ x^3 - y^3 = (x - y)(x^2 + xy + y^2) $

Final Solution: Through both direct algebraic expansion and derivation from the binomial cube identity, it is rigorously proven that $x^3 - y^3 = (x - y)(x^2 + xy + y^2)$. The intermediate terms $x^2y$ and $xy^2$ act as additive inverses and cancel out perfectly.

Initial Setup & Given Polynomial

We are tasked with factorising the following algebraic expression:

$P(a) = 27 - 125a^3 - 135a + 225a^2$

Step 1: Identifying the Relevant Algebraic Identity

By analyzing the polynomial, we observe the presence of perfect cubes ($27$ and $125a^3$) and alternating signs. This structural pattern strongly indicates the application of the standard cubic identity for the difference of two terms [Per the Binomial Theorem for exponent 3]:

$(x - y)^3 = x^3 - y^3 - 3x^2y + 3xy^2$

To utilize this identity, we must map the terms of our given polynomial to the terms of the expansion.

Step 2: Extracting the Base Terms (Cube Roots)

We isolate the perfect cube terms to determine our candidate values for $x$ and $y$.

• First term (Constant): The term $27$ is a perfect cube.
  $27 = (3)^3 \implies x = 3$
• Second term (Cubic): The term $125a^3$ is a perfect cube.
  $125a^3 = (5a)^3 \implies y = 5a$

Step 3: Verifying the Cross Terms

To rigorously confirm that the polynomial fits the identity $(x - y)^3$, we must evaluate the intermediate cross terms $-3x^2y$ and $+3xy^2$ using our candidate values $x = 3$ and $y = 5a$.

• Evaluating $-3x^2y$:
  $-3(3)^2(5a) = -3(9)(5a) = -27(5a) = -135a$
  [This perfectly matches the third term of the given polynomial].
• Evaluating $+3xy^2$:
  $+3(3)(5a)^2 = 9(25a^2) = 225a^2$
  [This perfectly matches the fourth term of the given polynomial].

Visual Mapping of the Algebraic Identity

The following diagram illustrates the exact one-to-one mapping between the given polynomial terms and the components of the cubic identity.

Step 4: Synthesizing the Factorised Form

Since all terms perfectly align with the expansion of $(x - y)^3$, we can rewrite the original polynomial in its expanded identity form:

$27 - 125a^3 - 135a + 225a^2 = (3)^3 - (5a)^3 - 3(3)^2(5a) + 3(3)(5a)^2$

Applying the identity $(x - y)^3$, where $x = 3$ and $y = 5a$, we condense the expression into a single perfect cube:

$= (3 - 5a)^3$

To express this as a product of its linear factors, we write the binomial three times.

Final Solution: The completely factorised form of the polynomial is $(3 - 5a)(3 - 5a)(3 - 5a)$.

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 & Expression Analysis

We are tasked with factorising the following algebraic polynomial:

$9x^2 + 6xy + y^2$

To factorise this expression efficiently, we must analyze its structural properties. The polynomial consists of three terms (a trinomial). Our primary objective is to determine if it conforms to the structure of a perfect square trinomial, which can be factorised using standard algebraic identities.

Step 1: Decomposing the Terms into Perfect Squares

We begin by examining the first and third terms to see if they can be expressed as perfect squares [Per the fundamental properties of exponents].

• First Term: The term $9x^2$ consists of a constant $9$ and a variable $x^2$. Since $3^2 = 9$, we can rewrite this term as the square of a single monomial:
  $9x^2 = (3x)^2$
• Third Term: The term $y^2$ is already a perfect square and can be explicitly written as:
  $y^2 = (y)^2$

Step 2: Verifying the Middle Term

For a trinomial to be classified as a perfect square trinomial, the middle term must be exactly twice the product of the bases of the squared terms identified in Step 1.

Let us define our base variables based on the standard identity structure:

• Let $a = 3x$
• Let $b = y$

Now, we calculate the theoretical middle term, $2ab$:

$2ab = 2 \cdot (3x) \cdot (y)$

$2ab = 6xy$

This calculated product perfectly matches the middle term of our given polynomial ($6xy$). Therefore, the expression is confirmed to be a perfect square trinomial.

Step 3: Applying the Standard Algebraic Identity

Having verified the structure, we apply the First Algebraic Identity for the square of a binomial [By the binomial expansion theorem]:

$a^2 + 2ab + b^2 = (a + b)^2$

Substituting our specific values ($a = 3x$ and $b = y$) into the identity:

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

Expanding the squared binomial into its linear factors yields:

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

Geometric Proof & Visualization

The algebraic identity $(a+b)^2 = a^2 + 2ab + b^2$ can be rigorously proven using area models in Euclidean geometry. Below is the geometric representation of our specific polynomial, where the total area of a square with side length $(3x + y)$ is equal to the sum of the areas of its four internal rectangular regions.

As demonstrated by the geometric model, the total area is the sum of the individual regions: $9x^2 + 3xy + 3xy + y^2 = 9x^2 + 6xy + y^2$, which perfectly corresponds to a square of side $(3x + y)$.

Final Solution: The completely factorised form of the polynomial $9x^2 + 6xy + y^2$ is $(3x + y)(3x + y)$, which is most concisely written as $(3x + y)^2$.

Given Expression & Initial Setup

We are tasked with analyzing the following algebraic expression to determine if it is a polynomial in one variable:

$P(x, y, t) = x^{10} + y^3 + t^{50}$

To rigorously evaluate this, we must test the expression against two distinct mathematical criteria:

• Criterion 1 (Polynomial Status): An algebraic expression is classified as a polynomial if and only if the exponents of all its variables are non-negative integers (whole numbers).
• Criterion 2 (Single Variable Status): A polynomial is "in one variable" if it is constructed using exactly one distinct variable symbol throughout the entire expression.

Step 1: Analyzing the Exponents (Checking for Polynomial Status)

We decompose the expression into its individual terms and inspect the exponent of each variable:

[Per the fundamental definition of polynomials, since all exponents ($10$, $3$, and $50$) belong to the set of whole numbers $\mathbb{W} = \{0, 1, 2, 3, \dots\}$, the given expression is mathematically verified to be a valid polynomial.]

Step 2: Analyzing the Variables (Checking for "One Variable" Status)

Next, we must determine the dimensionality of the polynomial by counting the number of unique variables present.

• The first term contains the variable $x$.
• The second term contains the variable $y$.
• The third term contains the variable $t$.

[By the axioms of algebraic notation, $x$, $y$, and $t$ represent three distinct, independent variables. Therefore, the expression is a multivariate polynomial, specifically a polynomial in three variables.]

Step 3: Synthesizing the Final Conclusion

While the expression successfully passes the test for being a polynomial (all exponents are non-negative integers), it fails the condition of being in "one variable." The presence of $x$, $y$, and $t$ makes it a polynomial in three variables.

Final Solution: The expression $x^{10} + y^3 + t^{50}$ is NOT a polynomial in one variable. While it is a valid polynomial (because all exponents are whole numbers), it is a polynomial in THREE variables ($x$, $y$, and $t$).

Given Polynomial & Initial Setup

We are given the following algebraic expression:

$P(t) = 3t$

Our objective is to classify this polynomial into one of three distinct algebraic categories: linear, quadratic, or cubic. The classification of a polynomial is strictly determined by its degree.

Step 1: Identify the Variable and its Exponents

In the given polynomial $P(t) = 3t$, the single independent variable present is $t$. To determine the nature of the polynomial, we must inspect the exponents attached to this variable.

The term $3t$ can be explicitly rewritten by revealing its hidden exponent:

$P(t) = 3t^1$

[Per the Fundamental Laws of Exponents, any variable $x$ written without a visible power is mathematically understood to have an exponent of $1$, i.e., $x = x^1$].

Step 2: Determine the Degree of the Polynomial

The degree of a polynomial in one variable is defined as the highest power (exponent) of the variable in that expression.

• The only term containing the variable is $3t^1$.
• The exponent of $t$ in this term is $1$.

Therefore, the highest power of the variable $t$ is $1$. This establishes that the degree of the polynomial $P(t) = 3t$ is exactly $1$.

Step 3: Apply Polynomial Classification Criteria

Polynomials are universally classified based on their degree [By the Fundamental Theorem of Algebra and standard algebraic nomenclature]. The classification matrix is as follows:

Because the degree of $3t$ is $1$, it perfectly satisfies the condition for a linear polynomial. It matches the standard linear form $at + b$ where the leading coefficient $a = 3$ and the constant term $b = 0$.

Step 4: Geometric Verification (Visualizing Linearity)

A defining characteristic of a linear polynomial is that its graph forms a perfectly straight line [Per Cartesian Coordinate Geometry]. Below is the precise geometric representation of $P(t) = 3t$, demonstrating a constant rate of change (slope $m = 3$).

Final Solution: The given polynomial $3t$ has a highest exponent (degree) of $1$. Therefore, it is classified as a linear polynomial.

Initial Setup & Given Expression

We are given the algebraic expression:

$P(x) = 1 + x$

Our objective is to classify this polynomial into one of three categories: linear, quadratic, or cubic. This classification is strictly determined by the degree of the polynomial.

Step 1: Standardizing the Polynomial

First, we rewrite the polynomial in its standard form, where the terms are ordered in descending powers of the variable $x$.

$P(x) = x + 1$

We can explicitly write the exponents for every term to ensure absolute clarity:

• The term $x$ is equivalent to $x^1$.
• The constant term $1$ is equivalent to $1 \cdot x^0$ [Per the Zero Exponent Rule, where $x^0 = 1$ for $x \neq 0$].

Thus, the fully expanded standard form is:

$P(x) = 1 \cdot x^1 + 1 \cdot x^0$

Step 2: Determining the Degree of the Polynomial

The degree of a polynomial in one variable is defined as the highest exponent (power) of the variable present in the expression with a non-zero coefficient.

Analyzing the exponents in $P(x) = x^1 + 1 \cdot x^0$:

• The exponent of the first term is $1$.
• The exponent of the second term is $0$.

Comparing these values, the maximum exponent is $1$. Therefore, the degree of the polynomial $P(x) = 1 + x$ is exactly $1$.

Step 3: Applying the Classification Theorem

Polynomials are classified by their degree according to the following universally accepted algebraic definitions:

Since we have rigorously determined that the degree of $1 + x$ is $1$, it maps directly to the definition of a linear polynomial.

Visual Verification: Geometric Interpretation

A linear polynomial, when graphed on a Cartesian coordinate system as a function $y = P(x)$, will always produce a perfectly straight line [Per the geometric definition of first-degree equations]. Below is the precise graphical representation of $y = x + 1$, demonstrating its constant slope ($m = 1$) and y-intercept ($c = 1$).

Final Solution: The polynomial $1 + x$ has a highest degree of $1$. Therefore, it is classified as a linear polynomial.

Step 1: Given Variables & Initial Setup

We are tasked with evaluating the sum of three cubes without performing direct cubing operations. The given mathematical expression is:

$(-12)^3 + (7)^3 + (5)^3$

Let us define the base of each cubic term as a distinct variable:

• $a = -12$
• $b = 7$
• $c = 5$

Step 2: Evaluating the Sum of the Base Variables

Before attempting to expand the expression, we must analyze the linear sum of the base variables. This is a critical diagnostic step in polynomial algebra to determine if a conditional identity applies.

$a + b + c = (-12) + 7 + 5$

$a + b + c = -12 + 12 = 0$

[Geometrically, this represents a closed vector loop in one dimension, where the displacement from the origin returns exactly to zero. See the precise vector visualization below.]

Step 3: Theoretical Justification & Algebraic Identity

We rely on the fundamental algebraic identity for the sum of three cubes:

$a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$

[Per the Zero Product Property], if the first factor on the right-hand side, $(a + b + c)$, is equal to zero, the entire right-hand side of the equation evaluates to zero, regardless of the values of the quadratic polynomial factor. Therefore:

$a^3 + b^3 + c^3 - 3abc = (0) \times (a^2 + b^2 + c^2 - ab - bc - ca)$

$a^3 + b^3 + c^3 - 3abc = 0$

Transposing $-3abc$ to the right side yields the conditional identity:

If $a + b + c = 0$, then $a^3 + b^3 + c^3 = 3abc$.

Step 4: Application of the Conditional Identity

Since we have rigorously proven in Step 2 that $(-12) + 7 + 5 = 0$, we can directly substitute our variables into the conditional identity:

$(-12)^3 + (7)^3 + (5)^3 = 3 \cdot (-12) \cdot (7) \cdot (5)$

Step 5: Final Arithmetic Calculation

We now perform the sequential multiplication of the terms on the right-hand side:

• First, multiply the constants: $3 \times (-12) = -36$
• Next, multiply the remaining terms: $7 \times 5 = 35$
• Finally, compute the product of the two results: $(-36) \times 35$

To calculate $(-36) \times 35$ mentally or systematically:

$(-36) \times (30 + 5) = (-36 \times 30) + (-36 \times 5)$

$= -1080 - 180 = -1260$

Final Solution: The value of $(-12)^3 + (7)^3 + (5)^3$ is $-1260$.

Initial Setup & Polynomial Analysis

We are given the polynomial function defined as:

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

Before evaluating the polynomial at specific discrete points, we can analyze its structure. The expression is factored into two binomials. [Per the algebraic identity for the difference of two squares: $(a - b)(a + b) = a^2 - b^2$], we can expand the polynomial into its standard quadratic form:

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

This represents a parabola opening upwards with its vertex at $(0, -1)$. We will evaluate the polynomial using both the factored form and the expanded form to ensure absolute mathematical rigor.

Step 1: Evaluating $p(0)$

To find the value of the polynomial when the independent variable $x$ is exactly $0$, we substitute $0$ into the function.

• Using the factored form:
  $p(0) = (0 - 1)(0 + 1)$
  $p(0) = (-1)(1)$
  $p(0) = -1$
• Using the expanded form [Verification]:
  $p(0) = (0)^2 - 1 = 0 - 1 = -1$

Geometrically, this represents the y-intercept of the polynomial curve at the coordinate $(0, -1)$.

Step 2: Evaluating $p(1)$

Next, we substitute $x = 1$ into the polynomial.

• Using the factored form:
  $p(1) = (1 - 1)(1 + 1)$
  $p(1) = (0)(2)$
  $p(1) = 0$
• Using the expanded form [Verification]:
  $p(1) = (1)^2 - 1 = 1 - 1 = 0$

Because $p(1) = 0$, $x = 1$ is a root (or zero) of the polynomial. [By the Factor Theorem, since $(x - 1)$ is a factor of $p(x)$, $p(1)$ must equal zero].

Step 3: Evaluating $p(2)$

Finally, we substitute $x = 2$ into the polynomial.

• Using the factored form:
  $p(2) = (2 - 1)(2 + 1)$
  $p(2) = (1)(3)$
  $p(2) = 3$
• Using the expanded form [Verification]:
  $p(2) = (2)^2 - 1 = 4 - 1 = 3$

This gives us the coordinate point $(2, 3)$ on the Cartesian plane.

Graphical Representation of the Polynomial

Below is the precise geometric plot of the quadratic function $p(x) = x^2 - 1$. The specific points we evaluated—$p(0)$, $p(1)$, and $p(2)$—are explicitly mapped to demonstrate the relationship between algebraic evaluation and spatial coordinates.

Final Solution: The evaluated values for the polynomial $p(x) = (x - 1)(x + 1)$ are $p(0) = -1$, $p(1) = 0$, and $p(2) = 3$.

Step 1: Initial Expression Analysis

We are given the quadratic trinomial:

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

Our objective is to factorise this expression by identifying it as a perfect square. A quadratic trinomial takes the general form $ax^2 + bx + c$. We must analyze the first and last terms to determine if they are perfect squares.

Step 2: Identifying the Appropriate Algebraic Identity

We observe that the signs of the terms alternate ($+, -, +$). This suggests the use of the standard binomial square identity for subtraction [Per the fundamental axioms of polynomial expansion]:

$(a - b)^2 = a^2 - 2ab + b^2$

Step 3: Term-by-Term Transformation & Verification

We must map the terms of our given polynomial to the components of the identity $a^2 - 2ab + b^2$.

• First Term ($a^2$): The term $4y^2$ can be rewritten as the square of a single monomial.
  $4y^2 = (2y)^2 \implies a = 2y$
• Third Term ($b^2$): The constant $1$ is a perfect square.
  $1 = (1)^2 \implies b = 1$
• Middle Term ($-2ab$): We must verify that the middle term of our polynomial matches $-2ab$ using our derived values for $a$ and $b$.
  $-2ab = -2(2y)(1) = -4y$

Since the middle term perfectly matches the given expression ($-4y$), the polynomial is confirmed to be a perfect square trinomial.

Step 4: Applying the Identity

By substituting $a = 2y$ and $b = 1$ back into the structural framework of the identity, we can rewrite the expanded polynomial as a squared binomial:

$4y^2 - 4y + 1 = (2y)^2 - 2(2y)(1) + (1)^2$

$4y^2 - 4y + 1 = (2y - 1)^2$

Step 5: Geometric Verification (Area Model)

To rigorously prove this factorization, we can use an area model. Consider a large square with a total side length of $2y$. Its total area is $(2y)^2 = 4y^2$. If we partition this square into segments of length $(2y - 1)$ and $1$, we can geometrically derive the identity.

From the geometric model, the total area of the square is the sum of its four internal regions:

$Area_{Total} = (2y - 1)^2 + (2y - 1) + (2y - 1) + 1$

$4y^2 = (2y - 1)^2 + 4y - 2 + 1$

$4y^2 = (2y - 1)^2 + 4y - 1$

Isolating the primary region $(2y - 1)^2$ yields:

$(2y - 1)^2 = 4y^2 - 4y + 1$

[This confirms our algebraic factorization is geometrically absolute].

Step 6: Final Factorisation

To express the polynomial fully in its factorised form, we write the squared binomial as the product of two identical linear binomials.

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

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

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

Given Variables & Initial Setup

We are given a linear polynomial in one variable:

$p(x) = cx + d$

Where the given conditions are:

• $c$ and $d$ are real numbers ($c, d \in \mathbb{R}$).
• $c \neq 0$ (This ensures the polynomial is strictly of degree 1, i.e., a linear polynomial).

Step 1: Applying the Definition of a Zero of a Polynomial

The "zero" (or root) of a polynomial is defined as the specific value of the variable $x$ for which the value of the entire polynomial becomes zero. [Per the Fundamental Theorem of Algebra and polynomial root definitions].

Therefore, to find the zero of $p(x)$, we must set the polynomial equal to zero:

$p(x) = 0$

Step 2: Setting up the Equation

Substitute the given expression for $p(x)$ into the equation:

$cx + d = 0$

Step 3: Algebraic Manipulation to Isolate $x$

To solve for $x$, we perform inverse operations to isolate the variable on one side of the equation.

First, subtract $d$ from both sides of the equation [By the Subtraction Property of Equality]:

$cx + d - d = 0 - d$

$cx = -d$

Next, divide both sides by $c$ [By the Division Property of Equality]. We are mathematically permitted to divide by $c$ because the initial problem explicitly stated the condition $c \neq 0$, thereby avoiding the undefined operation of division by zero:

$\frac{cx}{c} = \frac{-d}{c}$

$x = -\frac{d}{c}$

Graphical Interpretation (Visualizing the Zero)

Geometrically, the polynomial $p(x) = cx + d$ represents a straight line on the Cartesian coordinate plane. The "zero" of the polynomial corresponds to the $x$-intercept of this line—the exact point where the graph crosses the $x$-axis (where $y = 0$).

As demonstrated in the coordinate geometry above, the line intersects the horizontal axis precisely at the coordinate $(-\frac{d}{c}, 0)$.

Verification

To verify the accuracy of our derived zero, we substitute $x = -\frac{d}{c}$ back into the original polynomial $p(x)$:

$p\left(-\frac{d}{c}\right) = c\left(-\frac{d}{c}\right) + d$

$p\left(-\frac{d}{c}\right) = -d + d$

$p\left(-\frac{d}{c}\right) = 0$

Since the polynomial evaluates to $0$, the root is mathematically verified.

Final Solution: The zero of the polynomial $p(x) = cx + d$ is $x = -\frac{d}{c}$.

Given Expression & Algebraic Identity

We are tasked with expanding the following trinomial squared:

$(3a - 7b - c)^2$

To expand this expression systematically, we utilize the standard algebraic identity for the square of a trinomial [Derived from the distributive property of multiplication over addition]:

$(x + y + z)^2 = x^2 + y^2 + z^2 + 2xy + 2yz + 2zx$

Step 1: Variable Mapping

We must map the terms of our specific expression to the variables in the standard identity. By rewriting the given expression as a sum, we ensure that the negative signs are correctly associated with their respective terms:

$(3a - 7b - c)^2 = (3a + (-7b) + (-c))^2$

By direct comparison with $(x + y + z)^2$, we establish the following variable assignments:

• $x = 3a$
• $y = -7b$
• $z = -c$

Step 2: Substitution into the Identity

Substituting these mapped values into the expanded form of the identity yields:

$(3a + (-7b) + (-c))^2 = (3a)^2 + (-7b)^2 + (-c)^2 + 2(3a)(-7b) + 2(-7b)(-c) + 2(-c)(3a)$

Step 3: Term-by-Term Simplification

We now evaluate each term individually, applying the rules of exponents [$(ab)^n = a^n b^n$] and the rules of sign multiplication:

• Squaring the first term: $(3a)^2 = 3^2 \cdot a^2 = 9a^2$
• Squaring the second term: $(-7b)^2 = (-7)^2 \cdot b^2 = 49b^2$ [Note: The square of a negative number is positive]
• Squaring the third term: $(-c)^2 = (-1)^2 \cdot c^2 = c^2$
• First cross-product: $2(3a)(-7b) = 2 \cdot 3 \cdot (-7) \cdot a \cdot b = -42ab$
• Second cross-product: $2(-7b)(-c) = 2 \cdot (-7) \cdot (-1) \cdot b \cdot c = 14bc$ [Note: The product of two negative numbers is positive]
• Third cross-product: $2(-c)(3a) = 2 \cdot (-1) \cdot 3 \cdot c \cdot a = -6ca$

Step 4: Assembling the Final Polynomial

Combining all the simplified terms from Step 3, we construct the final expanded polynomial:

$9a^2 + 49b^2 + c^2 - 42ab + 14bc - 6ca$

Geometric Representation of the Trinomial Square Identity

The algebraic identity $(x+y+z)^2$ can be visualized as the area of a square with side length $(x+y+z)$, partitioned into 9 distinct rectangular regions. The sum of the areas of these regions corresponds exactly to the terms in our expanded formula.

Final Solution: $9a^2 + 49b^2 + c^2 - 42ab + 14bc - 6ca$

Step 1: Initial Setup & Identification of the Algebraic Identity

We are tasked with finding the product of the binomials $(3 - 2x)$ and $(3 + 2x)$.

By analyzing the structure of the expression, we observe that it consists of the product of the difference and the sum of the exact same two terms. This perfectly matches the standard algebraic identity for the Difference of Squares.

The Difference of Squares identity is defined as:

$(a - b)(a + b) = a^2 - b^2$

[Theoretical Justification: This identity is derived via the distributive property (FOIL method), where $(a - b)(a + b) = a^2 + ab - ab - b^2$. The middle terms $+ab$ and $-ab$ cancel each other out, leaving $a^2 - b^2$.]

Step 2: Variable Mapping

To apply the identity, we map the terms from our specific expression $(3 - 2x)(3 + 2x)$ to the general variables $a$ and $b$ in the identity:

• Let $a = 3$
• Let $b = 2x$

Step 3: Application of the Identity

Substituting the mapped variables into the right-hand side of the identity $(a^2 - b^2)$, we construct the following equation:

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

Step 4: Algebraic Simplification

Now, we evaluate the squares for both terms independently:

• First term: $(3)^2 = 3 \times 3 = 9$
• Second term: $(2x)^2 = (2)^2 \cdot (x)^2 = 4x^2$
  [Per the Power of a Product Rule in exponentiation: $(xy)^n = x^n y^n$]

Substituting these evaluated squares back into our equation yields:

$(3)^2 - (2x)^2 = 9 - 4x^2$

Geometric Representation of the Difference of Squares

The algebraic identity $(a - b)(a + b) = a^2 - b^2$ can be visualized geometrically. If we take a large square of area $a^2$ and remove a smaller square of area $b^2$, the remaining area can be rearranged into a rectangle with dimensions $(a - b)$ and $(a + b)$.

Final Solution: The product of $(3 - 2x)(3 + 2x)$ is $9 - 4x^2$.