Step 1: Initial Setup & Algebraic Transformation

The given algebraic expression is:

$y + \frac{2}{y}$

To rigorously analyze whether this expression is a polynomial, we must first express all terms such that the variable appears in the numerator. [Per the Laws of Exponents, specifically the negative exponent rule $\frac{1}{a^n} = a^{-n}$], we rewrite the second term of the expression:

$y^1 + 2y^{-1}$

Step 2: Variable Analysis

The problem requires us to verify two distinct conditions: whether the expression is in one variable, and whether it is a polynomial.

Looking at the transformed expression $y^1 + 2y^{-1}$, the only alphabetical symbol representing an unknown quantity is $y$. There are no other variables (such as $x$, $z$, etc.) present. Therefore, the expression strictly satisfies the condition of being in one variable.

Step 3: Theoretical Definition of a Polynomial

By mathematical definition, an algebraic expression is classified as a polynomial if and only if it can be written in the form:

$P(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0$

Where:

• $a_n, a_{n-1}, \dots, a_0$ are real number coefficients.
• The exponents $n, n-1, \dots$ must belong strictly to the set of non-negative integers (i.e., the set of whole numbers, $\mathbb{W} = \{0, 1, 2, 3, \dots\}$).

Step 4: Exponent Analysis

We now evaluate the exponents of the variable $y$ in our transformed expression $y^1 + 2y^{-1}$ against the theoretical definition:

• First Term ($y^1$): The exponent of $y$ is $1$. Because $1 \in \mathbb{W}$, this specific term satisfies the polynomial condition.
• Second Term ($2y^{-1}$): The exponent of $y$ is $-1$. Because $-1 \notin \mathbb{W}$ (it is a negative integer, not a whole number), this term violates the fundamental definition of a polynomial.

Step 5: Logical Conclusion

For an entire expression to be classified as a polynomial, every single term must possess a whole number exponent. The presence of even one term with a negative or fractional exponent immediately disqualifies the entire expression.

Final Solution: The expression $y + \frac{2}{y}$ is NOT a polynomial. While it is an expression in exactly one variable ($y$), the exponent of the variable in the second term is $-1$, which is not a whole number.

Initial Setup & Algebraic Identity

We are tasked with expanding the algebraic expression $(x + 2y + 4z)^2$. To execute this expansion systematically, we utilize the standard algebraic identity for the square of a trinomial.

[Per the distributive property of multiplication over addition, the square of a trinomial is given by the identity:]

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

Step 1: Variable Mapping

By comparing our given expression $(x + 2y + 4z)^2$ with the standard identity $(a + b + c)^2$, we establish a direct one-to-one mapping of the terms:

• $a = x$
• $b = 2y$
• $c = 4z$

Geometric Visualization of the Identity

The algebraic expansion of $(a + b + c)^2$ can be geometrically interpreted as the area of a square with side length $(a + b + c)$, partitioned into nine distinct rectangular and square regions. The sum of the areas of these nine regions perfectly mirrors the terms in our algebraic identity.

Step 2: Substitution into the Identity

Substituting the mapped variables into the right-hand side of the identity, we obtain the unsimplified expanded form:

$(x + 2y + 4z)^2 = (x)^2 + (2y)^2 + (4z)^2 + 2(x)(2y) + 2(2y)(4z) + 2(4z)(x)$

Step 3: Term-by-Term Simplification

We now apply the laws of exponents [specifically $(xy)^n = x^n y^n$] and basic arithmetic multiplication to simplify each term independently.

Step 4: Final Assembly

Combining all the simplified terms from Step 3 yields the fully expanded polynomial. By convention, we write the squared terms first, followed by the cross-product terms in cyclical order ($xy$, $yz$, $zx$).

$(x + 2y + 4z)^2 = x^2 + 4y^2 + 16z^2 + 4xy + 16yz + 8zx$

Final Solution: The expanded form of $(x + 2y + 4z)^2$ is $x^2 + 4y^2 + 16z^2 + 4xy + 16yz + 8zx$.

Step 1: Identify the Applicable Algebraic Identity

To expand the given expression $(\frac{3}{2}x + 1)^3$, we utilize the standard algebraic identity for the cube of a binomial. The expansion of a binomial sum cubed is derived from multiplying the binomial by itself three times: $(a+b)(a+b)(a+b)$.

The standard identity is defined as:

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

[Theoretical Justification: By the Binomial Theorem for $n=3$, the coefficients follow the sequence 1, 3, 3, 1 from Pascal's Triangle, representing the geometric decomposition of a cube into one $a^3$ volume, three $a^2b$ volumes, three $ab^2$ volumes, and one $b^3$ volume.]

Step 2: Map the Variables to the Given Expression

By comparing the given expression $(\frac{3}{2}x + 1)^3$ with the standard identity $(a + b)^3$, we establish the following variable assignments:

• $a = \frac{3}{2}x$
• $b = 1$

Step 3: Term-by-Term Expansion & Algebraic Manipulation

We will now substitute $a$ and $b$ into the identity and rigorously evaluate each of the four terms.

1. First Term ($a^3$):

$ a^3 = \left(\frac{3}{2}x\right)^3 $

$ a^3 = \frac{3^3}{2^3} \cdot x^3 = \frac{27}{8}x^3 $

[Applying the power of a product and quotient rules: $(\frac{p}{q} \cdot x)^n = \frac{p^n}{q^n} \cdot x^n$]

2. Second Term ($3a^2b$):

$ 3a^2b = 3 \cdot \left(\frac{3}{2}x\right)^2 \cdot (1) $

$ 3a^2b = 3 \cdot \left(\frac{9}{4}x^2\right) \cdot 1 $

$ 3a^2b = \frac{27}{4}x^2 $

3. Third Term ($3ab^2$):

$ 3ab^2 = 3 \cdot \left(\frac{3}{2}x\right) \cdot (1)^2 $

$ 3ab^2 = 3 \cdot \left(\frac{3}{2}x\right) \cdot 1 $

$ 3ab^2 = \frac{9}{2}x $

4. Fourth Term ($b^3$):

$ b^3 = (1)^3 = 1 $

Step 4: Synthesize the Final Polynomial

Combine the evaluated terms to form the expanded polynomial. It is standard mathematical convention to write polynomials in descending order of their degree (from the highest power of $x$ to the constant term).

$ \left(\frac{3}{2}x + 1\right)^3 = \frac{27}{8}x^3 + \frac{27}{4}x^2 + \frac{9}{2}x + 1 $

Final Solution: The expanded form of the cube is $\frac{27}{8}x^3 + \frac{27}{4}x^2 + \frac{9}{2}x + 1$.

Initial Setup & Theoretical Foundation

The area of a rectangle is geometrically defined as the product of its two adjacent spatial dimensions: length ($l$) and breadth ($b$). This relationship is expressed by the formula:

$\text{Area} = l \times b$

We are given the area of the rectangle in the form of a quadratic polynomial in terms of the variable $a$:

$\text{Area} = 25a^2 - 35a + 12$

[Per the Fundamental Theorem of Algebra and polynomial factorization principles], to find the possible expressions for the length and breadth, we must factorize this quadratic polynomial into the product of two linear binomials. These two resulting factors will represent the possible dimensions of the rectangle.

Step 1: Identifying the Factorization Method

We will factorize the quadratic polynomial $P(a) = 25a^2 - 35a + 12$ using the method of splitting the middle term. We compare the given polynomial to the standard quadratic form $Ax^2 + Bx + C$ to identify the coefficients:

• $A = 25$ (Coefficient of $a^2$)
• $B = -35$ (Coefficient of $a$)
• $C = 12$ (Constant term)

Step 2: Splitting the Middle Term

To split the middle term, we must find two real numbers, let us call them $p$ and $q$, that satisfy two specific conditions simultaneously:

1. Their product must equal $A \times C$:
   $p \times q = 25 \times 12 = 300$
2. Their sum must equal $B$:
   $p + q = -35$

Because the product ($300$) is positive and the sum ($-35$) is negative, [by the rules of integer arithmetic], both numbers $p$ and $q$ must be negative.

Let us evaluate the factor pairs of $300$ to find the correct combination:

The numbers $-15$ and $-20$ satisfy both conditions perfectly. Therefore, we will split the middle term $-35a$ into $-20a$ and $-15a$.

Step 3: Factorization by Grouping

We substitute the split terms back into the original polynomial:

$P(a) = 25a^2 - 20a - 15a + 12$

Next, we group the terms into pairs to extract the greatest common monomial from each group:

$P(a) = (25a^2 - 20a) - (15a - 12)$

Factor out the greatest common divisor (GCD) from the first group ($5a$) and the second group ($3$):

$P(a) = 5a(5a - 4) - 3(5a - 4)$

Notice that the binomial $(5a - 4)$ is now a common factor in both terms. We factor out $(5a - 4)$ [by the Distributive Property of Multiplication over Addition]:

$P(a) = (5a - 4)(5a - 3)$

Geometric Visualization

The factorization proves that a rectangle with an area of $25a^2 - 35a + 12$ can be constructed with sides measuring $(5a - 4)$ and $(5a - 3)$. Below is the geometric representation of this relationship.

Final Conclusion

Because multiplication is commutative ($l \times b = b \times l$), either of the two binomial factors can represent the length, and the remaining factor will represent the breadth.

Final Solution: The possible expressions for the length and breadth of the rectangle are $(5a - 3)$ and $(5a - 4)$.

Given Variables & Initial Setup

We are given a linear polynomial in one variable, $x$, defined as:

$p(x) = 3x + 1$

We are tasked with verifying whether the specific value $x = -\frac{1}{3}$ is a zero of this polynomial.

[Per the foundational definition in polynomial algebra, a real number $a$ is considered a zero (or root) of a polynomial $p(x)$ if and only if evaluating the polynomial at $x = a$ yields a result of zero. Mathematically, this is expressed as $p(a) = 0$.]

Step 1: Substitution of the Given Value

To test the condition $p(a) = 0$, we substitute $x = -\frac{1}{3}$ into the polynomial equation $p(x)$.

$p\left(-\frac{1}{3}\right) = 3\left(-\frac{1}{3}\right) + 1$

Step 2: Algebraic Evaluation

Next, we perform the multiplication operation. [By the properties of rational numbers, multiplying an integer by a fraction involves multiplying the integer by the numerator and dividing by the denominator].

$3 \times \left(-\frac{1}{3}\right) = \frac{3 \times -1}{3} = -1$

Substituting this product back into the polynomial expression:

$p\left(-\frac{1}{3}\right) = -1 + 1$

Step 3: Final Simplification and Logical Conclusion

Evaluating the arithmetic sum:

$p\left(-\frac{1}{3}\right) = 0$

Because the polynomial evaluates exactly to $0$, the condition for $x = -\frac{1}{3}$ being a zero of the polynomial is strictly satisfied.

Graphical Verification (Geometric Interpretation)

Geometrically, the zero of a linear polynomial $p(x) = mx + c$ corresponds to the x-intercept of the line $y = mx + c$ on a Cartesian coordinate plane. The graph below demonstrates the line $y = 3x + 1$ intersecting the x-axis precisely at the coordinate $\left(-\frac{1}{3}, 0\right)$.

Final Solution: Yes, $x = -\frac{1}{3}$ is a zero of the polynomial $p(x) = 3x + 1$ because $p\left(-\frac{1}{3}\right) = 0$.

Initial Setup & Theoretical Foundation

We are given the algebraic expression:

$P(t) = 5t - \sqrt{7}$

The degree of a polynomial in one variable is defined as the highest exponent (power) of the variable present in the polynomial with a non-zero coefficient. [Per the Fundamental Theorem of Algebra and standard polynomial definitions]. To find the degree, we must analyze the exponent of the variable $t$ in every term of the expression.

Step 1: Isolate and Analyze Each Term

A polynomial is constructed as a sum of individual terms. Let us break down the given polynomial $P(t)$ into its constituent terms:

• Term 1: $5t$
• Term 2: $-\sqrt{7}$

Step 2: Determine the Exponent of the Variable in Each Term

We rewrite each term to explicitly reveal the power of the variable $t$:

• Analyzing Term 1 ($5t$): The variable $t$ is written without an explicit exponent. By algebraic convention, any variable without a written exponent has an implicit exponent of $1$. Thus, it can be written as $5t^1$. The power of the variable in this term is $1$.
• Analyzing Term 2 ($-\sqrt{7}$): This is a constant term and contains no visible variable. By the laws of exponents, any non-zero variable raised to the power of $0$ equals $1$ (i.e., $t^0 = 1$). Thus, the constant $-\sqrt{7}$ can be mathematically expressed as $-\sqrt{7}t^0$. The power of the variable in this term is $0$.

Step 3: Identify the Highest Power (Degree)

We now compare the exponents of $t$ extracted from all terms:

• Power in Term 1: $1$
• Power in Term 2: $0$

The degree is the maximum value among these exponents: $\max(1, 0) = 1$. Therefore, the highest power of the variable $t$ is $1$.

Visual Representation: The Linear Nature of Degree 1 Polynomials

A polynomial of degree $1$ is classified geometrically as a linear polynomial. When graphed on a Cartesian coordinate system, a degree $1$ polynomial always forms a perfectly straight line. Below is the exact geometric representation of $P(t) = 5t - \sqrt{7}$, demonstrating its constant rate of change (slope = $5$) and its intercepts.

Final Solution: The degree of the polynomial $5t - \sqrt{7}$ is $1$.

Initial Algebraic Setup

We are tasked with factorising the following algebraic expression:

$8a^3 + b^3 + 12a^2b + 6ab^2$

Step 1: Identifying the Relevant Algebraic Identity

By analyzing the structure of the polynomial, we observe that it contains four terms, all of which are positive. Furthermore, the first and second terms ($8a^3$ and $b^3$) are perfect cubes. This structural signature strongly indicates the application of the standard binomial expansion identity for the cube of a sum [Per the Binomial Theorem for exponent 3].

The standard identity is defined as:

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

Our objective is to map the given polynomial to the right-hand side of this identity.

Step 2: Decomposing the Terms to Establish Base Variables

We must rewrite the perfect cube terms to identify the base variables $x$ and $y$.

• First Term (Cube): The term $8a^3$ can be rewritten as the cube of a single monomial. Since $2^3 = 8$, we have:
  $8a^3 = (2a)^3$
  [This establishes our $x = 2a$].
• Second Term (Cube): The term $b^3$ is already a perfect cube:
  $b^3 = (b)^3$
  [This establishes our $y = b$].

Step 3: Verifying the Cross-Terms

To rigorously prove that the identity applies, we must verify that the remaining terms in the polynomial ($12a^2b$ and $6ab^2$) perfectly match the $3x^2y$ and $3xy^2$ components of the identity using our established $x = 2a$ and $y = b$.

• Checking $3x^2y$:
  Substitute $x = 2a$ and $y = b$:
  $3(2a)^2(b) = 3(4a^2)(b) = 12a^2b$
  [This perfectly matches the third term of our given polynomial].
• Checking $3xy^2$:
  Substitute $x = 2a$ and $y = b$:
  $3(2a)(b)^2 = 3(2a)(b^2) = 6ab^2$
  [This perfectly matches the fourth term of our given polynomial].

Step 4: Visualizing the Algebraic Mapping

The following diagram illustrates the exact one-to-one mapping between the standard identity and our decomposed polynomial.

Step 5: Applying the Identity to Factorise

Since all terms perfectly satisfy the expansion of $(x + y)^3$, we can condense the expanded polynomial into its factorised binomial cube form by substituting $x = 2a$ and $y = b$:

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

Step 6: Expanding into Linear Factors

To express the polynomial fully factorised into irreducible linear polynomials, we expand the exponent. [By the fundamental definition of exponents, $z^3 = z \cdot z \cdot z$].

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

Final Solution: The completely factorised form of the polynomial $8a^3 + b^3 + 12a^2b + 6ab^2$ is $(2a + b)(2a + b)(2a + b)$.

Given Polynomial & Initial Setup

We are tasked with classifying the following algebraic expression based on its degree:

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

Step 1: Theoretical Foundation of Polynomial Classification

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

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

Step 2: Term-by-Term Analysis of $P(x)$

We decompose the given polynomial $P(x) = x^2 + x$ into its constituent terms to identify the exponents of the variable $x$:

• First Term ($x^2$): The variable $x$ is raised to the power of $2$.
• Second Term ($x$): The variable $x$ is implicitly raised to the power of $1$ (since $x = x^1$).

Comparing the exponents $\{2, 1\}$, the maximum value is $2$. Therefore, the highest power of the variable $x$ in the polynomial is $2$.

Step 3: Classification and Geometric Justification

Since the highest exponent is $2$, the degree of the polynomial $P(x) = x^2 + x$ is exactly $2$. By definition, a polynomial of degree $2$ is classified as a quadratic polynomial.

[Geometrically, a quadratic polynomial represents a parabola when graphed on a Cartesian plane. The presence of the $x^2$ term dictates this parabolic curvature, distinguishing it from the straight line of a linear polynomial or the inflection curve of a cubic polynomial.]

Final Solution: The highest power of the variable $x$ in the expression $x^2 + x$ is $2$. Therefore, it is classified as a quadratic polynomial.

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 $(x + 1)^2(x - 5)$.

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 algebraic expression:

$(2x - y + z)^2$

To expand a trinomial squared, we utilize the standard algebraic identity for the square of a trinomial [derived via the distributive property of multiplication over addition]:

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

Step 1: Variable Mapping

By comparing the given expression $(2x - y + z)^2$ with the standard identity $(a + b + c)^2$, we establish a direct mapping of the terms. It is critical to include the negative sign in the mapping to maintain algebraic equivalence.

• $a = 2x$
• $b = -y$
• $c = z$

Step 2: Term-by-Term Expansion

Substitute the mapped variables into the expanded form of the identity. We will compute the squared terms and the cross-product terms systematically.

1. Computing the Squared Terms ($a^2, b^2, c^2$):

• $a^2 = (2x)^2 = 4x^2$ [Applying the power of a product rule: $(xy)^n = x^n y^n$]
• $b^2 = (-y)^2 = y^2$ [The square of a negative quantity is positive]
• $c^2 = (z)^2 = z^2$

2. Computing the Cross-Product Terms ($2ab, 2bc, 2ca$):

• $2ab = 2(2x)(-y) = -4xy$
• $2bc = 2(-y)(z) = -2yz$
• $2ca = 2(z)(2x) = 4zx$

Step 3: Synthesis and Simplification

Now, we assemble all the computed components back into the framework of the identity:

$(2x - y + z)^2 = (4x^2) + (y^2) + (z^2) + (-4xy) + (-2yz) + (4zx)$

Simplifying the signs yields the final expanded polynomial:

$(2x - y + z)^2 = 4x^2 + y^2 + z^2 - 4xy - 2yz + 4zx$

Geometric Interpretation of the Identity

The algebraic identity $(a+b+c)^2$ can be visualized geometrically as the area of a square with side length $(a+b+c)$. The total area is the sum of the areas of the 9 smaller rectangular regions formed by partitioning the sides into segments of lengths $a$, $b$, and $c$.

As demonstrated in the area model, the total area consists of one $a^2$, one $b^2$, one $c^2$, two $ab$ rectangles, two $bc$ rectangles, and two $ac$ rectangles, perfectly mirroring our algebraic expansion.

Final Solution: The expanded form of $(2x - y + z)^2$ is $4x^2 + y^2 + z^2 - 4xy - 2yz + 4zx$.

Initial Setup & Theoretical Foundation

The geometric area of a rectangle is defined by the product of its two adjacent spatial dimensions, length and breadth. Mathematically, this is expressed as:

[$Area = Length \times Breadth$]

We are given the area of a rectangle as a quadratic polynomial in terms of the variable $y$:

$Area = 35y^2 + 13y - 12$

To find the possible expressions for the length and breadth, we must factorize this quadratic polynomial into the product of two linear binomials. Each resulting binomial will represent one of the dimensions of the rectangle.

Step 1: Coefficient Analysis and the Product-Sum Method

We will factorize the quadratic polynomial $ay^2 + by + c$ using the method of splitting the middle term. First, we identify the coefficients:

• Leading coefficient ($a$) = $35$
• Middle coefficient ($b$) = $13$
• Constant term ($c$) = $-12$

According to the product-sum factorization theorem, we must find two real numbers, let's call them $p$ and $q$, such that:

1. Their product equals $a \times c$:
$p \times q = 35 \times (-12) = -420$

2. Their sum equals $b$:
$p + q = 13$

Step 2: Prime Factorization to Determine Split Values

To systematically find the values of $p$ and $q$, we analyze the prime factorization of the absolute value of the product ($420$):

$420 = 2 \times 210 = 2^2 \times 105 = 2^2 \times 3 \times 35 = 2^2 \times 3 \times 5 \times 7$

We need to group these prime factors into two numbers whose difference is $13$ (since the product is negative, one number must be positive and the other negative). Let us test combinations:

• Combination 1: $(2^2 \times 5) = 20$ and $(3 \times 7) = 21$. Difference is $1$. (Incorrect)
• Combination 2: $(2^2 \times 7) = 28$ and $(3 \times 5) = 15$. Difference is $13$. (Correct)

Since the sum must be positive ($+13$), the larger number must be positive. Therefore, our two numbers are $28$ and $-15$.

[$28 \times (-15) = -420$ and $28 + (-15) = 13$]

Step 3: Splitting the Middle Term and Grouping

We substitute the middle term $13y$ with $28y - 15y$ in the original polynomial:

$35y^2 + 28y - 15y - 12$

Next, we apply factorization by grouping. We group the first two terms and the last two terms:

$(35y^2 + 28y) - (15y + 12)$

Extract the greatest common divisor (GCD) from each group:

For $(35y^2 + 28y)$, the GCD is $7y$:
$7y(5y + 4)$

For $-(15y + 12)$, the GCD is $-3$:
$-3(5y + 4)$

Now, substitute these back into the expression:

$7y(5y + 4) - 3(5y + 4)$

Notice that the binomial $(5y + 4)$ is a common factor. Factoring it out yields the final product of two linear expressions:

$(7y - 3)(5y + 4)$

Step 4: Geometric Representation and Dimensional Assignment

Because multiplication is commutative [by the Commutative Property of Multiplication, $A \times B = B \times A$], either of these binomial factors can represent the length, and the other will represent the breadth.

Final Solution: The possible expressions for the dimensions of the rectangle are Length = $(7y - 3)$ and Breadth = $(5y + 4)$, or vice versa.

Initial Setup & Theoretical Foundation

We are given the algebraic expression:

$P(t) = 3\sqrt{t} + t\sqrt{2}$

To determine whether this expression is a polynomial in one variable, we must evaluate it against the fundamental definition of a polynomial. [Per the algebraic definition of polynomials], an expression is classified as a polynomial in one variable if and only if:

• It contains exactly one variable.
• The exponent of the variable in every term is a non-negative integer (i.e., a whole number belonging to the set $\mathbb{W} = \{0, 1, 2, 3, \dots\}$).
• The coefficients of the variables are real numbers ($\mathbb{R}$).

Step 1: Algebraic Transformation of Radical Terms

To properly analyze the exponents, we must convert all radical notations into fractional exponents using the laws of indices. [By the definition of rational exponents, $\sqrt[n]{x^m} = x^{m/n}$].

Applying this to the given expression:

• The first term is $3\sqrt{t}$. The square root of $t$ can be rewritten as $t^{1/2}$. Thus, the term becomes $3t^{1/2}$.
• The second term is $t\sqrt{2}$. The variable $t$ is raised to the first power, and the coefficient is $\sqrt{2}$. This can be rewritten as $\sqrt{2}t^1$.

The transformed expression is:

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

Step 2: Exponent Analysis & Verification

We now isolate and examine the exponent of the variable $t$ in each term of the transformed expression.

Step 3: Logical Deduction

While the expression contains only one variable ($t$), the exponent of $t$ in the first term is $\frac{1}{2}$. Because $\frac{1}{2}$ is a rational number but not a non-negative integer (whole number), the expression violates the strict algebraic criteria required for polynomials.

Final Solution: The expression $3\sqrt{t} + t\sqrt{2}$ is NOT a polynomial in one variable. Reason: The exponent of the variable $t$ in the first term is $\frac{1}{2}$, which is not a whole number.

Initial Setup & Polynomial Definition

Let the given mathematical expression be defined as a polynomial function $P(x)$.

$P(x) = 5x - 4x^2 + 3$

[Per the standard form of a polynomial, it is conventionally written in descending order of the degree of its terms. Rearranging the terms yields:]

$P(x) = -4x^2 + 5x + 3$

Step 1: Substitution of the Variable

We are required to evaluate the polynomial at the specific coordinate $x = 0$. This process involves substituting the value $0$ for every instance of the variable $x$ in the polynomial expression.

$P(0) = -4(0)^2 + 5(0) + 3$

Step 2: Algebraic Evaluation

Apply the standard order of operations [PEMDAS/BODMAS: evaluating exponents first, followed by multiplication, and finally addition].

Summing the evaluated terms:

$P(0) = 0 + 0 + 3$

$P(0) = 3$

Graphical Interpretation (The Y-Intercept)

Geometrically, evaluating any polynomial $P(x)$ at $x = 0$ yields the y-intercept of its graph. The constant term of the polynomial represents the exact point where the curve crosses the y-axis. For the quadratic function $P(x) = -4x^2 + 5x + 3$, the graph is a downward-opening parabola that intersects the y-axis at the coordinate $(0, 3)$.

Final Solution: The value of the polynomial $5x - 4x^2 + 3$ at $x = 0$ is 3.

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 $x = 0$.

Step 1: Theoretical Foundation of Polynomials

To determine whether a given algebraic expression is a polynomial in one variable, we must evaluate it against two strict algebraic criteria:

• Single Variable Condition: The expression must contain exactly one distinct variable (e.g., only $x$, only $y$, or only $z$).
• Non-Negative Integer Exponent Condition: [Per the fundamental definition of polynomials], the exponent of the variable in every term must be a whole number (i.e., $0, 1, 2, 3, \dots$). Fractional, negative, or irrational exponents on the variable disqualify the expression from being a polynomial.

Step 2: Structural Analysis of the Given Expression

The given mathematical expression is:

$y^2 + \sqrt{2}$

We can rewrite this expression in its standard canonical form by explicitly showing the variable in the constant term. Any non-zero constant $c$ can be written as $c \cdot y^0$ because $y^0 = 1$ (for $y \neq 0$).

$y^2 + \sqrt{2} \cdot y^0$

Step 3: Evaluating the Variable Condition

Observing the expression $y^2 + \sqrt{2}$, the only alphabetical symbol representing an unknown quantity is $y$. There are no other variables (such as $x$ or $z$) present in the expression. Therefore, the expression satisfies the condition of being in one variable.

Step 4: Evaluating the Exponent Condition

We must now inspect the exponent of the variable $y$ in each term:

Note: The presence of $\sqrt{2}$ does not violate the polynomial definition. The restriction of having non-negative integer powers applies only to the variables, not to the numerical coefficients or constants. $\sqrt{2}$ is simply a real number acting as the constant term.

Step 5: Logical Conclusion

Since the expression contains exactly one variable ($y$) and all exponents of this variable ($2$ and $0$) are non-negative integers (whole numbers), it perfectly satisfies all algebraic axioms defining a polynomial in one variable.

Final Solution: Yes, the expression $y^2 + \sqrt{2}$ is a polynomial in one variable. The reason is that it contains only one variable ($y$), and the exponent of the variable in every term is a whole number.

Given Variables & Initial Setup

Let the given algebraic expression be denoted as a polynomial function $P(x)$:

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

Our objective is to determine the coefficient of the $x^2$ term. [A coefficient is defined mathematically as the numerical or constant multiplier of a specific variable within an algebraic term].

Step 1: Standardizing the Polynomial Form

To systematically analyze the polynomial, we first rearrange its terms in descending order of their degree to express it in standard form.

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

[Per the standard convention of polynomial representation, where terms are ordered from the highest exponent to the lowest exponent].

Step 2: Isolating the Target Term

We inspect the standardized polynomial to locate the specific term containing the variable $x$ raised to the power of $2$. The polynomial consists of three distinct terms separated by addition:

• First term: $x^2$ (Quadratic term, Degree 2)
• Second term: $x$ (Linear term, Degree 1)
• Third term: $2$ (Constant term, Degree 0)

The target term for our analysis is $+x^2$.

Step 3: Extracting the Coefficient

We must identify the constant numerical value that multiplies the variable $x^2$. In algebra, when a variable does not have an explicitly written numerical prefix, it is implicitly multiplied by $1$.

$x^2 = 1 \cdot x^2$

[By the Multiplicative Identity Property of Real Numbers, which states that any real number or variable multiplied by $1$ remains unchanged, i.e., $a \cdot 1 = a$].

Therefore, the numerical multiplier (coefficient) of the $x^2$ term is exactly $1$.

Final Solution: The coefficient of $x^2$ in the polynomial $2 + x^2 + x$ is $1$.

Initial Setup & Polynomial Definition

We are tasked with evaluating a given polynomial at a specific value of its variable. Let the given polynomial be denoted as $P(x)$.

The polynomial is defined as:

$P(x) = 5x - 4x^2 + 3$

For rigorous mathematical analysis, it is standard practice to rewrite the polynomial in descending order of its degree [Per the standard form of a polynomial $ax^2 + bx + c$]:

$P(x) = -4x^2 + 5x + 3$

Step 1: Substitution of the Variable

We must find the value of the polynomial at $x = -1$. This requires substituting $-1$ for every instance of the variable $x$ within the function $P(x)$.

$P(-1) = -4(-1)^2 + 5(-1) + 3$

Step 2: Evaluating the Exponent

According to the fundamental order of operations [PEMDAS/BODMAS], we must evaluate the exponent before performing multiplication.

The square of a negative number is positive [Since $(-a) \times (-a) = a^2$]:

$(-1)^2 = (-1) \times (-1) = 1$

Substituting this back into our equation yields:

$P(-1) = -4(1) + 5(-1) + 3$

Step 3: Executing the Multiplication

Next, we perform the scalar multiplications for each term:

• First term: $-4 \times 1 = -4$
• Second term: $5 \times (-1) = -5$

Substituting these products back into the polynomial expression:

$P(-1) = -4 - 5 + 3$

Step 4: Final Arithmetic Simplification

Finally, we perform addition and subtraction from left to right to find the scalar value of the polynomial.

Combine the negative terms:

$-4 - 5 = -9$

Add the constant term:

$P(-1) = -9 + 3$

$P(-1) = -6$

Graphical Verification

To provide a comprehensive understanding, the polynomial $P(x) = -4x^2 + 5x + 3$ represents a downward-opening parabola [Because the leading coefficient $a = -4$ is less than zero]. Evaluating the polynomial at $x = -1$ corresponds to finding the $y$-coordinate of the point on this curve where the $x$-coordinate is $-1$. As calculated, this point exists exactly at $(-1, -6)$.

Final Solution: The value of the polynomial $5x - 4x^2 + 3$ at $x = -1$ is $-6$.

Initial Setup & Theoretical Foundation

We are given the mathematical expression:

$P(x) = 3$

To determine the degree of this polynomial, we must first establish the formal definition of a polynomial's degree. [Per the Fundamental Theorem of Algebra and standard polynomial theory], the degree of a polynomial in a single variable is defined as the highest exponent (power) of the variable that possesses a non-zero coefficient.

Step 1: Algebraic Manipulation of the Constant

The given expression is a constant number, $3$. In algebra, any non-zero constant can be expressed as a product of that constant and a variable raised to the power of zero.

Let us introduce a variable, $x$. [By the Zero Exponent Rule of indices, $x^0 = 1$ for all $x \neq 0$]. Therefore, we can rewrite the constant polynomial as follows:

$P(x) = 3 \cdot 1$

$P(x) = 3 \cdot x^0$

Step 2: Identification of the Highest Power

A polynomial in one variable $x$ is generally written in the standard descending form:

$P(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x^1 + a_0 x^0$

By mapping our expression $P(x) = 3x^0$ to the standard form, we observe:

• The coefficient $a_0 = 3$.
• All other coefficients ($a_1, a_2, \dots, a_n$) are exactly $0$.

Since the only term with a non-zero coefficient is $3x^0$, the highest power of the variable $x$ present in the polynomial is $0$.

Step 3: Graphical Verification (Geometric Interpretation)

Geometrically, the degree of a polynomial $P(x)$ correlates with the maximum number of times its graph can intersect the x-axis (its roots). The graph of $P(x) = 3$ is a horizontal line parallel to the x-axis, maintaining a constant y-value of $3$. Because it is parallel to the x-axis and not coincident with it, it never intersects the x-axis. Zero intersections imply zero roots, which perfectly aligns with a polynomial of degree $0$.

Theoretical Exception Note

It is critical to distinguish between a non-zero constant polynomial (like $3$) and the zero polynomial ($P(x) = 0$). While the degree of any non-zero constant polynomial is strictly $0$, the degree of the zero polynomial is mathematically undefined because $0$ can be written as $0x^0$, $0x^1$, $0x^{100}$, etc., making it impossible to define a unique highest power.

Final Solution: The degree of the polynomial $3$ is $0$.