Initial Setup: The Polynomial Expression

We are tasked with factorising the following algebraic expression:

$P(a, b) = 8a^3 - b^3 - 12a^2b + 6ab^2$

Step 1: Identifying the Relevant Algebraic Identity

By observing the degree and the signs of the terms in the polynomial, we note that it consists of four terms: two perfect cubes ($8a^3$ and $-b^3$) and two cross-terms. This structure strongly suggests the expansion of the cube of a binomial difference. [Per the standard algebraic identities for polynomials], the relevant formula is:

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

Step 2: Structural Mapping and Term Transformation

To apply the identity, we must express each term of the given polynomial in the exact form of the identity's expansion. We will determine the base values for $x$ and $y$ by taking the cube roots of the perfect cube terms.

• First term ($x^3$): The term $8a^3$ can be rewritten as a perfect cube. Since $2^3 = 8$, we have $8a^3 = (2a)^3$. Thus, we set $x = 2a$.
• Second term ($-y^3$): The term $-b^3$ can be rewritten as $-(b)^3$. Thus, we set $y = b$.

Now, we must verify if the remaining terms ($-12a^2b$ and $6ab^2$) perfectly match the $-3x^2y$ and $+3xy^2$ components of the identity using our established values for $x$ and $y$.

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

Visualizing the Algebraic Mapping

The following diagram illustrates the structural equivalence between the given polynomial and the standard algebraic identity.

Step 3: Synthesizing the Factored Form

Since all terms of the polynomial $8a^3 - b^3 - 12a^2b + 6ab^2$ map flawlessly to the expansion of $(x - y)^3$ where $x = 2a$ and $y = b$, we can condense the expanded polynomial back into its factored binomial form.

$8a^3 - b^3 - 12a^2b + 6ab^2 = (2a)^3 - (b)^3 - 3(2a)^2(b) + 3(2a)(b)^2$

$= (2a - b)^3$

Step 4: Expanding into Linear Factors

Factorisation requires expressing the polynomial as a product of its irreducible linear factors. The exponent $3$ indicates that the binomial $(2a - b)$ is multiplied by itself three times.

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

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 Expression & Algebraic Identity

We are tasked with expanding the following trinomial squared:

$ \left( \frac{1}{4}a - \frac{1}{2}b + 1 \right)^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

By comparing our given expression $\left( \frac{1}{4}a - \frac{1}{2}b + 1 \right)^2$ with the standard identity $(x + y + z)^2$, we can establish a direct mapping of the terms. It is critical to include the negative sign in our mapping to maintain algebraic integrity.

• $ x = \frac{1}{4}a $
• $ y = -\frac{1}{2}b $
• $ z = 1 $

Step 2: Substitution into the Identity

Substituting the mapped variables into the expanded form of the identity, we get:

$ \left( \frac{1}{4}a \right)^2 + \left( -\frac{1}{2}b \right)^2 + (1)^2 + 2\left( \frac{1}{4}a \right)\left( -\frac{1}{2}b \right) + 2\left( -\frac{1}{2}b \right)(1) + 2(1)\left( \frac{1}{4}a \right) $

Step 3: Term-by-Term Simplification

We will now apply the exponent rules [specifically $(uv)^n = u^n v^n$] and perform scalar multiplication for each distinct term.

Step 4: Final Assembly

Combining all the simplified terms from Step 3 yields the fully expanded polynomial. We write the terms in descending order of degree where applicable, though standard expansion order is perfectly rigorous:

$ \frac{1}{16}a^2 + \frac{1}{4}b^2 + 1 - \frac{1}{4}ab - b + \frac{1}{2}a $

Final Solution: The expanded form of the given polynomial is $ \frac{1}{16}a^2 + \frac{1}{4}b^2 + 1 - \frac{1}{4}ab - b + \frac{1}{2}a $

Initial Setup & Theoretical Foundation

We are given the quadratic polynomial:

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

We must verify whether the given values, $x = -\frac{1}{\sqrt{3}}$ and $x = \frac{2}{\sqrt{3}}$, are zeroes of the polynomial.

[Theoretical Justification: By the definition of the zero of a polynomial, a real number $a$ is a zero of a polynomial $p(x)$ if and only if $p(a) = 0$. Therefore, we must substitute each given value of $x$ into the polynomial and evaluate the result.]

Step 1: Evaluating the Polynomial at $x = -\frac{1}{\sqrt{3}}$

Substitute $x = -\frac{1}{\sqrt{3}}$ into the polynomial $p(x)$:

$p\left(-\frac{1}{\sqrt{3}}\right) = 3\left(-\frac{1}{\sqrt{3}}\right)^2 - 1$

First, apply the exponent to the fraction. The square of a negative number is positive, and the square of a square root removes the radical:

$\left(-\frac{1}{\sqrt{3}}\right)^2 = \frac{(-1)^2}{(\sqrt{3})^2} = \frac{1}{3}$

Now, substitute this back into the equation:

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

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

Since $p\left(-\frac{1}{\sqrt{3}}\right) = 0$, the value $x = -\frac{1}{\sqrt{3}}$ satisfies the condition.

Step 2: Evaluating the Polynomial at $x = \frac{2}{\sqrt{3}}$

Substitute $x = \frac{2}{\sqrt{3}}$ into the polynomial $p(x)$:

$p\left(\frac{2}{\sqrt{3}}\right) = 3\left(\frac{2}{\sqrt{3}}\right)^2 - 1$

Apply the exponent to the numerator and the denominator:

$\left(\frac{2}{\sqrt{3}}\right)^2 = \frac{2^2}{(\sqrt{3})^2} = \frac{4}{3}$

Substitute this back into the equation:

$p\left(\frac{2}{\sqrt{3}}\right) = 3\left(\frac{4}{3}\right) - 1$

$p\left(\frac{2}{\sqrt{3}}\right) = 4 - 1 = 3$

Since $p\left(\frac{2}{\sqrt{3}}\right) = 3 \neq 0$, the value $x = \frac{2}{\sqrt{3}}$ does not satisfy the condition.

Graphical Analysis

To visualize this algebraically proven result, we can observe the graph of the parabola $y = 3x^2 - 1$. The zeroes of the polynomial correspond to the $x$-intercepts (where the graph crosses the horizontal axis, $y=0$).

As demonstrated by the graph, the point $x = -\frac{1}{\sqrt{3}}$ lies exactly on the $x$-axis (making it a zero), whereas the point $x = \frac{2}{\sqrt{3}}$ maps to a $y$-value of $3$, confirming it is not a zero.

Final Conclusion

Final Solution: $x = -\frac{1}{\sqrt{3}}$ is a zero of the polynomial $p(x) = 3x^2 - 1$, whereas $x = \frac{2}{\sqrt{3}}$ is not a zero of the polynomial.

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.

Given Variables & Initial Setup

We are given the following linear polynomial in one variable:

$p(x) = 5x - \pi$

We are tasked with verifying whether the following specific value of $x$ is a zero (root) of the polynomial:

$x = \frac{4}{5}$

Step 1: Theoretical Foundation

By the fundamental definition of the Zero of a Polynomial [Factor Theorem corollary], a real number $a$ is considered a zero of a polynomial $p(x)$ if and only if evaluating the polynomial at $x = a$ yields exactly zero. Mathematically, this is expressed as:

$p(a) = 0$

Therefore, to verify if $x = \frac{4}{5}$ is a zero, we must substitute $x = \frac{4}{5}$ into $p(x)$ and determine if the resulting value is equal to $0$.

Step 2: Substitution and Algebraic Evaluation

Substitute $x = \frac{4}{5}$ into the polynomial $p(x)$:

$p\left(\frac{4}{5}\right) = 5\left(\frac{4}{5}\right) - \pi$

Perform the multiplication. The factor of $5$ in the numerator and the denominator cancel each other out:

$p\left(\frac{4}{5}\right) = 4 - \pi$

Step 3: Analytical Verification

We must now evaluate the expression $4 - \pi$.

• The number $4$ is a rational integer.
• The number $\pi$ is an irrational mathematical constant representing the ratio of a circle's circumference to its diameter, where $\pi \approx 3.14159...$

Because $\pi$ is strictly less than $4$ (and is irrational), the difference $4 - \pi$ evaluates to a non-zero irrational number:

$4 - \pi \approx 4 - 3.14159 = 0.85841 \neq 0$

Step 4: Graphical Representation (Visual Verification)

Graphically, the zeroes of a polynomial $p(x)$ correspond to the $x$-intercepts of the graph $y = p(x)$. The graph below plots the linear function $y = 5x - \pi$. The true zero is located at $x = \frac{\pi}{5} \approx 0.628$, while our tested point $x = \frac{4}{5} = 0.8$ clearly yields a positive $y$-value, proving it does not intersect the $x$-axis at this coordinate.

Final Solution: Since $p\left(\frac{4}{5}\right) = 4 - \pi \neq 0$, the value $x = \frac{4}{5}$ is NOT a zero of the polynomial $p(x) = 5x - \pi$.

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 $x = \frac{2}{3}$.

Step 1: Initial Setup & Identification of the Algebraic Identity

We are tasked with finding the product of the given binomials:

$ \left(y^2 + \frac{3}{2}\right) \left(y^2 - \frac{3}{2}\right) $

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

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

[Theoretical Justification: The cross-terms in the expansion $(a)(a) - (a)(b) + (b)(a) - (b)(b)$ cancel out, leaving only the squared terms $a^2 - b^2$.]

Step 2: Variable Mapping and Substitution

To apply the identity rigorously, we establish a one-to-one correspondence between the variables in the identity and the terms in our specific expression.

• Let the first term $a = y^2$
• Let the second term $b = \frac{3}{2}$

Substituting these defined values into the right-hand side of the Difference of Squares identity ($a^2 - b^2$), we formulate the following equation:

$ \left(y^2\right)^2 - \left(\frac{3}{2}\right)^2 $

Step 3: Geometric Verification of the Identity (Visual Proof)

The algebraic identity $(a-b)(a+b) = a^2 - b^2$ can be geometrically proven by analyzing the area of a square of side $a$ with a smaller square of side $b$ removed. The remaining area can be rearranged into a rectangle with dimensions $(a+b)$ and $(a-b)$.

Step 4: Execution of the Algebraic Expansion & Simplification

We now simplify the expression $\left(y^2\right)^2 - \left(\frac{3}{2}\right)^2$ by applying the fundamental laws of exponents.

1. Simplifying the first term $\left(y^2\right)^2$:

According to the Power of a Power Property, $(x^m)^n = x^{m \cdot n}$. Therefore, we multiply the exponents:

$ \left(y^2\right)^2 = y^{2 \times 2} = y^4 $

2. Simplifying the second term $\left(\frac{3}{2}\right)^2$:

According to the Power of a Quotient Property, $\left(\frac{x}{y}\right)^n = \frac{x^n}{y^n}$. We distribute the square to both the numerator and the denominator:

$ \left(\frac{3}{2}\right)^2 = \frac{3^2}{2^2} = \frac{9}{4} $

Step 5: Final Polynomial Construction

By combining the simplified terms from Step 4 back into our expanded equation structure, we arrive at the final evaluated polynomial.

$ y^4 - \frac{9}{4} $

Final Solution: The product of the given binomials is $y^4 - \frac{9}{4}$.

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 $0$.

Initial Setup & Structural Analysis

We are tasked with factorising the following algebraic expression of six terms:

$P(x, y, z) = 4x^2 + 9y^2 + 16z^2 + 12xy - 24yz - 16xz$

[Per the fundamental theorems of polynomial algebra], an expression containing three perfect square terms and three cross-product terms strongly indicates the expansion of a squared trinomial. The governing algebraic identity is:

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

Step 1: Extraction of Base Variables (Magnitude)

We must map the first three terms of our polynomial to the squared terms of the identity ($a^2, b^2, c^2$) to find the magnitudes of $a, b,$ and $c$.

• First term: $a^2 = 4x^2 \implies |a| = \sqrt{4x^2} = 2x$
• Second term: $b^2 = 9y^2 \implies |b| = \sqrt{9y^2} = 3y$
• Third term: $c^2 = 16z^2 \implies |c| = \sqrt{16z^2} = 4z$

Step 2: Sign Determination via Cross-Product Analysis

The signs of $a, b,$ and $c$ are determined by analyzing the signs of the cross-product terms ($2ab, 2bc, 2ca$).

The given cross-product terms are:

• $2ab = +12xy$ (Positive)
• $2bc = -24yz$ (Negative)
• $2ca = -16xz$ (Negative)

[By the rules of integer multiplication], the product $2ab$ is positive, which dictates that $a$ and $b$ must share the same sign. Let us assume both $a$ and $b$ are positive:

$a = +2x$

$b = +3y$

The products $2bc$ and $2ca$ are both negative. Since $b$ is positive, for $2bc$ to be negative, $c$ must be negative. Similarly, since $a$ is positive, for $2ca$ to be negative, $c$ must be negative. This confirms that the negative sign originates exclusively from the $z$-term.

$c = -4z$

Step 3: Verification of the Identity

We substitute $a = 2x$, $b = 3y$, and $c = -4z$ back into the expanded identity to ensure absolute mathematical equivalence:

$a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$
$= (2x)^2 + (3y)^2 + (-4z)^2 + 2(2x)(3y) + 2(3y)(-4z) + 2(-4z)(2x)$
$= 4x^2 + 9y^2 + 16z^2 + 12xy - 24yz - 16xz$

The expanded form perfectly matches the original polynomial. Therefore, the expression can be written as the square of the trinomial $(2x + 3y - 4z)$.

Visualizing the Expansion (Tabular Area Model)

To rigorously prove the distribution of terms, we can use a tabular area model. The sum of all cells in the $3 \times 3$ matrix equals the original polynomial, demonstrating how the cross-terms combine.

Notice how the symmetric off-diagonal terms combine perfectly: $(6xy + 6xy = 12xy)$, $(-12yz - 12yz = -24yz)$, and $(-8xz - 8xz = -16xz)$.

Step 4: Final Synthesis

Having established the base terms and their respective signs, we write the expression in its fully factorised form. Since factorisation requires expressing the polynomial as a product of its irreducible factors, we write the squared binomial as the product of two identical brackets.

Final Solution: The factorised form of the given polynomial is $(2x + 3y - 4z)(2x + 3y - 4z)$, which can also be written as $(2x + 3y - 4z)^2$.

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 $(3 - 5a)(3 - 5a)(3 - 5a)$.

Given Polynomial & Theoretical Foundation

We are given the linear polynomial:

$p(x) = 3x$

Theoretical Definition: The "zero" or "root" of a polynomial $p(x)$ is defined as any real or complex number $c$ such that when $x$ is replaced by $c$, the value of the polynomial evaluates to zero. Mathematically, $c$ is a zero if and only if $p(c) = 0$ [Per the Fundamental Theorem of Algebra and the Factor Theorem].

Step 1: Setting up the Equation

To find the zero of the polynomial, we must equate the polynomial expression to zero. This transforms our polynomial expression into an algebraic equation.

$p(x) = 0$

Substituting the given expression for $p(x)$:

$3x = 0$

Step 2: Algebraic Manipulation

We now solve for the variable $x$. The term $3x$ represents the product of the constant $3$ and the variable $x$.

To isolate $x$, we apply the Multiplicative Property of Equality, dividing both sides of the equation by the coefficient $3$:

$\frac{3x}{3} = \frac{0}{3}$

Since any non-zero number dividing zero results in zero [Per the Zero Product Property and properties of real numbers]:

$x = 0$

Step 3: Verification of the Root

To ensure absolute mathematical rigor, we verify the solution by substituting $x = 0$ back into the original polynomial $p(x)$:

$p(0) = 3(0)$

$p(0) = 0$

Because the polynomial evaluates to $0$ when $x = 0$, the solution is verified as correct.

Graphical Representation of the Polynomial

Geometrically, the zero of a polynomial with real coefficients corresponds to the $x$-coordinate of the point where the graph of the function $y = p(x)$ intersects the $x$-axis. For $y = 3x$, this is a straight line passing through the origin.

As demonstrated in the Cartesian plane above, the line $y = 3x$ intersects the $x$-axis exactly at the origin $(0,0)$, visually confirming that the zero of the polynomial is $0$.

Final Conclusion

Final Solution: The zero of the polynomial $p(x) = 3x$ is $x = 0$.

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

Step 1: Initial Setup and Definition of the Polynomial

We are given the algebraic expression:

$P(x) = 7x^3$

To classify this mathematical expression, we must first verify that it satisfies the formal definition of a polynomial in one variable. A polynomial in a single variable $x$ is an expression consisting of variables and coefficients, where the exponents of the variables are non-negative integers.

• Coefficient: $7$ (a real number)
• Variable: $x$
• Exponent: $3$ (a non-negative integer)

Since the exponent is a non-negative integer, $P(x) = 7x^3$ is strictly a polynomial.

Step 2: Determining the Degree of the Polynomial

The classification of a polynomial is fundamentally determined by its degree. [By definition, the degree of a polynomial in one variable is the highest power (exponent) of the variable present in the expression with a non-zero coefficient].

Analyzing our given polynomial:

$P(x) = 7x^3$

This is a monomial (a polynomial with a single term). The only variable present is $x$, and its exponent is $3$. Therefore, the highest power of $x$ in this polynomial is $3$.

$ \text{Degree of } P(x) = 3 $

Step 3: Classifying the Polynomial

Polynomials are classified by their degree according to the following standard algebraic nomenclature:

Because the degree of $7x^3$ is exactly $3$, it falls into the category of a cubic polynomial.

Step 4: Geometric Interpretation (Visualizing the Cubic Polynomial)

A cubic polynomial of the form $P(x) = ax^3$ (where $a > 0$) produces a characteristic curve that passes through the origin $(0,0)$ and exhibits point symmetry about the origin. This is known as an inflection point, where the concavity of the graph changes. Below is the precise geometric representation of $y = 7x^3$.

Final Solution: The highest power of the variable $x$ in the expression $7x^3$ is $3$. Therefore, $7x^3$ is classified as a cubic polynomial.

Initial Setup & Algebraic Definition

Given the algebraic expression: $P(x) = 4x^2 - 3x + 7$

[Per the Fundamental Theorem of Algebra and Polynomial Definitions], an algebraic expression is classified as a polynomial in one variable if and only if it satisfies two strict mathematical conditions:

• Condition 1 (Single Variable): The expression must contain only one distinct variable symbol (e.g., only $x$, only $y$, or only $z$).
• Condition 2 (Non-Negative Integer Exponents): The exponent of the variable in every single term must be a whole number ($\mathbb{W} = \{0, 1, 2, 3, \dots\}$). Fractional, negative, or irrational exponents disqualify the expression from being a polynomial.

Step 1: Evaluating the Variable Count

By inspecting the expression $4x^2 - 3x + 7$, we observe that the only alphabetical symbol representing an unknown quantity is $x$. There are no secondary variables (such as $y$ or $t$) present in any of the terms.

Conclusion for Condition 1: The expression is strictly in one variable.

Step 2: Analyzing the Exponents of Each Term

To rigorously verify Condition 2, we must decompose the expression into its constituent terms and isolate the exponent of $x$ for each. Note that constants can be expressed as coefficients of $x^0$ [By the Zero Exponent Rule: $x^0 = 1$ for $x \neq 0$].

Step 3: Visualizing the Polynomial Structure

The following diagram maps the anatomical structure of the given expression, proving that all exponents belong to the set of whole numbers and only a single variable is utilized.

Step 4: Synthesizing the Findings

Because the expression contains exactly one variable ($x$) and every exponent of $x$ across all terms is a non-negative integer ($2, 1, \text{ and } 0$), the expression perfectly satisfies all algebraic axioms required to be classified as a polynomial in one variable.

Final Solution: The expression $4x^2 - 3x + 7$ is a polynomial in one variable. The reason is that it contains only a single variable ($x$), and the exponents of the variable in all terms ($2, 1, \text{ and } 0$) are whole numbers.

Initial Setup & Theoretical Foundation

We are given the linear polynomial:

$p(x) = x + 5$

Step 1: Applying the Definition of a Zero

The zero (or root) of a polynomial $p(x)$ is defined as the real number $c$ such that $p(c) = 0$. [Per the Fundamental Theorem of Algebra and basic polynomial theory, a linear polynomial of degree 1 will have exactly one real zero].

Step 2: Algebraic Evaluation

To find the zero, we equate the polynomial to zero:

$x + 5 = 0$

Subtracting $5$ from both sides of the equation [By the Subtraction Property of Equality]:

$x = 0 - 5$

$x = -5$

Step 3: Verification

Substitute $x = -5$ back into the original polynomial to ensure the condition $p(x) = 0$ is rigorously satisfied:

$p(-5) = (-5) + 5$

$p(-5) = 0$

The result is verified.

Step 4: Geometric Interpretation

Geometrically, the zero of a polynomial represents the x-coordinate of the point where the graph of the function $y = p(x)$ intersects the x-axis (the line where $y = 0$). As shown in the Cartesian plane below, the line $y = x + 5$ crosses the x-axis exactly at the coordinate point $(-5, 0)$.

Final Solution: The zero of the polynomial $p(x) = x + 5$ is $x = -5$.

Given Polynomial & Algebraic Foundation

We are tasked with factorising the following multivariable polynomial of degree 2:

$P(x,y,z) = 2x^2 + y^2 + 8z^2 - 2\sqrt{2}xy + 4\sqrt{2}yz - 8xz$

The structure of this expression—comprising three squared terms and three cross-product terms—directly corresponds to the algebraic identity for the square of a trinomial. [Per the standard algebraic expansion theorem]:

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

Step 1: Extracting the Base Magnitudes

We begin by equating the pure squared terms from the given polynomial to the squared terms in the identity to find the absolute values of $a$, $b$, and $c$.

• $a^2 = 2x^2 \implies |a| = \sqrt{2}x$
• $b^2 = y^2 \implies |b| = y$
• $c^2 = 8z^2 \implies |c| = \sqrt{8}z = 2\sqrt{2}z$ [Since $\sqrt{8} = \sqrt{4 \times 2} = 2\sqrt{2}$]

Step 2: Determining the Algebraic Signs

To assign the correct positive ($+$) or negative ($-$) signs to $a$, $b$, and $c$, we must analyze the signs of the cross-product terms in the given polynomial:

From the analysis above, $b$ and $c$ share the same sign, while $a$ has the opposite sign to both. We can conventionally set $b$ and $c$ as positive, which forces $a$ to be negative. Therefore, we define our terms as:

• $a = -\sqrt{2}x$
• $b = y$
• $c = 2\sqrt{2}z$

Note: Alternatively, setting $a$ as positive forces $b$ and $c$ to be negative. Both conventions are mathematically equivalent.

Step 3: Verifying the Cross-Product Terms

Before finalizing the factorization, we must rigorously verify that our chosen terms reconstruct the original polynomial exactly.

• Check $2ab$: $2(-\sqrt{2}x)(y) = -2\sqrt{2}xy$ (Matches the given term)
• Check $2bc$: $2(y)(2\sqrt{2}z) = 4\sqrt{2}yz$ (Matches the given term)
• Check $2ca$: $2(2\sqrt{2}z)(-\sqrt{2}x) = -4(\sqrt{2} \cdot \sqrt{2})xz = -4(2)xz = -8xz$ (Matches the given term)

Step 4: Constructing the Factored Form

Since all terms perfectly align with the identity $(a + b + c)^2$, we substitute our derived values of $a$, $b$, and $c$ into the factored form.

$(a + b + c)^2 = (-\sqrt{2}x + y + 2\sqrt{2}z)^2$

To express this as a product of its factors, we write the squared binomial as the product of two identical trinomials.

Final Solution: The factorised form of the polynomial is $(-\sqrt{2}x + y + 2\sqrt{2}z)(-\sqrt{2}x + y + 2\sqrt{2}z)$.

(Note: Factoring out a $-1$ yields the equally valid alternative form: $(\sqrt{2}x - y - 2\sqrt{2}z)(\sqrt{2}x - y - 2\sqrt{2}z)$).

Given Expression & Initial Setup

We are tasked with factorising the following algebraic expression:

$64m^3 - 343n^3$

Upon initial inspection, the expression consists of two terms separated by a minus sign. The variables $m$ and $n$ are both raised to the third power, which strongly indicates that we must evaluate the numerical coefficients to determine if they are also perfect cubes.

Step 1: Identifying Perfect Cubes

To apply the relevant algebraic identity, we must rewrite both terms entirely as perfect cubes in the form of $x^3$ and $y^3$.

• First Term ($64m^3$): The prime factorization of $64$ is $2^6$, which can be grouped as $(2^2)^3 = 4^3$. Therefore, the entire first term can be rewritten as:
  $64m^3 = (4m)^3$
• Second Term ($343n^3$): The prime factorization of $343$ is $7 \times 7 \times 7 = 7^3$. Therefore, the entire second term can be rewritten as:
  $343n^3 = (7n)^3$

Substituting these back into the original expression yields:

$(4m)^3 - (7n)^3$

Step 2: Stating the Relevant Algebraic Identity

The expression is now explicitly in the form of a difference of two cubes. [Per the fundamental algebraic identities of polynomials], the difference of two cubes is factored using the following formula:

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

Step 3: Substitution and Expansion

By mapping our specific terms to the identity, we establish the following equivalencies:

• $x = 4m$
• $y = 7n$

We now substitute these values directly into the right-hand side of the identity, $(x - y)(x^2 + xy + y^2)$:

$= (4m - 7n) \left[ (4m)^2 + (4m)(7n) + (7n)^2 \right]$

Step 4: Final Simplification

To achieve the final factorised form, we must expand the terms within the square brackets by applying the exponent rules [specifically $(ab)^n = a^n b^n$] and performing basic multiplication:

• Square the first term: $(4m)^2 = 4^2 \cdot m^2 = 16m^2$
• Multiply the two terms: $(4m)(7n) = 4 \cdot 7 \cdot m \cdot n = 28mn$
• Square the second term: $(7n)^2 = 7^2 \cdot n^2 = 49n^2$

Substituting these simplified components back into our factored expression yields:

$= (4m - 7n)(16m^2 + 28mn + 49n^2)$

Final Solution: $(4m - 7n)(16m^2 + 28mn + 49n^2)$

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

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