Initial Setup & Given Variables

We are given a single-variable polynomial function of degree 3 (a cubic polynomial). The polynomial is defined as:

$p(x) = x^3$

Our objective is to evaluate this polynomial at three specific discrete points in its domain: $x = 0$, $x = 1$, and $x = 2$. [Per the Substitution Principle of Functions, evaluating a polynomial $p(x)$ at a specific value $x = a$ requires replacing every instance of the variable $x$ in the expression with the constant $a$].

Step 1: Evaluating the Polynomial at $x = 0$

To find $p(0)$, we substitute $x = 0$ into the polynomial equation.

• $p(0) = (0)^3$
• $p(0) = 0 \times 0 \times 0$
• $p(0) = 0$

[Theoretical Justification: The zero product property and the definition of exponentiation state that zero raised to any positive real power is strictly zero. Geometrically, this indicates that the graph of the polynomial passes exactly through the origin $(0,0)$].

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

To find $p(1)$, we substitute $x = 1$ into the polynomial equation.

• $p(1) = (1)^3$
• $p(1) = 1 \times 1 \times 1$
• $p(1) = 1$

[Theoretical Justification: The number 1 is the multiplicative identity. Any real number multiplied by 1 remains unchanged, hence 1 raised to any real power is always 1].

Step 3: Evaluating the Polynomial at $x = 2$

To find $p(2)$, we substitute $x = 2$ into the polynomial equation.

• $p(2) = (2)^3$
• $p(2) = 2 \times 2 \times 2$
• $p(2) = 4 \times 2$
• $p(2) = 8$

[Theoretical Justification: Exponentiation represents repeated multiplication. The rapid growth from $p(1)=1$ to $p(2)=8$ demonstrates the non-linear, cubic expansion characteristic of degree-3 polynomials].

Graphical Analysis & Verification

Below is the precise Cartesian mapping of the polynomial $p(x) = x^3$. The specific points we evaluated—$(0,0)$, $(1,1)$, and $(2,8)$—are plotted to visually confirm the cubic curve's trajectory.

Summary Table of Evaluations

Final Solution: For the polynomial $p(x) = x^3$, the evaluated values are $p(0) = 0$, $p(1) = 1$, and $p(2) = 8$.

Given Variables & Initial Setup

We are given the linear polynomial:

$p(x) = lx + m$

We need to verify whether the given value of $x$ is a zero (or root) of the polynomial:

$x = -\frac{m}{l}$

[By the Fundamental Theorem of Algebra and the definition of polynomial roots, 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, i.e., $p(a) = 0$. Furthermore, for $p(x)$ to be a valid linear polynomial of degree 1, the leading coefficient $l$ must not be equal to zero ($l \neq 0$).]

Step 1: Substitution of the Given Value

To test the condition $p(a) = 0$, we substitute $x = -\frac{m}{l}$ into the polynomial expression $p(x)$.

$p\left(-\frac{m}{l}\right) = l\left(-\frac{m}{l}\right) + m$

Step 2: Algebraic Simplification

Next, we perform the multiplication. Since $l$ is in the numerator of the coefficient and the denominator of the substituted fraction, they cancel each other out [Per the multiplicative inverse property, assuming $l \neq 0$]:

$p\left(-\frac{m}{l}\right) = \left(l \cdot \frac{-m}{l}\right) + m$

$p\left(-\frac{m}{l}\right) = -m + m$

Step 3: Final Evaluation

Combining the terms yields:

$p\left(-\frac{m}{l}\right) = 0$

Because the polynomial evaluates to zero at this specific value of $x$, the condition for it being a zero of the polynomial is perfectly satisfied.

Graphical Representation & Geometric Verification

Geometrically, the zero of a polynomial corresponds to the $x$-intercept of its graph. For the linear function $y = lx + m$, the line crosses the $x$-axis exactly at the coordinate $\left(-\frac{m}{l}, 0\right)$. The SVG below illustrates this relationship (assuming $l > 0$ and $m > 0$ for visual representation).

Final Solution: Yes, $x = -\frac{m}{l}$ is a zero of the polynomial $p(x) = lx + m$, because substituting this value into the polynomial yields exactly $0$.

Given Variables & Initial Setup

We are given the following cubic polynomial in terms of the variable $t$:

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

To find the values of $p(0)$, $p(1)$, and $p(2)$, we must apply the Polynomial Evaluation Principle. This principle states that the value of a polynomial $p(t)$ at a specific real number $t = a$ is obtained by substituting $a$ for every instance of $t$ in the polynomial expression [Per the fundamental definition of a polynomial function].

Step 1: Evaluating $p(0)$

We substitute $t = 0$ into the polynomial equation. This operation isolates the constant term, as all terms containing the variable $t$ will evaluate to zero.

• $p(0) = 2 + (0) + 2(0)^2 - (0)^3$
• $p(0) = 2 + 0 + 2(0) - 0$
• $p(0) = 2$

Geometrically, this represents the $y$-intercept of the polynomial graph at the coordinate $(0, 2)$.

Step 2: Evaluating $p(1)$

Next, we substitute $t = 1$ into the polynomial. Evaluating a polynomial at $t = 1$ effectively yields the sum of its coefficients and the constant term.

• $p(1) = 2 + (1) + 2(1)^2 - (1)^3$
• $p(1) = 2 + 1 + 2(1) - 1$
• $p(1) = 3 + 2 - 1$
• $p(1) = 4$

This corresponds to the coordinate point $(1, 4)$ on the Cartesian plane.

Step 3: Evaluating $p(2)$

Finally, we substitute $t = 2$ into the polynomial. We must carefully follow the order of operations (PEMDAS/BODMAS), evaluating the exponents before multiplying by the coefficients.

• $p(2) = 2 + (2) + 2(2)^2 - (2)^3$
• $p(2) = 2 + 2 + 2(4) - 8$
• $p(2) = 4 + 8 - 8$
• $p(2) = 4$

This corresponds to the coordinate point $(2, 4)$ on the Cartesian plane.

Graphical Representation of $p(t)$

Below is the precise Cartesian plot of the polynomial $p(t) = -t^3 + 2t^2 + t + 2$. The specific points we evaluated—$p(0)$, $p(1)$, and $p(2)$—are explicitly marked to demonstrate their spatial relationship on the curve.

Summary of Results

Final Solution: The evaluated values for the polynomial are $p(0) = 2$, $p(1) = 4$, and $p(2) = 4$.

Given Variables & Initial Setup

We are given the linear polynomial:

$p(x) = 2x + 5$

In algebra, the zero of a polynomial is defined as the specific value of the variable (in this case, $x$) that makes the value of the entire polynomial equal to zero. Geometrically, this corresponds to the x-coordinate of the point where the graph of the polynomial intersects the x-axis (the x-intercept).

Step 1: Formulating the Equation

To find the zero of the polynomial, we must equate the polynomial function $p(x)$ to $0$. [Per the Fundamental Theorem of Algebra, a polynomial of degree 1 will have exactly one real zero].

$p(x) = 0$

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

$2x + 5 = 0$

Step 2: Algebraic Isolation of the Variable

We must isolate $x$ using standard algebraic operations.

• Step 2a: Subtract $5$ from both sides of the equation to isolate the term containing $x$ [By the Subtraction Property of Equality].
  $2x = -5$
• Step 2b: Divide both sides by $2$ to solve for $x$ [By the Division Property of Equality].
  $x = \frac{-5}{2}$

This fraction can also be expressed as the decimal $-2.5$.

Step 3: Geometric Representation (Visual Proof)

Graphing the linear equation $y = 2x + 5$ provides a visual confirmation of the zero. The line crosses the x-axis exactly at $x = -2.5$.

Step 4: Verification

To ensure absolute mathematical rigor, we substitute $x = -\frac{5}{2}$ back into the original polynomial to verify that it yields $0$.

$p\left(-\frac{5}{2}\right) = 2\left(-\frac{5}{2}\right) + 5$

$p\left(-\frac{5}{2}\right) = -5 + 5$

$p\left(-\frac{5}{2}\right) = 0$

Since the result is exactly zero, the calculated value is verified as correct.

Final Solution: The zero of the polynomial $p(x) = 2x + 5$ is $x = -\frac{5}{2}$.

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 & 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 & 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 & Theoretical Foundation

We are tasked with verifying whether the given value of $x$ is a zero (or root) of the provided linear polynomial. The given parameters are:

• Polynomial: $p(x) = 2x + 1$
• Value to verify: $x = \frac{1}{2}$

[Theoretical Justification: By the fundamental definition of polynomials, a real number $c$ is considered a "zero" of a polynomial $p(x)$ if and only if evaluating the polynomial at $x = c$ yields a result of zero. Mathematically, this is expressed as $p(c) = 0$.]

Step 1: Substitution of the Given Value

To test the condition $p(c) = 0$, we substitute $x = \frac{1}{2}$ into the polynomial $p(x)$.

$p\left(\frac{1}{2}\right) = 2\left(\frac{1}{2}\right) + 1$

Step 2: Algebraic Evaluation

Next, we perform the arithmetic operations following the standard order of operations (PEMDAS/BODMAS).

First, multiply the terms:

$2 \times \frac{1}{2} = 1$

Now, substitute this product back into the expression:

$p\left(\frac{1}{2}\right) = 1 + 1$

$p\left(\frac{1}{2}\right) = 2$

Step 3: Logical Verification

We must now compare our evaluated result with the required condition for a zero.

$p\left(\frac{1}{2}\right) = 2 \neq 0$

[Since the polynomial evaluates to $2$ rather than $0$, the value $x = \frac{1}{2}$ fails the condition $p(c) = 0$.]

Graphical Representation & Geometric Proof

Geometrically, the zeroes of a polynomial $p(x)$ correspond to the $x$-intercepts of its graph $y = p(x)$. The graph below plots the linear function $y = 2x + 1$. Notice that the true zero occurs where the line crosses the x-axis at $x = -\frac{1}{2}$, whereas evaluating at $x = \frac{1}{2}$ yields a y-coordinate of $2$.

Final Conclusion

Because the evaluation of the polynomial at the given value does not result in zero, the value is not a root of the equation.

Final Solution: No, $x = \frac{1}{2}$ is not a zero of the polynomial $p(x) = 2x + 1$, because $p\left(\frac{1}{2}\right) = 2 \neq 0$.

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 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 quadratic polynomial:

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

We must verify whether the given values, $x = 1$ and $x = -1$, are zeroes of the polynomial $p(x)$.

Theoretical Justification: [Per the Definition of a Zero of a Polynomial], a real number $c$ is considered a zero (or root) of a polynomial $p(x)$ if and only if evaluating the polynomial at $x = c$ yields zero. Mathematically, this is expressed as $p(c) = 0$. Furthermore, [By the Factor Theorem], if $p(c) = 0$, then $(x - c)$ is a factor of the polynomial.

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

To determine if $x = 1$ is a zero, we substitute $x = 1$ into the polynomial $p(x)$:

• $p(1) = (1)^2 - 1$
• $p(1) = 1 - 1$
• $p(1) = 0$

Conclusion for $x = 1$: Since the evaluation results in exactly $0$, $x = 1$ is a verified zero of the polynomial $p(x)$.

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

Next, we substitute $x = -1$ into the polynomial $p(x)$:

• $p(-1) = (-1)^2 - 1$
• $p(-1) = 1 - 1$ [Since the square of any negative real number is positive, $(-1) \times (-1) = 1$]
• $p(-1) = 0$

Conclusion for $x = -1$: Since the evaluation results in exactly $0$, $x = -1$ is also a verified zero of the polynomial $p(x)$.

Graphical Verification (Visualizing the Zeroes)

In coordinate geometry, the real zeroes of a polynomial $p(x)$ correspond precisely to the $x$-intercepts of the graph of the equation $y = p(x)$. By plotting the parabola $y = x^2 - 1$, we can visually confirm that the curve intersects the $x$-axis at exactly $x = -1$ and $x = 1$.

Alternative Algebraic Perspective (Factorization)

We can also verify the zeroes by factoring the polynomial completely. The expression $x^2 - 1$ is a classic "Difference of Squares", which follows the algebraic identity $a^2 - b^2 = (a - b)(a + b)$.

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

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

To find the zeroes, we set $p(x) = 0$:

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

[By the Zero Product Property], if the product of two factors is zero, at least one of the factors must be zero:

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

This algebraic derivation perfectly matches our initial substitution method.

Final Solution: Yes, both $x = 1$ and $x = -1$ are verified zeroes of the polynomial $p(x) = x^2 - 1$.

Initial Setup & Theoretical Foundation

We are given the polynomial function:

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

We must verify whether the given values, $x = -1$ and $x = 2$, are zeroes of the polynomial.

[By the Fundamental Definition of a Zero of a Polynomial: A real number $a$ is a zero of a polynomial $p(x)$ if and only if evaluating the polynomial at $x = a$ yields zero, i.e., $p(a) = 0$.]

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

To determine if $x = -1$ is a zero, we substitute $-1$ for every instance of $x$ in the polynomial $p(x)$.

• Substitute $x = -1$:
  $p(-1) = (-1 + 1)(-1 - 2)$
• Simplify the terms within the parentheses:
  $p(-1) = (0)(-3)$
• Apply the Zero Property of Multiplication [Any number multiplied by zero equals zero]:
  $p(-1) = 0$

Since $p(-1) = 0$, we mathematically confirm that $x = -1$ is a zero of the polynomial $p(x)$.

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

Next, we apply the same substitution methodology for $x = 2$.

• Substitute $x = 2$:
  $p(2) = (2 + 1)(2 - 2)$
• Simplify the terms within the parentheses:
  $p(2) = (3)(0)$
• Apply the Zero Property of Multiplication:
  $p(2) = 0$

Since $p(2) = 0$, we mathematically confirm that $x = 2$ is also a zero of the polynomial $p(x)$.

Graphical Verification

The polynomial $p(x) = (x + 1)(x - 2)$ expands to the quadratic form $p(x) = x^2 - x - 2$. Geometrically, the zeroes of a polynomial correspond to the $x$-intercepts of its graph. The precise Cartesian plot below demonstrates the parabola intersecting the $x$-axis exactly at the coordinates $(-1, 0)$ and $(2, 0)$.

Final Solution: Yes, both $x = -1$ and $x = 2$ are zeroes of the polynomial $p(x) = (x + 1)(x - 2)$, as evaluating the polynomial at both values yields exactly $0$.

Given Variables & Initial Setup

We are given the following polynomial function and a specific value of the independent variable $x$:

• Polynomial: $p(x) = x^2$
• Indicated Value: $x = 0$

Theoretical Foundation: [Per the Fundamental Theorem of Algebra and Polynomial Theory], a real number $c$ is defined as a "zero" (or root) of a polynomial $p(x)$ if and only if evaluating the polynomial at $x = c$ yields a result of zero. Mathematically, this is expressed as the condition: $p(c) = 0$.

Step 1: Substitution of the Indicated Value

To verify whether $x = 0$ is a zero of the given polynomial, we must substitute $0$ for every instance of $x$ in the function $p(x)$.

Substituting $x = 0$ into $p(x) = x^2$:

$p(0) = (0)^2$

Step 2: Algebraic Evaluation

Next, we evaluate the exponentiation. The square of zero is the product of zero multiplied by itself.

$p(0) = 0 \times 0$

$p(0) = 0$

[By the Zero Product Property], any real number multiplied by zero results in zero. Thus, the polynomial evaluates exactly to $0$.

Step 3: Graphical Verification

Graphically, the zeroes of a polynomial correspond to the $x$-intercepts of its graph (the points where the graph crosses or touches the $x$-axis). The function $p(x) = x^2$ represents a standard upward-opening parabola. Because $x = 0$ yields $p(0) = 0$, the vertex of the parabola lies exactly at the origin $(0,0)$, confirming that $x=0$ is a root (specifically, a repeated root of multiplicity 2).

Final Conclusion

Because the evaluation of the polynomial at the given value satisfies the condition $p(0) = 0$, the indicated value is indeed a valid zero of the polynomial.

Final Solution: Yes, $x = 0$ is a zero of the polynomial $p(x) = x^2$.

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 $x = 0$.

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

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 & 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 $-6$.

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

Given Variables & Initial Setup

We are given a polynomial in a single variable, $x$. Let us define the polynomial as a function $P(x)$:

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

The objective is to evaluate this polynomial at the specific domain value $x = 2$. [Per the fundamental theorem of polynomial evaluation, finding the value of a polynomial at a given point requires the direct substitution of that point into the variable, followed by simplification according to the standard order of operations (PEMDAS/BODMAS)].

Step 1: Substitution of the Variable

Substitute $x = 2$ into every instance of $x$ within the polynomial $P(x)$:

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

Step 2: Resolution of Exponents

According to the order of operations, exponentiation must be resolved before multiplication. We evaluate the quadratic term $(2)^2$:

$2^2 = 2 \times 2 = 4$

Substituting this back into the equation yields:

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

Step 3: Execution of Multiplication

Next, perform the scalar multiplications for both the linear and quadratic terms:

• Linear term: $5 \times 2 = 10$
• Quadratic term: $4 \times 4 = 16$

Updating the polynomial expression:

$P(2) = 10 - 16 + 3$

Step 4: Sequential Addition and Subtraction

Finally, evaluate the arithmetic expression from left to right:

$P(2) = (10 - 16) + 3$

$P(2) = -6 + 3$

$P(2) = -3$

Graphical Verification

The polynomial $P(x) = -4x^2 + 5x + 3$ represents a downward-opening parabola [since the leading coefficient $a = -4$ is less than zero]. Evaluating the polynomial at $x = 2$ corresponds to finding the $y$-coordinate of the point on this parabola where the $x$-coordinate is $2$. As calculated, this point exists exactly at $(2, -3)$ on the Cartesian plane.

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

Given Polynomial & Theoretical Foundation

We are tasked with finding the zero of the following 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 specific real value of the variable $x$ for which the polynomial evaluates to zero. [Per the fundamental definition of polynomial roots]. Therefore, to find the zero, we must set the polynomial equal to zero:

$p(x) = 0$

Step 2: Algebraic Substitution and Manipulation

Substituting the given algebraic expression for $p(x)$ into our equation, we obtain:

$x - 5 = 0$

To isolate the variable $x$ and solve the linear equation, we add $5$ to both sides of the equation [Per the Addition Property of Equality, which states that adding the same number to both sides of an equation maintains its balance]:

$x - 5 + 5 = 0 + 5$

$x = 5$

Step 3: Rigorous Verification

To ensure absolute mathematical accuracy, we verify the solution by substituting $x = 5$ back into the original polynomial to check if it yields zero:

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

$p(5) = 0$

Since the evaluation results in exactly zero, $x = 5$ is confirmed as the correct zero of the polynomial.

Step 4: Graphical Interpretation

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 (where $y = 0$). For the linear function $y = x - 5$, the $x$-intercept occurs exactly at the Cartesian coordinate $(5, 0)$.

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