Given: A quadratic polynomial $p(s) = 4s^2 - 4s + 1$.

To Find:
1. The zeroes of the polynomial $p(s)$.
2. Verification of the relationship between the zeroes and the coefficients of the polynomial.

Step 1: Finding the zeroes of the polynomial

To find the zeroes of $p(s)$, we set $p(s) = 0$.
$4s^2 - 4s + 1 = 0$

We use the splitting the middle term method. We look for two numbers whose product is $4 \times 1 = 4$ and whose sum is $-4$. These numbers are $-2$ and $-2$.

$4s^2 - 2s - 2s + 1 = 0$

Now, factor by grouping:
$2s(2s - 1) - 1(2s - 1) = 0$
$(2s - 1)(2s - 1) = 0$

Setting each factor to zero:
$2s - 1 = 0 \implies s = \frac{1}{2}$
$2s - 1 = 0 \implies s = \frac{1}{2}$

Thus, the zeroes of the polynomial are $\alpha = \frac{1}{2}$ and $\beta = \frac{1}{2}$.

Step 2: Identifying coefficients

Comparing $4s^2 - 4s + 1$ with the standard form $as^2 + bs + c$:
$a = 4$
$b = -4$
$c = 1$

Step 3: Verification of the relationship between zeroes and coefficients

The relationship states:
1. Sum of zeroes ($\alpha + \beta$) = $-\frac{b}{a}$
2. Product of zeroes ($\alpha \cdot \beta$) = $\frac{c}{a}$

Verification of Sum of Zeroes:
$\alpha + \beta = \frac{1}{2} + \frac{1}{2} = 1$
$-\frac{b}{a} = -\frac{(-4)}{4} = \frac{4}{4} = 1$
Since $1 = 1$, the sum of zeroes is verified.

Verification of Product of Zeroes:
$\alpha \cdot \beta = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$
$\frac{c}{a} = \frac{1}{4}$
Since $\frac{1}{4} = \frac{1}{4}$, the product of zeroes is verified.

Final Answer: The zeroes of the polynomial $4s^2 - 4s + 1$ are $\frac{1}{2}$ and $\frac{1}{2}$. The relationship between the zeroes and coefficients is verified as the sum of zeroes is $1$ and the product of zeroes is $\frac{1}{4}$.

Given: A quadratic polynomial $p(u) = 4u^2 + 8u$.

To Find: The zeroes of the polynomial and verify the relationship between the zeroes and the coefficients of the polynomial.

Step 1: Finding the zeroes of the polynomial

To find the zeroes of the polynomial $p(u)$, we set $p(u) = 0$.

$4u^2 + 8u = 0$

We factor out the greatest common factor, which is $4u$:

$4u(u + 2) = 0$ [Using the distributive property of multiplication over addition]

For the product to be zero, either $4u = 0$ or $u + 2 = 0$ [Zero Product Property].

Case 1: $4u = 0 \implies u = 0$

Case 2: $u + 2 = 0 \implies u = -2$

Thus, the zeroes of the polynomial are $\alpha = 0$ and $\beta = -2$.

Step 2: Identifying coefficients

Comparing the given polynomial $4u^2 + 8u$ with the standard form $au^2 + bu + c$, we have:

$a = 4$

$b = 8$

$c = 0$

Step 3: Verifying the relationship between zeroes and coefficients

The relationships to verify are:

1. Sum of zeroes ($\alpha + \beta$) = $-\frac{b}{a}$

2. Product of zeroes ($\alpha \cdot \beta$) = $\frac{c}{a}$

Verification of Sum of Zeroes:

Sum of zeroes = $\alpha + \beta = 0 + (-2) = -2$

$-\frac{b}{a} = -\frac{8}{4} = -2$

Since $-2 = -2$, the relationship $\alpha + \beta = -\frac{b}{a}$ is verified.

Verification of Product of Zeroes:

Product of zeroes = $\alpha \cdot \beta = 0 \cdot (-2) = 0$

$\frac{c}{a} = \frac{0}{4} = 0$

Since $0 = 0$, the relationship $\alpha \cdot \beta = \frac{c}{a}$ is verified.

Final Answer: The zeroes of the polynomial $4u^2 + 8u$ are $0$ and $-2$. The relationship between the zeroes and coefficients is verified as the sum of zeroes is $-2$ and the product of zeroes is $0$.

Given:

The sum of the zeroes of the quadratic polynomial ($\alpha + \beta$) = $0$.

The product of the zeroes of the quadratic polynomial ($\alpha \cdot \beta$) = $\sqrt{5}$.

To Find:

A quadratic polynomial $p(x)$ that satisfies the given conditions.

Step 1: Understanding the Relationship between Zeroes and Coefficients

For any quadratic polynomial of the form $ax^2 + bx + c$, where $a \neq 0$, the relationship between its zeroes ($\alpha$ and $\beta$) and its coefficients is given by:

Sum of zeroes ($\alpha + \beta$) = $-\frac{b}{a}$

Product of zeroes ($\alpha \cdot \beta$) = $\frac{c}{a}$

Step 2: Formulating the General Quadratic Polynomial

A quadratic polynomial can be expressed in terms of the sum and product of its zeroes using the following standard formula:

$p(x) = k[x^2 - (\text{sum of zeroes})x + (\text{product of zeroes})]$

where $k$ is a non-zero real constant.

Step 3: Substituting the Given Values

Substitute the given values into the formula:

Sum of zeroes = $0$

Product of zeroes = $\sqrt{5}$

$p(x) = k[x^2 - (0)x + (\sqrt{5})]$

Step 4: Simplifying the Expression

Perform the arithmetic operations inside the brackets:

$p(x) = k[x^2 - 0 + \sqrt{5}]$

$p(x) = k[x^2 + \sqrt{5}]$

Step 5: Determining the Polynomial

By choosing the simplest constant $k = 1$, we obtain the quadratic polynomial:

$p(x) = x^2 + \sqrt{5}$

Verification:

For $p(x) = x^2 + 0x + \sqrt{5}$:

Sum of zeroes = $-\frac{b}{a} = -\frac{0}{1} = 0$ (Matches given)

Product of zeroes = $\frac{c}{a} = \frac{\sqrt{5}}{1} = \sqrt{5}$ (Matches given)

Final Answer: The required quadratic polynomial is $x^2 + \sqrt{5}$.

Given: A quadratic polynomial $p(x) = 3x^2 - x - 4$.

To Find: The zeroes of the polynomial and verify the relationship between the zeroes and the coefficients of the polynomial.

Step 1: Finding the zeroes of the polynomial
To find the zeroes, we set $p(x) = 0$.
$3x^2 - x - 4 = 0$

We use the splitting the middle term method. We need two numbers whose product is $3 \times (-4) = -12$ and whose sum is $-1$. These numbers are $-4$ and $3$.

$3x^2 - 4x + 3x - 4 = 0$ [Splitting the middle term $-x$ into $-4x + 3x$]
$x(3x - 4) + 1(3x - 4) = 0$ [Factoring by grouping]
$(3x - 4)(x + 1) = 0$ [Taking $(3x - 4)$ as a common factor]

Setting each factor to zero:
1) $3x - 4 = 0 \implies 3x = 4 \implies x = \frac{4}{3}$
2) $x + 1 = 0 \implies x = -1$

Thus, the zeroes are $\alpha = \frac{4}{3}$ and $\beta = -1$.

Step 2: Identifying coefficients
Comparing $3x^2 - x - 4$ with the standard form $ax^2 + bx + c$:
$a = 3$
$b = -1$
$c = -4$

Step 3: Verifying the relationship between zeroes and coefficients
The relationships to verify are:
1) Sum of zeroes ($\alpha + \beta$) = $-\frac{b}{a}$
2) Product of zeroes ($\alpha \cdot \beta$) = $\frac{c}{a}$

Verification of Sum of Zeroes:
LHS: $\alpha + \beta = \frac{4}{3} + (-1) = \frac{4}{3} - \frac{3}{3} = \frac{1}{3}$
RHS: $-\frac{b}{a} = -\frac{(-1)}{3} = \frac{1}{3}$
Since LHS = RHS, the relationship is verified.

Verification of Product of Zeroes:
LHS: $\alpha \cdot \beta = \left(\frac{4}{3}\right) \cdot (-1) = -\frac{4}{3}$
RHS: $\frac{c}{a} = \frac{-4}{3} = -\frac{4}{3}$
Since LHS = RHS, the relationship is verified.

Final Answer: The zeroes of the polynomial $3x^2 - x - 4$ are $\frac{4}{3}$ and $-1$. The relationship between the zeroes and coefficients is verified as the sum of zeroes is $\frac{1}{3}$ and the product of zeroes is $-\frac{4}{3}$.

Given: A quadratic polynomial $p(t) = t^2 - 15$.

To Find: The zeroes of the polynomial and verify the relationship between the zeroes and the coefficients of the polynomial.

Step 1: Finding the zeroes of the polynomial
To find the zeroes of the polynomial $p(t)$, we set $p(t) = 0$.
$t^2 - 15 = 0$
We can rewrite this equation using the algebraic identity $a^2 - b^2 = (a - b)(a + b)$.
Since $15 = (\sqrt{15})^2$, we have:
$t^2 - (\sqrt{15})^2 = 0$
$(t - \sqrt{15})(t + \sqrt{15}) = 0$
[By the Zero Product Property, if $ab = 0$, then $a=0$ or $b=0$]
Therefore, $t - \sqrt{15} = 0$ or $t + \sqrt{15} = 0$.
$t = \sqrt{15}$ or $t = -\sqrt{15}$.
The zeroes of the polynomial are $\alpha = \sqrt{15}$ and $\beta = -\sqrt{15}$.

Step 2: Identifying coefficients
Comparing the given polynomial $t^2 - 15$ with the standard quadratic form $at^2 + bt + c$:
$a = 1$
$b = 0$ (since there is no $t$ term)
$c = -15$

Step 3: Verifying the relationship between zeroes and coefficients
The relationships to verify are:
1. Sum of zeroes ($\alpha + \beta$) = $-\frac{b}{a}$
2. Product of zeroes ($\alpha \cdot \beta$) = $\frac{c}{a}$

Verification of Sum of Zeroes:
Sum of zeroes = $\alpha + \beta = \sqrt{15} + (-\sqrt{15}) = 0$
Coefficient ratio = $-\frac{b}{a} = -\frac{0}{1} = 0$
Since $0 = 0$, the relationship is verified.

Verification of Product of Zeroes:
Product of zeroes = $\alpha \cdot \beta = (\sqrt{15}) \cdot (-\sqrt{15}) = -(\sqrt{15})^2 = -15$
Coefficient ratio = $\frac{c}{a} = \frac{-15}{1} = -15$
Since $-15 = -15$, the relationship is verified.

Final Answer: The zeroes of the polynomial $t^2 - 15$ are $\sqrt{15}$ and $-\sqrt{15}$. The relationship between the zeroes and coefficients is verified as the sum of zeroes is $0$ and the product of zeroes is $-15$.

Given:

The sum of the zeroes of the quadratic polynomial ($\alpha + \beta$) = $4$.

The product of the zeroes of the quadratic polynomial ($\alpha \cdot \beta$) = $1$.

To Find:

A quadratic polynomial $p(x)$ that satisfies the given conditions.

Step 1: Understanding the Relationship between Zeroes and Coefficients

For any quadratic polynomial of the form $ax^2 + bx + c$, where $a \neq 0$, the relationship between the zeroes ($\alpha, \beta$) and the coefficients is given by the following standard formulas:

Sum of zeroes: $\alpha + \beta = -\frac{b}{a}$

Product of zeroes: $\alpha \cdot \beta = \frac{c}{a}$

Step 2: Formulating the General Quadratic Polynomial

A quadratic polynomial can be expressed in terms of the sum and product of its zeroes using the following identity:

$p(x) = k[x^2 - (\text{sum of zeroes})x + (\text{product of zeroes})]$

where $k$ is any non-zero real constant.

Step 3: Substituting the Given Values

Substitute the given values into the identity:

Sum of zeroes = $4$

Product of zeroes = $1$

$p(x) = k[x^2 - (4)x + (1)]$

Step 4: Simplifying the Expression

By choosing the simplest case where $k = 1$, we obtain the polynomial:

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

Justification:

If we verify the zeroes of $p(x) = x^2 - 4x + 1$:

Sum of zeroes = $-\frac{b}{a} = -\frac{-4}{1} = 4$ [Matches the given sum]

Product of zeroes = $\frac{c}{a} = \frac{1}{1} = 1$ [Matches the given product]

Final Answer: The required quadratic polynomial is $x^2 - 4x + 1$.

Given:

The sum of the zeroes of the quadratic polynomial ($\alpha + \beta$) = $1$.

The product of the zeroes of the quadratic polynomial ($\alpha \cdot \beta$) = $1$.

To Find:

A quadratic polynomial $p(x)$ of the form $ax^2 + bx + c$, where $a \neq 0$.

Step 1: Understanding the Relationship between Zeroes and Coefficients

For any quadratic polynomial $p(x) = ax^2 + bx + c$ with zeroes $\alpha$ and $\beta$, the following relationships hold true:

Sum of zeroes: $\alpha + \beta = -\frac{b}{a}$

Product of zeroes: $\alpha \cdot \beta = \frac{c}{a}$

Step 2: Formulating the General Quadratic Polynomial

A quadratic polynomial can be expressed in terms of the sum and product of its zeroes using the following identity:

$p(x) = k[x^2 - (\text{sum of zeroes})x + (\text{product of zeroes})]$

where $k$ is a non-zero real constant.

Step 3: Substituting the Given Values

Substitute the given values into the identity:

Sum of zeroes = $1$

Product of zeroes = $1$

$p(x) = k[x^2 - (1)x + (1)]$

$p(x) = k[x^2 - x + 1]$

Step 4: Determining the Polynomial

To obtain the simplest form of the polynomial, we assume $k = 1$ (as $k$ can be any non-zero real number, choosing $1$ provides the standard representation).

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

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

Verification (Optional):

For $p(x) = x^2 - x + 1$, here $a=1, b=-1, c=1$.

Sum of zeroes = $-\frac{b}{a} = -\frac{-1}{1} = 1$ (Matches given).

Product of zeroes = $\frac{c}{a} = \frac{1}{1} = 1$ (Matches given).

Final Answer: The required quadratic polynomial is $x^2 - x + 1$.

Given:

The sum of the zeroes of the quadratic polynomial ($\alpha + \beta$) = $\frac{1}{4}$

The product of the zeroes of the quadratic polynomial ($\alpha \cdot \beta$) = $-1$

To Find:

A quadratic polynomial $p(x)$ that satisfies the given conditions.

Step 1: Understanding the Relationship between Zeroes and Coefficients

A quadratic polynomial in the variable $x$ is generally expressed in the form $p(x) = ax^2 + bx + c$, where $a \neq 0$.

According to the relationship between the zeroes and coefficients of a quadratic polynomial:

Sum of zeroes ($\alpha + \beta$) = $-\frac{b}{a}$

Product of zeroes ($\alpha \cdot \beta$) = $\frac{c}{a}$

Step 2: Formulating the Polynomial

Any quadratic polynomial can be expressed in terms of its sum and product of zeroes using the following identity:

$p(x) = k[x^2 - (\text{sum of zeroes})x + (\text{product of zeroes})]$

where $k$ is a non-zero real constant.

Step 3: Substituting the Given Values

Substitute the given values into the identity:

$p(x) = k[x^2 - (\frac{1}{4})x + (-1)]$

$p(x) = k[x^2 - \frac{1}{4}x - 1]$

Step 4: Simplifying the Expression

To obtain a polynomial with integer coefficients, we can choose a suitable value for the constant $k$. Let $k = 4$ (the denominator of the fraction):

$p(x) = 4[x^2 - \frac{1}{4}x - 1]$

Distribute the $4$ across the terms inside the bracket:

$p(x) = 4(x^2) - 4(\frac{1}{4}x) - 4(1)$

$p(x) = 4x^2 - 1x - 4$

$p(x) = 4x^2 - x - 4$

Step 5: Verification (Optional but Recommended)

For $p(x) = 4x^2 - x - 4$:

Sum of zeroes = $-\frac{b}{a} = -(\frac{-1}{4}) = \frac{1}{4}$ [Matches the given sum]

Product of zeroes = $\frac{c}{a} = \frac{-4}{4} = -1$ [Matches the given product]

Final Answer: The required quadratic polynomial is $4x^2 - x - 4$.

Given: The sum of the zeroes ($\alpha + \beta$) is $-\frac{1}{4}$ and the product of the zeroes ($\alpha \cdot \beta$) is $\frac{1}{4}$.

To Find: A quadratic polynomial $p(x)$ that satisfies the given conditions.

Step 1: Understanding the Relationship between Zeroes and Coefficients

For any quadratic polynomial of the form $ax^2 + bx + c$, where $a \neq 0$, the relationship between the zeroes ($\alpha, \beta$) and the coefficients is given by the following formulas:

Sum of zeroes: $\alpha + \beta = -\frac{b}{a}$

Product of zeroes: $\alpha \cdot \beta = \frac{c}{a}$

Step 2: Formulating the General Quadratic Polynomial

A quadratic polynomial can be expressed in terms of the sum and product of its zeroes using the following identity:

$p(x) = k[x^2 - (\text{sum of zeroes})x + (\text{product of zeroes})]$

where $k$ is a non-zero real constant.

Step 3: Substituting the Given Values

Substitute the given values $\alpha + \beta = -\frac{1}{4}$ and $\alpha \cdot \beta = \frac{1}{4}$ into the identity:

$p(x) = k\left[x^2 - \left(-\frac{1}{4}\right)x + \left(\frac{1}{4}\right)\right]$

[Simplifying the signs: $-(- \frac{1}{4}) = +\frac{1}{4}$]

$p(x) = k\left[x^2 + \frac{1}{4}x + \frac{1}{4}\right]$

Step 4: Simplifying the Expression

To obtain a polynomial with integer coefficients, we can choose $k = 4$ (or any multiple of the denominator):

$p(x) = 4\left[x^2 + \frac{1}{4}x + \frac{1}{4}\right]$

Distributing the $4$ across the terms inside the brackets:

$p(x) = 4(x^2) + 4\left(\frac{1}{4}x\right) + 4\left(\frac{1}{4}\right)$

$p(x) = 4x^2 + x + 1$

Final Answer: The required quadratic polynomial is $4x^2 + x + 1$.

Given: A quadratic polynomial $p(x) = x^2 - 2x - 8$.

To Find: The zeroes of the polynomial and verify the relationship between the zeroes and the coefficients of the polynomial.

Step 1: Finding the zeroes of the polynomial by factorization.

To find the zeroes of $p(x)$, we set $p(x) = 0$. Thus, we solve the quadratic equation:

We use the splitting the middle term method. We look for two numbers whose product is the product of the coefficient of $x^2$ ($1$) and the constant term ($-8$), which is $-8$, and whose sum is the coefficient of $x$ ($-2$).

The factors of $-8$ are: $(1, -8), (-1, 8), (2, -4), (-2, 4)$.

The pair that sums to $-2$ is $(2, -4)$.

Rewriting the middle term:

Grouping the terms:

Factoring out the common terms from each group:

Setting each factor to zero:

Therefore, the zeroes of the polynomial are $\alpha = 4$ and $\beta = -2$.

Step 2: Verifying the relationship between zeroes and coefficients.

For a quadratic polynomial of the form $ax^2 + bx + c$, the relationships are:

1. Sum of zeroes ($\alpha + \beta$) = $-\frac{b}{a}$

2. Product of zeroes ($\alpha \cdot \beta$) = $\frac{c}{a}$

Comparing $x^2 - 2x - 8$ with $ax^2 + bx + c$, we have:

Verification of Sum of Zeroes:

Sum of zeroes = $\alpha + \beta = 4 + (-2) = 2$

$-\frac{b}{a} = -\frac{-2}{1} = 2$

Since $2 = 2$, the relationship is verified.

Verification of Product of Zeroes:

Product of zeroes = $\alpha \cdot \beta = 4 \cdot (-2) = -8$

$\frac{c}{a} = \frac{-8}{1} = -8$

Since $-8 = -8$, the relationship is verified.

Final Answer: The zeroes of the polynomial $x^2 - 2x - 8$ are $4$ and $-2$. The relationship between the zeroes and coefficients is verified as the sum of zeroes is $2$ and the product of zeroes is $-8$.

Given:

The sum of the zeroes of the quadratic polynomial ($\alpha + \beta$) = $\sqrt{2}$

The product of the zeroes of the quadratic polynomial ($\alpha \cdot \beta$) = $\frac{1}{3}$

To Find:

A quadratic polynomial $p(x)$ that satisfies the given conditions.

Step 1: Understanding the Relationship between Zeroes and Coefficients

For any quadratic polynomial of the form $ax^2 + bx + c$, where $a \neq 0$, the relationship between the zeroes ($\alpha$ and $\beta$) and the coefficients is given by the following formulas:

Sum of zeroes: $\alpha + \beta = -\frac{b}{a}$

Product of zeroes: $\alpha \cdot \beta = \frac{c}{a}$

Step 2: Formulating the General Quadratic Polynomial

A quadratic polynomial can be expressed in terms of the sum and product of its zeroes as follows:

$p(x) = k[x^2 - (\text{sum of zeroes})x + (\text{product of zeroes})]$

where $k$ is a non-zero real constant.

Step 3: Substituting the Given Values

Substitute the given values into the formula:

$p(x) = k[x^2 - (\sqrt{2})x + (\frac{1}{3})]$

Step 4: Simplifying the Expression

To express the polynomial with integer coefficients, we can choose a value for $k$ that clears the denominator. Let $k = 3$:

$p(x) = 3[x^2 - \sqrt{2}x + \frac{1}{3}]$

Distribute the $3$ across the terms inside the brackets:

$p(x) = 3 \cdot x^2 - 3 \cdot \sqrt{2}x + 3 \cdot \frac{1}{3}$

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

Step 5: Verification (Optional but Recommended)

For the polynomial $3x^2 - 3\sqrt{2}x + 1$:

Sum of zeroes = $-\frac{b}{a} = -\frac{-3\sqrt{2}}{3} = \sqrt{2}$ [Matches the given sum]

Product of zeroes = $\frac{c}{a} = \frac{1}{3}$ [Matches the given product]

Final Answer: The required quadratic polynomial is $3x^2 - 3\sqrt{2}x + 1$.

Given: A quadratic polynomial $p(x) = 6x^2 - 3 - 7x$.

To Find: The zeroes of the polynomial and verify the relationship between the zeroes and the coefficients.

Step 1: Standardizing the Polynomial
The standard form of a quadratic polynomial is $ax^2 + bx + c$. We rearrange the given polynomial in descending order of the powers of $x$:
$p(x) = 6x^2 - 7x - 3$

Step 2: Finding the Zeroes by Factorization
To find the zeroes, we set $p(x) = 0$.
$6x^2 - 7x - 3 = 0$
We use the splitting the middle term method. We look for two numbers whose product is $a \times c = 6 \times (-3) = -18$ and whose sum is $b = -7$.
The factors of $-18$ that sum to $-7$ are $-9$ and $2$.
$6x^2 - 9x + 2x - 3 = 0$
Group the terms:
$(6x^2 - 9x) + (2x - 3) = 0$
Factor out the common terms:
$3x(2x - 3) + 1(2x - 3) = 0$
$(3x + 1)(2x - 3) = 0$

Step 3: Solving for $x$
Setting each factor to zero:
1) $3x + 1 = 0 \implies 3x = -1 \implies x = -\frac{1}{3}$
2) $2x - 3 = 0 \implies 2x = 3 \implies x = \frac{3}{2}$
Thus, the zeroes are $\alpha = \frac{3}{2}$ and $\beta = -\frac{1}{3}$.

Step 4: Verifying the Relationship between Zeroes and Coefficients
For the polynomial $6x^2 - 7x - 3$, the coefficients are $a = 6$, $b = -7$, and $c = -3$.

Verification of Sum of Zeroes:
Formula: $\alpha + \beta = -\frac{b}{a}$
LHS: $\alpha + \beta = \frac{3}{2} + (-\frac{1}{3}) = \frac{9 - 2}{6} = \frac{7}{6}$
RHS: $-\frac{b}{a} = -\frac{-7}{6} = \frac{7}{6}$
Since LHS = RHS, the relationship is verified.

Verification of Product of Zeroes:
Formula: $\alpha \cdot \beta = \frac{c}{a}$
LHS: $\alpha \cdot \beta = (\frac{3}{2}) \cdot (-\frac{1}{3}) = -\frac{3}{6} = -\frac{1}{2}$
RHS: $\frac{c}{a} = \frac{-3}{6} = -\frac{1}{2}$
Since LHS = RHS, the relationship is verified.

Final Answer: The zeroes of the polynomial are $\frac{3}{2}$ and $-\frac{1}{3}$. The relationship between the zeroes and coefficients is verified as the sum of zeroes is $\frac{7}{6}$ and the product of zeroes is $-\frac{1}{2}$.