What should be subtracted from the polynomial , so that 15 is the zero of the resulting polynomial?

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 ($\alpha + \beta$) = $1$.

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: 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:

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 $x^2 - 4x + 1$.