What should be added to the polynomial , so that 3 is the zero of the resulting polynomial?

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 $0$.