Question 1

We will find the zeroes by factorizing each polynomial. Then, we will verify the sum and product of zeroes with the coefficients.

Step 1 — Find Zeroes for (i)

Let's set the polynomial to zero. We factorize the quadratic expression.

$x^2 - 2x - 8 = 0$

$x^2 - 4x + 2x - 8 = 0$

$x(x - 4) + 2(x - 4) = 0$

$(x - 4)(x + 2) = 0$

$x - 4 = 0 \text{ or } x + 2 = 0$

$x = 4 \text{ or } x = -2$

$\boxed{\text{Zeroes are } 4, -2}$

Step 2 — Verify Relationship for (i)

We identify the coefficients. For $x^2 - 2x - 8$, $a = \mathbf{1}$, $b = \mathbf{-2}$, $c = \mathbf{-8}$. Let the zeroes be $\alpha = \mathbf{4}$ and $\beta = \mathbf{-2}$.

Let's find the sum of the zeroes.

$\alpha + \beta = 4 + (-2)$

$= 2$

Now, let's find $-b/a$.

$-b/a = -(-2)/1$

$= 2$

The sum of zeroes matches.

Let's find the product of the zeroes.

$\alpha \cdot \beta = 4 \cdot (-2)$

$= -8$

Now, let's find $c/a$.

$c/a = -8/1$

$= -8$

$\boxed{\text{Sum of zeroes } = 2, \text{ Product of zeroes } = -8}$

Step 1 — Find Zeroes for (ii)

Let's set the polynomial to zero. We factorize the quadratic expression.

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

$(2s - 1)^2 = 0$

$2s - 1 = 0$

$2s = 1$

$s = 1/2$

$\boxed{\text{Zeroes are } 1/2, 1/2}$

Step 2 — Verify Relationship for (ii)

We identify the coefficients. For $4s^2 - 4s + 1$, $a = \mathbf{4}$, $b = \mathbf{-4}$, $c = \mathbf{1}$. Let the zeroes be $\alpha = \mathbf{1/2}$ and $\beta = \mathbf{1/2}$.

Let's find the sum of the zeroes.

$\alpha + \beta = 1/2 + 1/2$

$= 1$

Now, let's find $-b/a$.

$-b/a = -(-4)/4$

$= 1$

The sum of zeroes matches.

Let's find the product of the zeroes.

$\alpha \cdot \beta = (1/2) \cdot (1/2)$

$= 1/4$

Now, let's find $c/a$.

$c/a = 1/4$

$\boxed{\text{Sum of zeroes } = 1, \text{ Product of zeroes } = 1/4}$

Step 1 — Find Zeroes for (iii)

First, let's write the polynomial in standard form. $6x^2 - 7x - 3$. We set the polynomial to zero. We factorize the quadratic expression.

$6x^2 - 7x - 3 = 0$

$6x^2 - 9x + 2x - 3 = 0$

$3x(2x - 3) + 1(2x - 3) = 0$

$(3x + 1)(2x - 3) = 0$

$3x + 1 = 0 \text{ or } 2x - 3 = 0$

$3x = -1 \text{ or } 2x = 3$

$x = -1/3 \text{ or } x = 3/2$

$\boxed{\text{Zeroes are } -1/3, 3/2}$

Step 2 — Verify Relationship for (iii)

We identify the coefficients. For $6x^2 - 7x - 3$, $a = \mathbf{6}$, $b = \mathbf{-7}$, $c = \mathbf{-3}$. Let the zeroes be $\alpha = \mathbf{-1/3}$ and $\beta = \mathbf{3/2}$.

Let's find the sum of the zeroes.

$\alpha + \beta = -1/3 + 3/2$

$= -2/6 + 9/6$

$= 7/6$

Now, let's find $-b/a$.

$-b/a = -(-7)/6$

$= 7/6$

The sum of zeroes matches.

Let's find the product of the zeroes.

$\alpha \cdot \beta = (-1/3) \cdot (3/2)$

$= -3/6$

$= -1/2$

Now, let's find $c/a$.

$c/a = -3/6$

$= -1/2$

$\boxed{\text{Sum of zeroes } = 7/6, \text{ Product of zeroes } = -1/2}$

Step 1 — Find Zeroes for (iv)

Let's set the polynomial to zero. We factorize the quadratic expression.

$4u^2 + 8u = 0$

$4u(u + 2) = 0$

$4u = 0 \text{ or } u + 2 = 0$

$u = 0 \text{ or } u = -2$

$\boxed{\text{Zeroes are } 0, -2}$

Step 2 — Verify Relationship for (iv)

We identify the coefficients. For $4u^2 + 8u$, $a = \mathbf{4}$, $b = \mathbf{8}$, $c = \mathbf{0}$. Let the zeroes be $\alpha = \mathbf{0}$ and $\beta = \mathbf{-2}$.

Let's find the sum of the zeroes.

$\alpha + \beta = 0 + (-2)$

$= -2$

Now, let's find $-b/a$.

$-b/a = -8/4$

$= -2$

The sum of zeroes matches.

Let's find the product of the zeroes.

$\alpha \cdot \beta = 0 \cdot (-2)$

$= 0$

Now, let's find $c/a$.

$c/a = 0/4$

$= 0$

$\boxed{\text{Sum of zeroes } = -2, \text{ Product of zeroes } = 0}$

Step 1 — Find Zeroes for (v)

Let's set the polynomial to zero. We factorize the quadratic expression using difference of squares.

$t^2 - 15 = 0$

$t^2 - (\sqrt{15})^2 = 0$

$(t - \sqrt{15})(t + \sqrt{15}) = 0$

$t - \sqrt{15} = 0 \text{ or } t + \sqrt{15} = 0$

$t = \sqrt{15} \text{ or } t = -\sqrt{15}$

$\boxed{\text{Zeroes are } \sqrt{15}, -\sqrt{15}}$

Step 2 — Verify Relationship for (v)

We identify the coefficients. For $t^2 - 15$, $a = \mathbf{1}$, $b = \mathbf{0}$, $c = \mathbf{-15}$. Let the zeroes be $\alpha = \mathbf{\sqrt{15}}$ and $\beta = \mathbf{-\sqrt{15}}$.

Let's find the sum of the zeroes.

$\alpha + \beta = \sqrt{15} + (-\sqrt{15})$

$= 0$

Now, let's find $-b/a$.

$-b/a = -0/1$

$= 0$

The sum of zeroes matches.

Let's find the product of the zeroes.

$\alpha \cdot \beta = (\sqrt{15}) \cdot (-\sqrt{15})$

$= -15$

Now, let's find $c/a$.

$c/a = -15/1$

$= -15$

$\boxed{\text{Sum of zeroes } = 0, \text{ Product of zeroes } = -15}$

Step 1 — Find Zeroes for (vi)

Let's set the polynomial to zero. We factorize the quadratic expression.

$3x^2 - x - 4 = 0$

$3x^2 - 4x + 3x - 4 = 0$

$x(3x - 4) + 1(3x - 4) = 0$

$(x + 1)(3x - 4) = 0$

$x + 1 = 0 \text{ or } 3x - 4 = 0$

$x = -1 \text{ or } 3x = 4$

$x = -1 \text{ or } x = 4/3$

$\boxed{\text{Zeroes are } -1, 4/3}$

Step 2 — Verify Relationship for (vi)

We identify the coefficients. For $3x^2 - x - 4$, $a = \mathbf{3}$, $b = \mathbf{-1}$, $c = \mathbf{-4}$. Let the zeroes be $\alpha = \mathbf{-1}$ and $\beta = \mathbf{4/3}$.

Let's find the sum of the zeroes.

$\alpha + \beta = -1 + 4/3$

$= -3/3 + 4/3$

$= 1/3$

Now, let's find $-b/a$.

$-b/a = -(-1)/3$

$= 1/3$

The sum of zeroes matches.

Let's find the product of the zeroes.

$\alpha \cdot \beta = (-1) \cdot (4/3)$

$= -4/3$

Now, let's find $c/a$.

$c/a = -4/3$

$\boxed{\text{Sum of zeroes } = 1/3, \text{ Product of zeroes } = -4/3}$

Answer

(i) Zeroes are $4, -2$. Sum of zeroes $2 = -(-2)/1$. Product of zeroes $-8 = -8/1$. (ii) Zeroes are $1/2, 1/2$. Sum of zeroes $1 = -(-4)/4$. Product of zeroes $1/4 = 1/4$. (iii) Zeroes are $-1/3, 3/2$. Sum of zeroes $7/6 = -(-7)/6$. Product of zeroes $-1/2 = -3/6$. (iv) Zeroes are $0, -2$. Sum of zeroes $-2 = -8/4$. Product of zeroes $0 = 0/4$. (v) Zeroes are $\sqrt{15}, -\sqrt{15}$. Sum of zeroes $0 = -0/1$. Product of zeroes $-15 = -15/1$. (vi) Zeroes are $-1, 4/3$. Sum of zeroes $1/3 = -(-1)/3$. Product of zeroes $-4/3 = -4/3$.