Sure, let's find the quadratic polynomials for the given zeroes step by step.
### Case (i): Zeroes are [tex]\( \sqrt{3} \)[/tex] and [tex]\( -\sqrt{3} \)[/tex]
1. Form the polynomial from roots:
The roots of the polynomial are [tex]\( \sqrt{3} \)[/tex] and [tex]\( -\sqrt{3} \)[/tex]. A polynomial with these roots can be written in the factored form:
[tex]\[
(x - \sqrt{3})(x + \sqrt{3})
\][/tex]
2. Expand the factors:
To find the quadratic polynomial, we need to expand the product of these binomials:
[tex]\[
(x - \sqrt{3})(x + \sqrt{3}) = x^2 - (\sqrt{3})^2
\][/tex]
3. Simplify the expression:
Simplify the quadratic term:
[tex]\[
x^2 - (\sqrt{3})^2 = x^2 - 3
\][/tex]
Thus, the quadratic polynomial for the zeroes [tex]\( \sqrt{3} \)[/tex] and [tex]\( -\sqrt{3} \)[/tex] is:
[tex]\[
x^2 - 3
\][/tex]
### Case (ii): Zeroes are [tex]\( \frac{1}{2} \)[/tex] and [tex]\( \frac{3}{2} \)[/tex]
1. Form the polynomial from roots:
The roots of the polynomial are [tex]\( \frac{1}{2} \)[/tex] and [tex]\( \frac{3}{2} \)[/tex]. A polynomial with these roots can be written in the factored form:
[tex]\[
(x - \frac{1}{2})(x - \frac{3}{2})
\][/tex]
2. Convert to common factor form:
To facilitate expansion, we can rewrite the factors by converting to a common factor:
[tex]\[
(2x - 1)(2x - 3)
\][/tex]
3. Expand the factors:
Expand the product of the binomials:
[tex]\[
(2x - 1)(2x - 3) = 4x^2 - 6x - 2x + 3
\][/tex]
4. Combine like terms:
Combine the terms to get the standard quadratic form:
[tex]\[
4x^2 - 8x + 3
\][/tex]
Thus, the quadratic polynomial for the zeroes [tex]\( \frac{1}{2} \)[/tex] and [tex]\( \frac{3}{2} \)[/tex] is:
[tex]\[
4x^2 - 8x + 3
\][/tex]
### Summary
The resulting quadratic polynomials for the given roots are:
1. For zeroes [tex]\( \sqrt{3} \)[/tex] and [tex]\( -\sqrt{3} \)[/tex]:
[tex]\[
x^2 - 3
\][/tex]
2. For zeroes [tex]\( \frac{1}{2} \)[/tex] and [tex]\( \frac{3}{2} \)[/tex]:
[tex]\[
4x^2 - 8x + 3
\][/tex]
