Question 2

A quadratic polynomial can be formed using the sum and product of its zeroes.

Step 1 — Polynomial for $\frac{1}{4}, -1$ We use the general form for a quadratic polynomial. The general form is $P(x) = x^2 - (\text{sum})x + (\text{product})$. Let's substitute the given values.

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

$= x^2 - \frac{1}{4}x - 1$ We can multiply by a constant to remove fractions. Let's multiply by 4.

$P(x) = 4 \left(x^2 - \frac{1}{4}x - 1\right)$

$= 4x^2 - x - 4$

$\boxed{4x^2 - x - 4}$

Step 2 — Polynomial for $\sqrt{2}, \frac{1}{3}$ We use the general form for a quadratic polynomial. The general form is $P(x) = x^2 - (\text{sum})x + (\text{product})$. Let's substitute the given values.

$P(x) = x^2 - (\sqrt{2})x + \left(\frac{1}{3}\right)$

$= x^2 - \sqrt{2}x + \frac{1}{3}$ We can multiply by a constant to remove fractions. Let's multiply by 3.

$P(x) = 3 \left(x^2 - \sqrt{2}x + \frac{1}{3}\right)$

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

$\boxed{3x^2 - 3\sqrt{2}x + 1}$

Step 3 — Polynomial for $0, \sqrt{5}$ We use the general form for a quadratic polynomial. The general form is $P(x) = x^2 - (\text{sum})x + (\text{product})$. Let's substitute the given values.

$P(x) = x^2 - (0)x + (\sqrt{5})$

$= x^2 + \sqrt{5}$ This is our polynomial.

$\boxed{x^2 + \sqrt{5}}$

Step 4 — Polynomial for $1, 1$ We use the general form for a quadratic polynomial. The general form is $P(x) = x^2 - (\text{sum})x + (\text{product})$. Let's substitute the given values.

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

$= x^2 - x + 1$ This is our polynomial.

$\boxed{x^2 - x + 1}$

Step 5 — Polynomial for $-\frac{1}{4}, \frac{1}{4}$ We use the general form for a quadratic polynomial. The general form is $P(x) = x^2 - (\text{sum})x + (\text{product})$. Let's substitute the given values.

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

$= x^2 + \frac{1}{4}x + \frac{1}{4}$ We can multiply by a constant to remove fractions. Let's multiply by 4.

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

$= 4x^2 + x + 1$

$\boxed{4x^2 + x + 1}$

Step 6 — Polynomial for $4, 1$ We use the general form for a quadratic polynomial. The general form is $P(x) = x^2 - (\text{sum})x + (\text{product})$. Let's substitute the given values.

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

$= x^2 - 4x + 1$ This is our polynomial.

$\boxed{x^2 - 4x + 1}$

Answer

(i) $4x^2 - x - 4$ (ii) $3x^2 - 3\sqrt{2}x + 1$ (iii) $x^2 + \sqrt{5}$ (iv) $x^2 - x + 1$ (v) $4x^2 + x + 1$ (vi) $x^2 - 4x + 1$