Q1
If the zeroes of polynomial $x^3 – 3x^2 + x + 1$ are $a – b$, $a$, $a + b$, find $a$ and $b$.
Step 1: Identify Coefficients and Zeroes
Let the polynomial be $p(x) = x^3 – 3x^2 + x + 1$.
Comparing with $Ax^3 + Bx^2 + Cx + D$:
$$A = 1, \quad B = -3, \quad C = 1, \quad D = 1$$
Given zeroes: $\alpha = a – b$, $\beta = a$, $\gamma = a + b$.
Step 2: Use Sum of Zeroes
Sum of zeroes ($\alpha + \beta + \gamma$) = $-\frac{B}{A}$
$$(a – b) + a + (a + b) = -\frac{-3}{1}$$
$$3a = 3 \implies a = 1$$
Step 3: Use Product of Zeroes
Product of zeroes ($\alpha \beta \gamma$) = $-\frac{D}{A}$
$$(a – b)(a)(a + b) = -\frac{1}{1}$$
$$a(a^2 – b^2) = -1$$
Substitute $a = 1$:
$$1(1^2 – b^2) = -1 \implies 1 – b^2 = -1$$
$$b^2 = 2 \implies b = \pm \sqrt{2}$$
✔️ $a = 1$ and $b = \pm \sqrt{2}$