Question

3. If α and β are the zeroes of a quadratic polynomial x2 + x – 2 then find the value of α 3+ β3

Answer 1

$p(x) = {x}^{2} + x - 2$

$p(1) = {(1)}^{2} + (1) - 2 \\ = > 1 + 1 - 2 \\ = > 2 - 2 \\ = > 0$

$q(x) = (x - 1)$

$p(x) \div q(x)$

$= > \\ x - 1) {x}^{2} + x - 2(x + 2\\ \: \: \: \: \: \: \: \: \: \: \: \: {x}^{2} - x \\ \: \: \: \: \: \: \: \: \: \: \: \: - - - - - \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 2x - 2 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 2x - 2 \\ \: \: \: \: \: \: \: \: \: \: \: \: - - - - - \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 0$

$\alpha = 1 \: and \: \beta = - 2$

$\: \: \: \: \: { \alpha }^{3} + { \beta }^{3} \\ = {( 1)}^{3} + {( - 2)}^{3} \\ = 1 - 8 \\ = - 7$

I think this will help you