Question

if the zeroes of the polynomial x^3-3x^2+x+1 are a-b, a, a+b, find a and b.​

Answer 1

       

Hi!!!

Please look below ↓↓↓↓↓↓

   

As zeroes of p(x) = x³ - 3x² + x + 1 = 0 are a - b, a, and a + b, then (x - a), (x - (a - b)) and (x - (a + b)) are the factors.

 

Factorization of p(x),

   

$\mapsto\ x^3-3x^2+x+1 \\ \\ \mapsto\ x^3-x^2-2x^2+2x-x+1 \\ \\ \mapsto\ x^2(x-1)-2x(x-1)-1(x-1) \\ \\ \mapsto\ (x-1)(x^2-2x-1) \\ \\ \mapsto\ (x-1)(x^2+(\sqrt{2}-1)x-(\sqrt{2}+1)x-1) \\ \\ \mapsto\ (x-1)(x(x+\sqrt{2}-1)-(\sqrt{2}+1)(x+\sqrt{2}-1)) \\ \\ \mapsto\ (x-1)(x+\sqrt{2}-1)(x-(\sqrt{2}+1)) \\ \\ \mapsto\ (x-(1))(x-(1-\sqrt{2}))(x-(1+\sqrt{2})) \\ \\ \mapsto\ Seems\ like\ \ (x-a)(x-(a-b))(x-(a+b))$

$OR\\ \\ \mapsto\ (x-1)(x+\sqrt{2}-1)(x-(\sqrt{2}+1)) \\ \\ \mapsto\ (x-(1))(x-(1+(-\sqrt{2})))(x-(1-(-\sqrt{2})))} \\ \\ \mapsto\ Seems\ like\ \ (x-a)(x-(a+b))(x-(a-b))$

 

∴ a = 1   ;   b = ±√2

 

Plz ask me if you have any doubt.

Thank you...

$\mathfrak{\#adithyasajeevan}$