If 3+i and 2- 3 are the roots of the equation f ( x )= a ₀+ a ₁ x + a ₂ x ^2+ .+ a _ n x ^ n ; a ₀, a ₁ . . a _ n Z , then the least value of n and value of a ₀ are respectively.

Step-by-step solution

Given f(x)=a₀+a₁ x+a₂ x^2+ +a_n x^n, a₀, a₁, a₂ , a_n Z . ...(i) Since (3+i) and (2- 3 ) are roots of equation (i) Hence (3-i) and (2+ 3 ) will also be the roots of given equation. Therefore, now we have at least 4 roots of given equation. In other words, then the least value of n will be 4 . n=4 . If 4 roots are there, then the correspondingly equation (i) will reduce to: array r x^4-( + + + ) x^3+( + + + + + ) x^2 +( + + + ) x+ =0 (ii) array Clearly, is the constant term of equation Hence a₀= =(3+i)(3-i)(2- 3 )(2