Question

CAN someone help me about this?

Answer 1

Explanation:

`i=√(-1) \ \ \ \ \ \ \ \ i^2=(√(-1))^2=-1`

a) ±2, 3

`(x-2)(x-(-2))(x-3)=\\\\(x-2)(x+2)(x-3)=\\\\(x^2-4)(x-3)=\\\\x^3-4x-3x^2+12=\\\\x^3-3x^2-4x+12`

b) -2, ±i

`(x-(-2))(x-i)(x-(-i))=\\\\(x+2)(x-i)(x+i)=\\\\(x+2)(x^2-i^2)=\\\\(x+2)(x^2-(-1))=\\\\(x+2)(x^2+1)=\\\\`

`x^3+x+2x^2+2=\\\\x^3+2x^2+x+2`

c) 3, -1 ±i

`(x-3)(x-(-1-i))(x-(-1+i)=\\\\(x-3)(x+1+i)(x+1-i)=\\\\(x-3)((x+1)+i)((x+1)-i)=\\\\(x-3)((x+1)^2-i^2)=\\\\(x-3)((x+1)^2-(-1))=\\\\(x-3)((x+1)^2+1)=\\\\(x-3)(x+1)^2+(x-3)(1)=\\\\(x-3)(x^2+2x+1)+(x-3)=\\\\x^3+2x^2+x-3x^2-6x-3+x-3=\\\\x^3-x^2-4x-6`

d) -1, -2±√2

`(x-(-1))(x-(-2-√(2)))(x-(-2+√(2)))=\\\\ (x+1)(x+2+√(2))(x+2-√(2))=\\\\(x+1)((x+2)+√(2))((x+2)-√(2))=\\\\ (x+1)((x+2)^2-(√(2))^2)=\\\\(x+1)((x+2)^2-2)=\\\\(x+1)(x+2)^2-2(x+1)=\\\\(x+1)(x^2+4x+4)-2x-2=\\\\x^3+4x^2+4x+x^2+4x+4-2x-2=\\\\x^3+5x^2+6x+2`