what is the equation of lowest degree with real coefficients if two of its roots are -1 and 1+i?

Question

A. x^3+x^2+2=0
B. x^-x^2-2=0
C. x^3-x+2=0
D. x^3-x^2+2=0
e. none of the above

i don't get it is it (x+1) (1+i)

Answer 1

1)

x ^ 3 + a x ^ 2 + b x + c = ( x - x1 ) ( x - x2 ) ( x - x3 )

2)

For an equation with real coefficients, imaginary roots ( if any ) occur in conjugate pairs,

i.e. f ( x ) = 0 a polynomial equation with real coefficients an

n + i * m
an imaginary root of it , then

n - i * m

is also a root of it.

In this case :

x1 = - 1

x2 = 1 + i

x3 = 1 - i

Your equation becomes :

( x - 1) [ x - ( 1 + i ) ] [ x - ( 1 - i ) ]

_____________________________________
Remark:

[ x - ( 1 + i ) ] [ x - ( 1 - i ) ] = x ^ 2 - 2 x + 2
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( x - 1) [ x - ( 1 + i ) ] [ x - ( 1 - i ) ] =

( x - 1 ) ( x ^ 2 - 2 x + 2 ) =

x ^ 3 - x ^ 2 + 2

Answer : D

Answer 2

For any polynomial equation with rational coefficients, complex roots must occur as conjugate pairts, that is,
if one root is 1+i, there has to be another one 1 - i
so, including the root of -1, the equation could have been
(x+1)(x - (1+i) )(x - (1-i) ) = 0
(x+1)(x-1-i)(x-1+i) = 0
(x+1)(x^2 - 2x + 2) = 0
x^3 - 2x^2 + 2x + x^2 - 2x + 2 =0
x^3 - x^2 + 2 = 0

looks like D

Answer 3

Correction :

Your equation becomes :

( x + 1) [ x - ( 1 + i ) ] [ x - ( 1 - i ) ]

( x + 1) [ x - ( 1 + i ) ] [ x - ( 1 - i ) ] =

( x + 1 ) ( x ^ 2 - 2 x + 2 ) =

x ^ 3 - x ^ 2 + 2

Answer D