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