Let the roots of the polynomial P ( x ) = x 3 </msup> &#x2212; x + 1

hyprkathknmk

hyprkathknmk

Answered question

2022-05-12

Let the roots of the polynomial P ( x ) = x 3 x + 1 be a, b, c. Construct the polynomial with roots a 4 , b 4 , c 4 and degree 3

Answer & Explanation

tomatoland45wt8wm

tomatoland45wt8wm

Beginner2022-05-13Added 19 answers

You are on the right track, if you write u 1 4 = x and form an equation in u which can be changed into the form of a polynomial.
So,
u 3 4 u 1 4 + 1 = 0 u 1 4 ( u 1 2 1 ) = 1
u 1 2 1 = 1 u 1 4
Now squaring both sides,
u 2 u 1 2 + 1 = 1 u 1 2
u + 1 = 1 u 1 2 + 2 u 1 2
Squaring again,
u 2 + 2 u + 1 = 1 u + 4 + 4 u
So the polynomial is
u 3 2 u 2 3 u 1 = 0
Derick Richard

Derick Richard

Beginner2022-05-14Added 4 answers

Maybe I was wrong in calculations, one needs to check.
Vieta: a + b + c = 0, a b + a c + b c = 1, a b c = 1
Let's do some algebraic manipulations to obtain the same for a 4 , b 4 , c 4
( a + b + c ) 2 = a 2 + b 2 + c 2 + 2 ( a b + b c + a c ) a 2 + b 2 + c 2 = 2
( a b + a c + b c ) 2 = a 2 b 2 + a 2 c 2 + b 2 c 2 + 2 a b c ( a + b + c ) a 2 b 2 + a 2 c 2 + b 2 c 2 = 1
( a 2 + b 2 + c 2 ) 2 = a 4 + b 4 + c 4 + 2 ( a 2 b 2 + a 2 c 2 + b 2 c 2 ) a 4 + b 4 + c 4 = 2
a 2 b 2 c 2 = ( a b c ) 2 = 1
( a 2 b 2 + a 2 c 2 + b 2 c 2 ) 2 = a 4 b 4 + a 4 c 4 + b 4 c 4 + 2 a 2 b 2 c 2 ( a 2 + b 2 + c 2 ) a 4 b 4 + a 4 c 4 + b 4 c 4 = 3
It's ok, because there are complex roots in a , b , c
a 4 b 4 c 4 = ( a b c ) 4 = 1
Reverse Vieta: x 3 2 x 2 + 3 x 1 is polynomial with roots a 4 , b 4 , c 4

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?