I want to show that the polynomial, ( 2 c + 1 ) ( b &#x2212;<!-- − --> 2 c

veneciasp

veneciasp

Answered question

2022-07-01

I want to show that the polynomial,
( 2 c + 1 ) ( b 2 c ) ( b 2 c + d 1 ) ( a / 2 + d 2 c ) ( 2 + a / 2 d + 2 c ) ( b 2 c + 2 d 1 ) ( 2 c + 1 d ) 2

Answer & Explanation

Leslie Rollins

Leslie Rollins

Beginner2022-07-02Added 25 answers

( 2 c + 1 ) ( b 2 c ) ( b 2 c + d 1 ) ( a / 2 + d 2 c ) ( 2 + a / 2 d + 2 c ) ( b 2 c + 2 d 1 ) ( 2 c + 1 d ) 2
We want
( 2 c + 1 ) ( b 2 c ) ( b 2 c + d 1 ) ( a 2 + d 2 c ) ( 2 + a 2 d + 2 c ) ( b 2 c + 2 d 1 ) ( 2 c + 1 d ) 2
as c a 4 1 , we get 2 c a 2 2 , with 2 c 2 a 2 . Since b a 2 . we reach b 2 c 2.
It follows that
( b 2 c ) ( b 2 c + d 1 ) 2 ( b 2 c ) + 2 d 2 ,
( b 2 c ) ( b 2 c + d 1 ) ( b 2 c ) + 2 d + ( b 2 c ) 2 ,
( b 2 c ) ( b 2 c + d 1 ) ( b 2 c ) + 2 d ( b 2 c + 2 d 1 ) .
( b 2 c ) ( b 2 c + d 1 ) ( b 2 c + 2 d 1 ) .
Next we see
( 2 c + 1 ) ( 2 c + 1 d ) ,
( 2 + a 2 d + 2 c ) ( 2 c + 1 d ) .
Mix in our three inequalities, we find
( 2 c + 1 ) ( b 2 c ) ( b 2 c + d 1 ) ( a 2 + d 2 c ) ( 2 + a 2 d + 2 c ) ( b 2 c + 2 d 1 ) ( 2 c + 1 d ) 2

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?