How to find all the real solutions of the polynomial equation 2y^4+7y^3 -26y^2 +23y-6 =0?

kukuhushlq3p

kukuhushlq3p

Answered question

2023-02-18

How to find all the real solutions of the polynomial equation 2 y 4 + 7 y 3 - 26 y 2 + 23 y - 6 = 0 ?

Answer & Explanation

Adrienne Clark

Adrienne Clark

Beginner2023-02-19Added 9 answers

2 y 4 + 7 y 3 - 26 y 2 + 23 y - 6
Keep in mind that 0 is the total of the coefficients. Which is:
2 + 7 - 26 + 23 - 6 = 0
So y = 1 is a zero and ( y - 1 ) a factor:
2 y 4 + 7 y 3 - 26 y 2 + 23 y - 6 = ( y - 1 ) ( 2 y 3 + 9 y 2 - 17 y + 6 )
The sum of the coefficients of the remaining cubic factor also sum to 0 . That is:
2 + 9 - 17 + 6 = 0
So y = 1 is a zero again and ( y - 1 ) a factor:
2 y 3 + 9 y 2 - 17 y + 6 = ( y - 1 ) ( 2 y 2 + 11 y - 6 )
We can factor the remaining quadratic using an AC method:
Find a pair of factors of A C = 2 6 = 12 which differ by B = 11
The pair 12 , 1 works.
Split the middle term using that pair, then factor by grouping:
2 y 2 + 11 y - 6
= 2 y 2 + 12 y - y - 6
= ( 2 y 2 + 12 y ) - ( y + 6 )
= 2 y ( y + 6 ) - 1 ( y + 6 )
= ( 2 y - 1 ) ( y + 6 )
So the remaining zeros are y = 1 2 and y = - 6

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