If zero of polynomial is (6-√5) sum of zeroes is 12 then what p( x)?
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Answer: 1st zero = (6-√5)
Let 2nd zero be 'A'
Sum of zeroes = 12
1st zero+2nd zero = 12
(6-√5) + A = 12
A = 12-(6-√5)
A = 12-6+√5
A = 6+√5
P(x) = (x-1st zero)(x-2nd zero)
= [x+(6-√5)][x-(6+√5)
= x^2 - x(6+√5) + x(6-√5) + (6)^2
- (√5)^2
= x^2 -6x +√5x + 6x -√5x +36-5
= X^2 + 31
Step-by-step explanation:
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