Math, asked by BibonBeing01, 9 months ago


p(x) =  {x}^{2}  + x + k

Answers

Answered by Anonymous
1

P(x) = x² + x + k

P(1) = 1² + 1 + k

= 1 + 1 + k = 0

= 2 + k = 0

→ K = -2

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