Question

find the value of k for which [ x+1] is a factor of the polynomial x cube + x square + x +k

Answer 1

annyoonghaseyo.

Given that :- ( X - 1 ) is a factor of the given polynomial.

So,

( X - 1 ) = 0

X = 1

P(X) = X³+KX²+142X-120

P(1) = (1)³ + K × (1)² + 142 × 1 - 120

=> 1 + K + 142 - 120 = 0

=> K + 23 = 0

=> K = -23

✴hope it's help✴