Question

For what value of k the polynomial y²+KY+2k-2 is exactly divisible by y+2

Answer 1

hlo

here is your answer

given  divisor = y+2

and dividend =y²+KY+2k-2

we first find the zero of the   divisor

so the zero of  the divisor is ⇒ y+2=0

                                                  ⇒y =-2

so that ht  p(-2) ⇒ 2²+K*2+2*2-2 =0

                             ⇒4+ 2k+4-2=0

                                ⇒6 +2k =0

                                    ⇒  2k =-6 ⇒ k = -3

hence the value of the k = -3

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