Question

find the value of k,if x-1 is a factorof p(x) in each of the following cases
(i) p(x)=x²+x+k
(ii) p(x)=2x²+kx+root2​

Answer 1

Answer:

This is for answer 1

According to remainder theorem p(1) = 0 we get

According to remainder theorem p(1) = 0 we getPlug x = 1 we get

According to remainder theorem p(1) = 0 we getPlug x = 1 we get => k(1)2 + 1+ 1 =0

According to remainder theorem p(1) = 0 we getPlug x = 1 we get => k(1)2 + 1+ 1 =0=>k +1 + 1 =0

According to remainder theorem p(1) = 0 we getPlug x = 1 we get => k(1)2 + 1+ 1 =0=>k +1 + 1 =0=> k + 2 = 0

According to remainder theorem p(1) = 0 we getPlug x = 1 we get => k(1)2 + 1+ 1 =0=>k +1 + 1 =0=> k + 2 = 0=> k = - 2

This is for answer 2

According to remainder theorem p(1) = 0 we get

According to remainder theorem p(1) = 0 we getPlug x = 1 we get

According to remainder theorem p(1) = 0 we getPlug x = 1 we get p(1) = 2(1)2 + k(1) + √2

According to remainder theorem p(1) = 0 we getPlug x = 1 we get p(1) = 2(1)2 + k(1) + √2p(1) =2 + k + √2

According to remainder theorem p(1) = 0 we getPlug x = 1 we get p(1) = 2(1)2 + k(1) + √2p(1) =2 + k + √20 = 2 + √2 + k

According to remainder theorem p(1) = 0 we getPlug x = 1 we get p(1) = 2(1)2 + k(1) + √2p(1) =2 + k + √20 = 2 + √2 + k-2 - √2 = k

According to remainder theorem p(1) = 0 we getPlug x = 1 we get p(1) = 2(1)2 + k(1) + √2p(1) =2 + k + √20 = 2 + √2 + k-2 - √2 = k- (2 + √2) = k

According to remainder theorem p(1) = 0 we getPlug x = 1 we get p(1) = 2(1)2 + k(1) + √2p(1) =2 + k + √20 = 2 + √2 + k-2 - √2 = k- (2 + √2) = k Answer is k = - (2 + √2)