Given that, (x + 1) is a factor f(x) = 2x5+ax2+2bx+1 then f(-1) = 0
[if (x + α) is a factor of f(x) = ax2 + bx + c, then f(-) = 0]
⇒2(−1)3+a(−1)2+2b(−1)+1=0
⇒ -2 + a – 2b + 1 = 0
⇒ a – 2b – 1 = 0
⇒ 2a – 3b = 4
⇒ 3b = 2a – 4
⇒b=(2a−43)
Now, put the value of b in Eq. (i), we get
a−2(2a−43)−1=0
⇒ 3a – 2(2a – 4) – 3 = 0
⇒ 3a – 4a + 8 – 3 = 0
⇒ - a + 5 = 0
⇒ a = 5
Now, put the value of b in Eq. (i), we get
a−22a−43−1=0
⇒ 3a – 2(2a – 4) – 3 = 0
⇒ - a + 5 = 0
⇒ a = 5
Now, put the value of a in Eq (i), we get
5 – 2b – 1 = 0
⇒ 2b = 4
⇒ b = 2
Hence, the required values of a and b are 5 and 2, respectively