Question

If f(x)=x4 - 2x + 3x^2 - ax + b is a polynomial such that
when it is divided by x - 1 and x + 1, the remainders are
respectively 5 and 19. Determine the remainder When )
is divided by (x-2)​

Answer 1

Answer:

$p(x) = x {}^{4} - 2x + 3x {}^{2} - ax + b \\ f(x) = x - 1and \: g(x) = x + 1 \\ when \: p(x)divided \: \: by \: \: (x - 1)then \: rem. = 5 \\ x - 1 = 0 \\ x = 1 \\ by \: rem. \: th. \\ p(x) = x {}^{4} - 2 \times 1 + 3 \times 1 {}^{2} - a \times 1 + b \\ = 1 - 2 + 3 - a +b \\ = 2 - a + b = 5 \\ = - a + b = 5 - 2 \\ = - a + b = 3.........(1) \\ when \: p(x) \: divided \: by \: (x + 1)then \: rem. = 19 \\ x + 1 = 0 \\ x = - 1 \\ by \: rem. \: th. \\ p(x) = - 1 {}^{4} - 2 \times - 1 + 3 \times - 1 {}^{2} - a \times - 1 + b \\ 1 - 2 \times - 1 + 3 + a + b \\ 1 + 2 + 3 + a + b \\ 6 + a + b = 19 \\ a + b = 13.........(2) \\ from \: eq \: (1)and \: (2) \\ \: \: - a + b = 3 \\ \: a + b = 13 \: adding \\ - - - - - - - - \\ 2b = 16 \\ - - - - - - - - \\ b = 8 \\ from \: eq \: ......(2) \\ a + 8 = 13 \\ a = 13 - 8 \\ a = 5 \\ p(x) = x {}^{4} - 2 x + 3x { }^{2} - ax + b \\ x {}^{4} - 2x + 3x { }^{2} - 5x + 8 \\ g(x) = x - 2 = 0 \\ x = 2 \\ by \: rem \: . \: th . \\ p(x) = x {}^{4} - 2x + 3x {}^{2} - ax + b \\ 2 {}^{4} - 2 \times 2 + 3 \times 2 {}^{2} - 5 \times 2 + 8 \\ 16 - 4 + 3 \times 4 - 5 \times 2 + 8 \\ = 16 - 4 + 12 - 10 + 8 \\ = 36 - 14 \\ = 22......$