Question

Consider the cubic function f(x) = x^3 + ax^2 + bx + 4 where (a) and (b) are constants. When f(x) is divided by x-3, the remainder is 10. When f (x) is divided by x+1, the remainder is 6.

Show that a = -1 and b = -4

Answer 1

see explanation

Explanation:

Using the Remainder theorem

If f(x) is divided by (x - h) then f(h) = remainder, thus

f(3) = 3³ + a(3)² + b(3) + 4 = 10, that is

27 + 9a + 3b + 4 = 10

9a + 3b + 31 = 10 ( subtract 31 from both sides )

9a + 3b = - 21 → (1)

f(- 1) = (- 1)³ + a(- 1)² + b(- 1) + 4 = 6, that is

- 1 + a - b + 4 = 6

a - b + 3 = 6 ( subtract 3 from both sides )

a - b = 3 → (2)

Rearrange (2) expressing a in terms of b

a = 3 + b → (3)

Substitute a = 3 + b into (1)

9(3 + b) + 3b = - 21 ← distribute and simplify left side

27 + 9b + 3b = - 21

12b + 27 = - 21 ( subtract 27 from both sides )

12b = - 48 ( divide both sides by 12 )

b = - 4

Substitute b = - 4 into (3)

a = 3 - 4 = - 1

Hence a = - 1 and b = - 4