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

Question

Esperanza Gregory

Mathematics

24 July, 11:24

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

+2

Answers (1)

Know the Answer?

Houston Walsh · Answer 1

see explanation

Step-by-step 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

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

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