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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

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  1. 24 July, 11:37
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    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
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