a The expression ax²+bx+c is divisible by x-1, has a remainder of 2 when divided by x+1 and has a remainder of 8 and when divided

Question

a
The expression ax²+bx+c is divisible by x-1, has a
remainder of 2 when divided by x+1 and has a remainder of 8 and when divided by x-2 find the values of a, b, and c and
factorise completely.​

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Trung Dũng 3 days 2021-07-22T14:25:18+00:00 1 Answers 0 views 0

Answers ( )

    0
    2021-07-22T14:26:18+00:00

    Answer:

    (x – 1)(3x + 2)

    Step-by-step explanation:

    By the Remainder theorem

    Given expression is divisible by x – 1 then f(1) = 0

    when divided by (x + 1) then f(- 1) = 2

    when divided by x – 2 then f(2) = 8

    Substitute x = 1, x = – 1, x = 2 into the expression , that is

    a + b + c = 0 → (1)

    a – b + c = 2 → (2)

    4a + 2b + c = 8 → (3)

    Subtract (1) from (2) to eliminate a and c

    – 2b = 2 ( divide both sides by – 2 )

    b = – 1

    Subtract (1) from (3) to eliminate c

    3a + b = 8 ← substitute b = – 1

    3a – 1 = 8 ( add 1 to both sides )

    3a = 9 ( divide both sides by 3 )

    a = 3

    Substitute a = 3, b = – 1 into (1) and evaluate for c

    3 – 1 + c = 0

    2 + c = 0 ( subtract 2 from both sides )

    c = – 2

    Then a = 3, b = – 1, c = – 2

    Thus

    ax² + bx + c

    = 3x² – x – 2

    = (x – 1)(3x + 2) ← in factored form

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Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )