How do I solve this?

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  • 1 month ago
    Favorite Answer

    When a polynomial p(x) is divided by the binomial (x - a), the remainder is p(a).

    So, with p(x) = x^4 + ax^3 + bx + c, the given remainders mean that:

    p(1) = 14

    p(-1) = 0

    p(-2) = -16

    Evaluate p for those values of x on the left:

    1 + a + b + c = 14

    1 - a - b + c = 0

    16 - 8a - 2b + c = -16

    Three linear equations in three unknowns.  Add the first two and both b and c cancel out:

    2 + 2c = 14 = 0

    2c = 12

    c = 6

    Subtract them instead and get one reduced equation:

    2a + 2b = 14; a + b = 7

    Plug c=6 into the 3rd equation to get another reduced equation:

    16 - 8a - 2b + 6 = -16

    16 + 16 + 6 = 8a + 2b

    8a + 2b = 38

    4a + b = 19

    Subtract a + b = 7 and get

    3a = 19 - 7 = 12

    a = 4

    ...and b = 7 - a = 3

    I just typed all that on the run to show the math.  You should carry out those steps yourself in case I made a typo along the way.

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  • 1 month ago

    Ruffini

    (x⁴+ax³+bx+c)/(x+1)=x³+(a-1)x²-(a-1)x+(a-1+b) with remaind (1-a-b+c)=0

    (x⁴+ax³+bx+c)/(x-1)=x³+(a+1)x²+(a+1)x+(a+1+b) with remaind (1+a+b+c)=14

    (x⁴+ax³+bx+c)/(x+2)=x³+(a-2)x²+(4-2a)x+(4a-8+b) with remaind (16-8a-2b+c)=-16

    Now

    -a-b+c=-1

    a+b+c=13

    -8a-2b+c=-32

    Solve

    a=4

    b=3

    c=6

    Try

    (x⁴+4x³+3x+6)/(x+1)=(x³+3x²-3x+6)(x+1)+0

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