Please help!?

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  • 1 month ago
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    Let f(x)=2x^3+ax^2+bx-3, then

    f(1)=2+a+b-3=8

    =>

    a+b=9-------(1)

    f(-2)=-16+4a-2b-3=-7

    =>

    2a-b+6------(2)

    according to the remainder theorem.

    Solving the system (1) & (2) for a, b, get

    a=5

    b=4

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  • 1 month ago

    You have two unknowns, so you need to set up a system of two equations. You can make this system of equations by dividing the polynomials by (x-1) and setting the remainder equal to 8, and then dividing by (x+2) and setting that remainder equal to -7.

    You can do long division of polynomials to get remainders. Look up how to do long division of polynomials if you do not know how to do it.

    Dividing 2x^3 + ax^2 + bx - 3 by (x-1), you get a remainder of -3-b+a-2 = -5-b+a. So you have:

    -5 -b + a = 8 or a + b = 13

    Dividing 2x^3 + ax^2 + bx - 3 by (x+2), you get a remainder of -2b+4a-19. So you have:

    -2b+4a-19 = -7 or -2b +4a = 12

    So your system of equations is:

        a + b = 13

    -2b +4a = 12

    -2b +4(13-b) = 12

    -2b -4b  +52 = 12

    -6b = -40

    b = 40/6 = 20/3

    a = 13 - 20/3 = 19/3   

    Note: This is the process you use to find a and b, but there is a possibility that I made a mistake somewhere, so you should do the problem out yourself to check. 

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  • Pope
    Lv 7
    1 month ago

    Let p(x) = 2x³ + ax² + bx - 3.

    By the remainder theorem of polynomials, p(1) = 8 and p(-2) = -7.

    p(1) = 8

    2 + a + b - 3 = 8

    a + b = 9

    p(-2) = -7

    -16 + 4a - 2b - 3 = -7

    4a - 2b = 12

    2a - b = 6

    It is now a system of two linear equations in a and b:

    a + b = 9

    2a - b = 6

    a = 5

    b = 4

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