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    x4 + x3 + x - 10 = (ax2 + bx + c) (x2 + 2x - 3) + Dx + e

    I am supposed to solve for a,b,c,d and e.

    i think the first step is to sub in -3 where ever there is an x to find d?

    Thank you!!
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    (Original post by JamesNeedHelp2)
    x4 + x3 + x - 10 = (ax2 + bx + c) (x2 + 2x - 3) + Dx + e

    I am supposed to solve for a,b,c,d and e.

    i think the first step is to sub in -3 where ever there is an x to find d?

    Thank you!!
    You could indeed find out -3d+e that way, and if you substitute x=1 you'll get another expression which is d+e. Those you can solve simultaneously.

    To get the a,b,c terms, it's easiest just to expand the brackets of the right-hand side and compare coefficients, in my opinion.
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    (Original post by JamesNeedHelp2)
    x4 + x3 + x - 10 = (ax2 + bx + c) (x2 + 2x - 3) + Dx + e

    I am supposed to solve for a,b,c,d and e.

    i think the first step is to sub in -3 where ever there is an x to find d?

    Thank you!!
    I presume the numbers after the x's are supposed to be indices?

    How far have you got - you should be able to write down the value of a straight away just by looking!
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    (Original post by Smaug123)
    You could indeed find out -3d+e that way, and if you substitute x=1 you'll get another expression which is d+e. Those you can solve simultaneously.

    To get the a,b,c terms, it's easiest just to expand the brackets of the right-hand side and compare coefficients, in my opinion.
    Thank you sir.
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    (Original post by davros)
    I presume the numbers after the x's are supposed to be indices?

    How far have you got - you should be able to write down the value of a straight away just by looking!
    Is it true that i am supposed to compare the coefficients of x to find a? if so, i think a= 1?
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    (Original post by JamesNeedHelp2)
    Is it true that i am supposed to compare the coefficients of x to find a? if so, i think a= 1?
    Quite correct.
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    (Original post by JamesNeedHelp2)
    Is it true that i am supposed to compare the coefficients of x to find a? if so, i think a= 1?
    You can compare coefficients to get simultaneous equations that will give you all the unknown constants; you can also do things like sub in various x values like 1 and -3 which I think will make one of your brackets vanish and let you find d and e quite simply
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    (Original post by davros)
    You can compare coefficients to get simultaneous equations that will give you all the unknown constants; you can also do things like sub in various x values like 1 and -3 which I think will make one of your brackets vanish and let you find d and e quite simply

    Thanks again!!:cool:
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    (Original post by brianeverit)
    Quite correct.
    Thank you!!
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    (Original post by Smaug123)
    You could indeed find out -3d+e that way, and if you substitute x=1 you'll get another expression which is d+e. Those you can solve simultaneously.

    To get the a,b,c terms, it's easiest just to expand the brackets of the right-hand side and compare coefficients, in my opinion.
    Am i over thinking it or is it really difficult to find -3d + e?

    So i subbed in -3;

    (a(-3)2 - 3b + c) ((-3)2 + 2(-3) - 3) + dx +e

    9 - 3b + c -3d + e

    Is this correct? if so, and how do i proceed from here to find -3d + e?

    Thank you!!
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    (Original post by JamesNeedHelp2)
    Am i over thinking it or is it really difficult to find -3d + e?

    So i subbed in -3;

    (a(-3)2 - 3b + c) ((-3)2 + 2(-3) - 3) + dx +e

    9 - 3b + c -3d + e

    Is this correct? if so, and how do i proceed from here to find -3d + e?

    Thank you!!
    I try to avoid coming into a thread where others (perhaps too many) are trying to help you, but ....

    ... is there something wrong in doing a long division which takes 30 or so seconds and read off the answers for all the constants?
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    (Original post by TeeEm)
    I try to avoid coming into a thread where others (perhaps too many) are trying to help you, but ....

    ... is there something wrong in doing a long division which takes 30 or so seconds and read off the answers for all the constants?
    I didnt really know that was possible. There seems to be quite a few methods, and i am baffled by all of them. But i will try this method out.
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    (Original post by JamesNeedHelp2)
    I didnt really know that was possible. There seems to be quite a few methods, and i am baffled by all of them. But i will try this method out.
    in my humble opinion what you are being told is correct but certainly a quick long division is the most effective method
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    (Original post by TeeEm)
    in my humble opinion what you are being told is correct but certainly a quick long division is the most effective method
    Thank you. In this problem, x4 + x3 + x - 10 = (ax2 + bx + c) (x2 + 2x - 3) + Dx + e, am i dividing x4 + x3 + x - 10 by ax2 + bx + c?

    I am just trying to figure out, what i need to divide by what to find the constants a, b, c, d an e?

    Thanks again!
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    (Original post by JamesNeedHelp2)
    Thank you. In this problem, x4 + x3 + x - 10 = (ax2 + bx + c) (x2 + 2x - 3) + Dx + e, am i dividing x4 + x3 + x - 10 by ax2 + bx + c?

    I am just trying to figure out, what i need to divide by what to find the constants a, b, c, d an e?

    Thanks again!
    x4 + x3 + x - 10 by (x2 + 2x - 3)
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    (Original post by JamesNeedHelp2)
    Am i over thinking it or is it really difficult to find -3d + e?

    So i subbed in -3;

    (a(-3)2 - 3b + c) ((-3)2 + 2(-3) - 3) + dx +e

    9 - 3b + c -3d + e

    Is this correct? if so, and how do i proceed from here to find -3d + e?

    Thank you!!
    (-3)^2 + 2 \times (-3) - 3 is 0. That entire bracket vanishes.
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    (Original post by Smaug123)
    (-3)^2 + 2 \times (-3) - 3 is 0. That entire bracket vanishes.
    ok. Which leaves me with 9 - 3b + c -3d + e:confused:
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    (Original post by davros)
    You can compare coefficients to get simultaneous equations that will give you all the unknown constants; you can also do things like sub in various x values like 1 and -3 which I think will make one of your brackets vanish and let you find d and e quite simply

    (Original post by JamesNeedHelp2)
    Am i over thinking it or is it really difficult to find -3d + e?

    So i subbed in -3;

    (a(-3)2 - 3b + c) ((-3)2 + 2(-3) - 3) + dx +e

    9 - 3b + c -3d + e

    Is this correct? if so, and how do i proceed from here to find -3d + e?

    Thank you!!

    So baffled
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    (Original post by JamesNeedHelp2)
    ok. Which leaves me with 9 - 3b + c -3d + e:confused:
    No it doesn't!

    If one of the brackets is 0 then you just get 0 for the 1st term on the RHS leaving you with -3d + e.

    You can plug in x = 1 to get a similar equation, or plug in lots of other small values of x to get a set of simultaneous equations e.g. x = 0 or x = -1.
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    (Original post by davros)
    No it doesn't!

    If one of the brackets is 0 then you just get 0 for the 1st term on the RHS leaving you with -3d + e.

    You can plug in x = 1 to get a similar equation, or plug in lots of other small values of x to get a set of simultaneous equations e.g. x = 0 or x = -1.
    aah i see what you guys are saying, my bad. Silly mistake on my part
 
 
 
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