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  1. MHB Master
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    #1
    what is a b c and d so that all values of s are true

    \begin{align}\displaystyle
    &f_{15}=\\
    &-17d+11s^2-4s+10as^3=(b+2)s+90s^3+(3c-1)s^2+85\\
    &-17d+11s^2-6s+10as^3=bs+90s^3+3cs^2-s^2+85\\
    &(s=0)\\
    &-17d=85 \therefore d=-5\\
    &11s^2-6s+10as^3=bs+90s^3+3cs^2-s^2\\
    &(s=-1)\\
    &11+6-10a=-b-90+3c-1\\
    &-10a+b-3c=-18 \\
    &(s=1) \\
    &11-6+10a=b+90+3c-1\\
    &10a-b-3c=-6 \\
    &--3c=-24\therefore c=4
    \end{align}

    what i couldn't get is a and b

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    #2
    From the first equation, we have be equating like coefficients:

    $ \displaystyle 10a=90$

    $ \displaystyle 11=3c-1$

    $ \displaystyle -4=b+2$

    $ \displaystyle -17d=85$

    Solve each of these to find the values in question.

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    #3
    Or: you have four unknowns, a, b, c, and d so you need four equations, not just three. Take one more values for s, say s= 2, to get one more equation.

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    #4 Thread Author
    mahalo much

    I tried earlier to do this by some factoring
    and a thot a matrix could be used

    but ran into fog banks

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    #5
    Quote Originally Posted by karush View Post
    mahalo much

    I tried earlier to do this by some factoring
    and a thot a matrix could be used

    but ran into fog banks
    You could use an augmented matrix:

    $ \displaystyle \left[\begin{array}{cccc|c}10 & 0 & 0 & 0 & 90 \\ 0 & 1 & 0 & 0 & -6 \\ 0 & 0 & 3 & 0 & 12 \\ 0 & 0 & 0 & -17 & 85 \\ \end{array}\right]$

    Now perform:

    $ \displaystyle \frac{1}{10}R_1,\,\frac{1}{3}R_3,\,-\frac{1}{17}R_4$

    to obtain:

    $ \displaystyle \left[\begin{array}{cccc|c}1 & 0 & 0 & 0 & 9 \\ 0 & 1 & 0 & 0 & -6 \\ 0 & 0 & 1 & 0 & 4 \\ 0 & 0 & 0 & 1 & -5 \\ \end{array}\right]$

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