Solve Polynomial Scale: Find a,b,c,d

[karush]

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

 

[MarkFL]

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. :)

 

[HallsofIvy]

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.

 

[karush]

mahalo much

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

but ran into fog banks

 

[MarkFL]

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:

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