A polynomial in x of third …
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Sunil C N 8 years, 4 months ago
equation is{tex} ax^3+bx^2+cx+d{/tex}
when x=1
{tex} a+b+c+d=0{/tex} when {tex} x=1{/tex}
{tex} -8a+4b-2c+d=0{/tex} when {tex} x=-2{/tex}
{tex} -a+b-c+d=4{/tex} when {tex} x=-1{/tex}
{tex} 8a+4b+2c+d=28{/tex} when {tex} x=2{/tex}
{tex}\begin{pmatrix} 1 & 1 & 1 & 1 \\ -8 & 4 & -2 & 1 \\ -1 & 1 & -1& 1\\ 8 & 4 &2 & 1\end{pmatrix} \ \begin{pmatrix} a \\ b \\ c\\ d\end{pmatrix} \ =\begin{pmatrix} 0 \\ 0 \\ 4\\ 28\end{pmatrix} \ {/tex}
<pre> {tex} \begin{pmatrix} a \\ b \\ c\\ d\end{pmatrix} \ =inverse\begin{pmatrix} 1 & 1 & 1 & 1 \\ -8 & 4 & -2 & 1 \\ -1 & 1 & -1& 1\\ 8 & 4 &2 & 1\end{pmatrix}\begin{pmatrix} 0 \\ 0 \\ 4\\ 28\end{pmatrix} \ {/tex} </pre>0Thank You