Mara Cook

2022-06-22

Find the remainder of f divided by $g\left(x\right)={x}^{4}+{x}^{2}+1$ if the remainder of f divided by ${h}_{1}\left(x\right)={x}^{2}+x+1$ is $-x+1$ and the remainder of f divided by ${h}_{2}\left(x\right)={x}^{2}-x+1$ is $3x+5$

Jovan Wong

Let the polynomial be $P\left(x\right)$. It is given that for some polynomials $Q\left(x\right),{Q}_{1}\left(x\right)$
$P\left(x\right)=\left({x}^{2}+x+1\right)Q\left(x\right)+1-x$
$P\left(x\right)=\left({x}^{2}-x+1\right){Q}_{1}\left(x\right)+3x+5$
Now it is well known that ${x}^{2}+x+1$ has the zeros as $\omega ,{\omega }^{2}$ So We get
$P\left(\omega \right)=1-\omega$
$P\left({\omega }^{2}\right)=1-{\omega }^{2}$
$P\left(-\omega \right)=3\omega +5$
$P\left(-{\omega }^{2}\right)=-3{\omega }^{2}+5$
Now let us assume for some polynomial ${Q}_{2}\left(x\right)$ we have
$P\left(x\right)=\left({x}^{4}+{x}^{2}+1\right){Q}_{2}\left(x\right)+A{x}^{3}+B{x}^{2}+Cx+D$
Using the fact that ${x}^{4}+{x}^{2}+1$ has the zeros $\omega ,{\omega }^{2},-\omega ,-{\omega }^{2}$ we get four linear equations as:
$\left[\begin{array}{cccc}1& {\omega }^{2}& \omega & 1\\ 1& \omega & {\omega }^{2}& 1\\ -1& {\omega }^{2}& -\omega & 1\\ -1& \omega & -{\omega }^{2}& 1\end{array}\right]\left[\begin{array}{c}A\\ B\\ C\\ D\end{array}\right]=\left[\begin{array}{c}1-\omega \\ 1-{\omega }^{2}\\ 5-3\omega \\ 5-3{\omega }^{2}\end{array}\right]$
Now let us solve by Cramer's rule. The determinant of the matrix is $\mathrm{\Delta }=12$, ${\mathrm{\Delta }}_{1}=-24$, ${\mathrm{\Delta }}_{2}=24$, ${\mathrm{\Delta }}_{3}=12$ and ${\mathrm{\Delta }}_{4}=60$. Thus the values of A,B,C,D are
$\begin{array}{rl}& A=\frac{{\mathrm{\Delta }}_{1}}{\mathrm{\Delta }}=-2\\ & B=\frac{{\mathrm{\Delta }}_{2}}{\mathrm{\Delta }}=2\\ & C=\frac{{\mathrm{\Delta }}_{3}}{\mathrm{\Delta }}=1\\ & D=\frac{{\mathrm{\Delta }}_{4}}{\mathrm{\Delta }}=5\end{array}$
So the required remainder is
$-2{x}^{3}+2{x}^{2}+x+5$

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