Mara Cook

2022-06-22

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

Jovan Wong

Beginner2022-06-23Added 23 answers

Let the polynomial be $P(x)$. It is given that for some polynomials $Q(x),{Q}_{1}(x)$

$P(x)=({x}^{2}+x+1)Q(x)+1-x$

$P(x)=({x}^{2}-x+1){Q}_{1}(x)+3x+5$

Now it is well known that ${x}^{2}+x+1$ has the zeros as $\omega ,{\omega}^{2}$ So We get

$P(\omega )=1-\omega $

$P({\omega}^{2})=1-{\omega}^{2}$

$P(-\omega )=3\omega +5$

$P(-{\omega}^{2})=-3{\omega}^{2}+5$

Now let us assume for some polynomial ${Q}_{2}(x)$ we have

$P(x)=({x}^{4}+{x}^{2}+1){Q}_{2}(x)+A{x}^{3}+B{x}^{2}+Cx+D$

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

$-2{x}^{3}+2{x}^{2}+x+5$

$P(x)=({x}^{2}+x+1)Q(x)+1-x$

$P(x)=({x}^{2}-x+1){Q}_{1}(x)+3x+5$

Now it is well known that ${x}^{2}+x+1$ has the zeros as $\omega ,{\omega}^{2}$ So We get

$P(\omega )=1-\omega $

$P({\omega}^{2})=1-{\omega}^{2}$

$P(-\omega )=3\omega +5$

$P(-{\omega}^{2})=-3{\omega}^{2}+5$

Now let us assume for some polynomial ${Q}_{2}(x)$ we have

$P(x)=({x}^{4}+{x}^{2}+1){Q}_{2}(x)+A{x}^{3}+B{x}^{2}+Cx+D$

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

$-2{x}^{3}+2{x}^{2}+x+5$