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Divide the polynomial $3 x^{4}-4 x^{3}-3 x-1$ by $x-1$.
Solution
By long division, we have :
$\overset{3{{x}^{3}}-{{x}^{2}}-x-4}{\mathop{\begin{align}
& x-1\sqrt{\begin{align}
& 3{{x}^{4}}-4{{x}^{3}}-3x-1 \\
& 3{{x}^{4}}-3{{x}^{3}} \\
\end{align}} \\
& \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\
\end{align}}}\,$
$-{{x}^{3}}-3x-1$
$\mp \,\,{{x}^{3}}\,\,\pm \,\,\,{{x}^{2}}$
$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_$
$-x^{2}-3 x-1$
$\mp \,{{x}^{2}}\pm \,\,x$
$\_\_\_\_\_\_\_\_\_\_\_\_\_\_$
$-4 x-1$
$\mp \,\,4x\pm 1$
$\_\_\_\_\_\_\_\_\_\_\_\_\_\_$
$-5$
Here, the remainder is $-5 .$ Now, the zero of $x-1$ is $1 .$ So, putting $x=1$ in $p(x),$ we see that
$p(1)=3(1)^{4}-4(1)^{3}-3(1)-1$
$=3-4-3-1$
$=-\,5,$ which is the remainder.