Question

The polynomial $x^{4a}+x^{4b+1}+x^{4c+2}+x^{4d-1}$ where $a,b,c\&d\in N$ is divisible by

• A. $x(x-1)$
• B. $(x^2-1)$
• C. $x^3+x^2+x+1$
• D. $x^4-1$

Answer 1

The correct option is C $x^3+x^2+x+1$

$f\left(x\right)=x^{4a}+x^{4b+1}+x^{4c+2}+x^{4d-1}x(x-1)=0\Rightarrow x=0,1f\left(0\right)=0\text{but}f\left(1\right)\ne 0$
$\Rightarrow f\left(x\right)$ is not divisible by $x(x-1)$
$\text{For}(x^2-1)f(-1)=0\text{But}f\left(1\right)\ne 0$
$\Rightarrow f\left(x\right)$ is not divisible by $(x^2-1)$
$\text{For}{x}^3+x^2+x+1=0x=i,-i,-1f\left(i\right)=0,f(-1)=0,f(-i)=0$
$\Rightarrow f\left(x\right)$ is completely divisible by $x^3+x^2+x+1$
$\text{For}{x}^4-1=0x=1,i,-i-1$
$\text{But}f\left(1\right)\ne 0$
$\Rightarrow f\left(x\right)$ is not divisible by $x^4-1$

Alternate solution:
We know that $f\left(1\right)\ne 0$ so,
$f\left(x\right)$ will not be divisible by $x(x-1),(x^2-1),(x^4-1)$

OR,
If we put $a=b=c=d=1$
$f\left(x\right)=x^4+x^5+x^6+x^3\Rightarrow f\left(x\right)=x^3(x^3+x^2+x+1)$
So it is divisible by $x^3+x^2+x+1$