Question

If, Q(x) = $4x^4+3x^3+2x^2+x+1$. What is the expression that is to be subtracted from it to make it divisible by $(3x-1)$?

• A. $\frac{139}{80}$
  
• B. $\frac{139}{81}$
  
• C. $\frac{140}{81}$
  
• D. $\frac{140}{80}$
  
  

Answer 1

The correct option is B $\frac{139}{81}$


$p\left(x\right)=g\left(x\right)+q\left(x\right)+r\left(x\right)$
On subtracting $r\left(x\right)$
$p\left(x\right)=g\left(x\right)+q\left(x\right)+r\left(x\right)-r\left(x\right)$
$p\left(x\right)=g\left(x\right)+q\left(x\right)$
Thus on subtracting remainder, $p\left(x\right)$ is completely divisible by $g\left(x\right)$
Given : $p\left(x\right)=4x^4+3x^3+2x^2+x+1$
$p\left(x\right)$ is divided by (3x-1) or $g\left(x\right)$
$g\left(x\right)$ must be equated to zero (By remainder theorem)
$(3x-1)=0\Rightarrow (x=1/3)$
So, by remainder theorem; Remainder = P(1/3)
$r\left(x\right)=P\left(1/3\right)=4(1/3{)}^4+3(1/3{)}^3+2(1/3{)}^2+1/3+1$
$=4/81+3/27+2/9+1/3+1$
$=\frac{4+9+18+27+81}{81}=\frac{139}{81}$
Thus $r\left(x\right)=\frac{139}{81}$ has to be subtracted from $p\left(x\right)$ for it to be completely divisible by $3x-1.$