Question

Consider two polynomials $f\left(x\right)=x^3-2x^2+\frac{1}{2}$, $g\left(x\right)=x^3-3x+1$.

$f\left(x\right)$ is divided by $x+1$, to get remainder $R_1$ and $g\left(x\right)$ is divided by $x+2$, to obtain remainder $R_2$. Find $(R_1^2+R_2^2-R_1{R}_2)$.

• A. $\frac{16}{19}$
  
• B. $\frac{17}{4}$
  
• C. $\frac{19}{16}$
  
• D. $\frac{19}{4}$
  

Answer 1

The correct option is D $\frac{19}{4}$


$f\left(x\right)=x^3-2x^2+\frac{1}{2},g\left(x\right)=x^3-3x+1$
When $f\left(x\right)$ is divided by $x+1$, then by Remainder theorem,
$R_1=f(-1)=(-1{)}^3-2(-1{)}^2+\frac{1}{2}$
$\Rightarrow R_1=-1-2+\frac{1}{2}=-3+\frac{1}{2}=\frac{-5}{2}$
$\therefore R_1=\frac{-5}{2}$
When $g\left(x\right)$ is divided by $x+2$ then by Remainder theorem,
$R_2=g(-2)=(-2{)}^3-3(-2)+1$
$=-8+6+1$
$\therefore R_2=-1$
$R_1^2+R_2^2-R_1{R}_2$
$={(-\frac{5}{2})}^2+(-1{)}^2-\left(\frac{-5}{2}\right)(-1)$
$=\frac{25}{4}+1-\frac{5}{2}=\frac{25+4-10}{4}$
$=\frac{19}{4}$
So, the correct answer is option (d).