Question

Consider a polynomial, $f\left(x\right)=a{x}^3+b{x}^2+x+\frac{2}{3}.$ If $x+3$ is a factor of $f\left(x\right)$ and $f\left(x\right)$ is divided by $x+2$, then we get remainder as 4. Find the values of a and b.

• A. $a=\frac{29}{27},b=\frac{94}{27}$
  
• B. $a=\frac{25}{27},b=\frac{90}{27}$
  
• C. $a=\frac{29}{27},b=\frac{90}{27}$
  
• D. $a=\frac{9}{27},b=\frac{55}{27}$
  

Answer 1

The correct option is A $a=\frac{29}{27},b=\frac{94}{27}$


$f\left(x\right)=a{x}^3+b{x}^2+x+\frac{2}{3}$
$\therefore (x+3)$ is factor of $f\left(x\right)$.
By factor theorem,
$f(-3)=0$
$\Rightarrow a(-3{)}^3+b(-3{)}^2+(-3)+\frac{2}{3}=0$
$\Rightarrow -27a+9b-\frac{7}{3}=0$
$\Rightarrow -27a+9b=\frac{7}{3}$
$\Rightarrow 3(-9a+3b)=\frac{7}{3}$
$\Rightarrow -9a+3b=\frac{7}{9}$
$\therefore 3b=\frac{7}{9}+9a$.......(1)

When $f\left(x\right)$ is divided by $(x+2)$, remainder is 4.
By remainder theorem,
$f(-2)=4$
$\Rightarrow a(-2{)}^3+b(-2{)}^2+(-2)+\frac{2}{3}=4$
$\Rightarrow -8a+4b-\frac{4}{3}=4$
$\Rightarrow -24a+12b-4=12$
$\Rightarrow -24a+12b=16\Rightarrow -6a+3b=4$
$\therefore 3b=4+6a$ ...........(2)

From (1) and (2), we get
$3b=\frac{7}{9}+9a$ and $3b=6a+4$
$\frac{7}{9}+9a=6a+4$
$\Rightarrow 9a-6a=4-\frac{7}{9}\Rightarrow 3a=\frac{29}{9}$
$\therefore a=\frac{29}{27}$

from (1), we get
$3b=\frac{7}{9}+9a=\frac{7}{9}+9\times \frac{29}{27}$
$\Rightarrow 3b=\frac{7+87}{9}\Rightarrow 3b=\frac{94}{9}$
$\therefore b=\frac{94}{27}$

So, the correct answer is option (a).