Using remainder theorem find the remainder when $4x ^3-12x^2+11x-3$ is divided by $x+\frac{1}{2}$ without using long division. Don't use long division.

Given: $4x ^3-12x^2+11x-3$ is divided by $x+\frac{1}{2}$.

To do: To find the remainder by using remainder theorem when $4x ^3-12x^2+11x-3$ is divided by $x+\frac{1}{2}$.

Solution:

Let $f( x)=4x ^3-12x^2+11x-3$ and $g( x)=x+\frac{1}{2}$.

Let $g( x)=0$

$\Rightarrow x+\frac{1}{2}=0$

$\Rightarrow x=-\frac{1}{2}$, on substituting this value in $f( x)$.

$\Rightarrow f( -\frac{1}{2})=4( -\frac{1}{2})^3-12( -\frac{1}{2})^2+11( -\frac{1}{2})-3$

$\Rightarrow f( -\frac{1}{2})=4( -\frac{1}{2}\times-\frac{1}{2}\times-\frac{1}{2})-12( -\frac{1}{2}\times-\frac{1}{2})+11( -\frac{1}{2})-3$

$\Rightarrow f( -\frac{1}{2})=4( -\frac{1}{8})-12( \frac{1}{4})-\frac{11}{2}-3$

$\Rightarrow f( -\frac{1}{2})=-\frac{1}{2}-3-\frac{11}{2}-3$

$\Rightarrow f( -\frac{1}{2})=-\frac{12}{2}-3$

$\Rightarrow f( -\frac{1}{2})=-6-3$

$\Rightarrow f( -\frac{1}{2})=-9$

Thus, the remainder is $-9$.