If polynomials ax $^{3}+3 x^{2}-3$ and $2 x^{3}-5 x+a$ when divided by $(x-4)$ leave the remainders as $R_{1}$ and $R_{2}$ respectively. Find the values of a in each of the following cases, if
1. $R_{1}=R_{2}$
2. $R_{1}+R_{2}=0$
3. $2 R_{1}-R_{2}=0$
Here, the polynomials are
$f(x)=a x_{3}+3 x_{2}-3$
$p(x)=2 x^{3}-5 x+a$
Let,
R1 is the remainder when f(x) is divided by x - 4
$\Rightarrow R_{1}=f(4)$
$\Rightarrow \mathrm{R}_{1}=\mathrm{a}(4)^{3}+3(4)^{2}-3$
$=64 \mathrm{a}+48-3$
$=64 \mathrm{a}+45 \ldots .1$
Now, let
$R_{2}$ is the remainder when $p(x)$ is divided by $x-4$
$\Rightarrow R_{2}=p(4)$
$\Rightarrow R_{2}=2(4)^{3}-5(4)+a$
$=128-20+a$
$=108+a \ldots .2$
1. Given, $R_{1}=R_{2}$
$\Rightarrow 64 a+45=108+a$
$\Rightarrow 63 a=63$
$\Rightarrow a=1$
2. Given, $R_{1}+R_{2}=0$
$\Rightarrow 64 a+45+108+a=0$
$\Rightarrow 65 a+153=0$
$\Rightarrow a=-153 / 65$
3. Given, $2 R_{1}-R_{2}=0$
$\Rightarrow 2(64 a+45)-108-a=0$
$\Rightarrow 128 a+90-108-a=0$
⟹ 127a - 18 = 0
⟹ a = 18/127