# If polynomials ax3 + 3x2 − 3

Question:

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. $2 R_{1}-R_{2}=0$

Solution:

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