Question

The polynomials $(2x^3-5x^2+x+a)$ and $(a{x}^3+2x^3-3)$ when divided by $(x-2)$ leave the remainders $R_1$ and $R_2$ respectively. Find the value of ${}^{\prime }{a}^{\prime }$ in the following case, if
$2R_1+R_2=0$

Answer 1

The Remainder Theorem states that when we divide a polynomial $p\left(x\right)$ by any factor $(x-a)$; which is not necessarily a factor of the polynomial; we will obtain a new smaller polynomial and a remainder, and this remainder is the value of $p\left(x\right)$ at $x=a$, that is $p\left(a\right)$.

Let $p\left(x\right)=2x^3-5x^2+x+a$ and $q\left(x\right)=a{x}^3+2x^2-3$ and the factor given is $g\left(x\right)=x-2$, therefore, by remainder theorem, the remainders are $p\left(2\right)$ and $q\left(2\right)$ respectively and thus,

$p\left(2\right)=(2\times 2^3)-(5\times 2^2)+2+a=(2\times 8)-(5\times 4)+2+a=16-20+2+a=a-2q\left(2\right)=(a\times 2^3)+(2\times 2^2)-3=(a\times 8)+(2\times 4)-3=8a+8-3=8a+5$

Now since it is given that both the polynomials $p\left(x\right)=2x^3-5x^2+x+a$ and $q\left(x\right)=a{x}^3+2x^2-3$ leave the remainders $R_1$ and $R_2$ and $2R_1+R_2=0$ when divided by $(x-2)$, therefore $p\left(2\right)=q\left(2\right)$ that is:

$2(a-2)+(8a+5)=0\Rightarrow 2a-4+8a+5=0\Rightarrow 10a+1=0\Rightarrow 10a=-1\Rightarrow a=-\frac{1}{10}$

Hence, $a=-\frac{1}{10}$.