Question

If the polynomials $(2x^3+a{x}^2+3x-5)$ and $(x^3+x^2-4x-a)$ leave the same remainder when divided by $(x-1)$, find the value of $a$

Answer 1

The Remainder Theorem states that when you divide a polynomial $p\left(x\right)$ by any factor $(x-a)$; which is not necessarily a factor of the polynomial; you 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 $p\left(x\right)=2x^3+a{x}^2+3x-5$ and $q\left(x\right)=x^3+x^2-4x-a$ and the factor given is $g\left(x\right)=x-1$, therefore, by remainder theorem, the remainders are $p\left(1\right)$ and $q\left(1\right)$ respectively and thus,

$p\left(1\right)=(2\times 1^3)+(a\times 1^2)+(3\times 1)-5=(2\times 1)+(a\times 1)+3-5=2+a-2=aq\left(1\right)=1^3+1^2-(4\times 1)-a=1+1-4-a=-2-a$

Now since it is given that both the polynomials $p\left(x\right)=2x^3+a{x}^2+3x-5$ and $q\left(x\right)=x^3+x^2-4x-a$ leave the same remainder when divided by $(x-1)$, therefore $p\left(1\right)=q\left(1\right)$ that is:

$a=-a-2\Rightarrow a+a=-2\Rightarrow 2a=-2\Rightarrow a=-\frac{2}{2}\Rightarrow a=-1$

Hence, $a=-1$.