Question

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

Answer 1

Given $a{x}^3+4x^2+3x-4$ and $x^3-4x+a$ leave the same remainder when divided by $x-3$.

Let p(x)= $a{x}^3+4x^2+3x-4$ and g(x)=$x^3-4x+a$

By remainder theorem, if f(x) is divided by (x − a) then the remainder is f(a).
Here, when p(x) and g(x) are divided by (x − 3) the remainders are p(3) and g(3) respectively.

Put x = 3 in both p(x) and g(x), so we get p(3)=g(3) as they are equal

$\Rightarrow a(3{)}^3+4(3{)}^2+3\left(3\right)-4=(3{)}^3-4(3)+a$

$\Rightarrow 27a+36+9-4=27-12+a$

$\Rightarrow 27a-a=15-41$

$\Rightarrow 26a=-26$

$\therefore a=-1$