Question

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

• A. composite number
• B. prime number
• C. neither composite nor prime
• D. not an integer

Answer 1

The correct option is C neither composite nor prime

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.
Also given that $p^3=q^3$ → (1)
Put $x=3$ in both $p\left(x\right)$ and $g\left(x\right)$

Hence equation (1) becomes,
$a(3{)}^3+4(3{)}^2+3\left(3\right)-4=(3{)}^3-4(3)+a$
⇒ 27a + 36 + 9 − 4 = 27 − 12 + a
⇒ 27a + 41 = 15 + a
⇒ 26a = 15 − 41 = − 26
∴ $a=-1\Rightarrow -a=1$

Then the value is neither composite nor prime