The polynomials $ax^3 + 3x^2 - 3$ and $2x^3 - 5x + a$ when divided by $(x - 4)$ leave the remainders $R_1$ and $R_2$, respectively. Find the values of $a$ in each case of the following cases, if $R_1 = R_2$.

Given:

The polynomials $ax^3 + 3x^2 - 3$ and $2x^3 - 5x + a$ when divided by $(x - 4)$ leave the remainders $R_1$ and $R_2$, respectively.

$R_1 = R_2$.

To do:

We have to find the value of $a$.

Solution:

The remainder theorem states that when a polynomial, $p(x)$ is divided by a linear polynomial, $x - a$ the remainder of that division will be equivalent to $p(a)$.

Let $f(x) = ax^3 + 3x^2 - 3$ and $g(x) = 2x^3 - 5x + a$

$p(x) = x-4$

So, the remainders will be $f(4)$ and $g(4)$.

$f(4) = a(4)^3+3(4)^2 -3$

$= a(64) + 3(16) -3$

$=64a+48-3$

$=64a+45$

$g(4) = 2(4)^3-5(4) +a$

$= 2(64) -20 +a$

$=128-20+a$

$=a+108$

$R_1=R_2$

This implies,

$64a+45=a+108$

$64a-a=108-45$

$63a=63$

$a=1$

Therefore, the value of $a$ is $1$.    

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Updated on: 10-Oct-2022

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