If the polynomials $2x^3 + ax^2 + 3x - 5$ and $x^3 + x^2 - 4x + a$ leave the same remainder when divided by $x - 2$, find the value of $a$.

Given:

The polynomials $2x^3 + ax^2 + 3x - 5$ and $x^3 + x^2 - 4x + a$ leave the same remainder when divided by $x - 2$.

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 $f(x) = 2x^3 + ax^2 + 3x - 5$ and $g(x) = x^3 + x^2 - 4x + a$

$p(x) = x-2$

So, the remainders will be $f(2)$ and $g(2)$.

$f(2) = 2(2)^3+a(2)^2+3(2) -5$

$= 2(8) + a(4) +6-5$

$=16+4a+1$

$=4a+17$

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

$= 8 + 4 -8 +a$

$=4+a$

This implies,

$4a+17=4+a$

$4a-a=4-17$

$3a=-13$

$a=\frac{-13}{3}$

Therefore, the value of $a$ is $\frac{-13}{3}$.  

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Updated on: 10-Oct-2022

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