If the polynomial $(2x^3+a{x}^2+3x-5)$ and $(x^3+x^2-2x+a)$ leave the same remainder when divided by $(x-2)$, find the remainder.

The correct answer is 5.

Step-by-step Explanation:

Let the two given polynomials be:
$P\left(x\right)=2x^3+a{x}^2+3x-5$
and
$Q\left(x\right)=x^3+x^2-2x+a$

According to the Remainder Theorem, when a polynomial f(x) is divided by $(x-c)$, the remainder is f(c).
Here, both polynomials are divided by $(x-2)$. Therefore, the remainders are $P\left(2\right)$ and $Q\left(2\right)$ respectively.

Since it is given that both polynomials leave the same remainder:
$P\left(2\right)=Q\left(2\right)$

First, let's find the value of $P\left(2\right)$ by substituting x = 2 into P(x):
$P\left(2\right)=2{\left(2\right)}^3+a{\left(2\right)}^2+3\left(2\right)-5$
$P\left(2\right)=2\left(8\right)+4a+6-5$
$P\left(2\right)=16+4a+1$
$P\left(2\right)=17+4a$

Next, let's find the value of $Q\left(2\right)$ by substituting x = 2 into Q(x):
$Q\left(2\right)={\left(2\right)}^3+{\left(2\right)}^2-2\left(2\right)+a$
$Q\left(2\right)=8+4-4+a$
$Q\left(2\right)=8+a$

Equating the two expressions for the remainders:
$17+4a=8+a$

Subtract a from both sides:
$17+3a=8$

Subtract 17 from both sides:
$3a=8-17$
$3a=-9$
$a=-3$

Now, substitute the value of a back into either remainder expression to find the final remainder value:
$\text{Remainder}=8+a$
$\text{Remainder}=8+(-3)=5$

Thus, the remainder is 5.