Question

If the polynomials \[2{x^3} + a{x^2} + 3x - 5\] and \[{x^3} + {x^2} - 4x + a\] , leave the same remainder when divided by \[x - 2\], find the value of \[a\]

Answer 1

Hint: Here the question is related to the polynomial, where we have to find the value of a. They have given the one common factor of the both polynomials. So by equating the value of the factor to both polynomials we are going to determine the value of a.

Complete step-by-step answer:
In mathematics, a polynomial is an expression consisting of variables and coefficients, that involves only the operations of addition, subtraction, multiplication, and non-negative integer exponentiation of variables.
Now consider the given polynomials, we denote these polynomials as \[f(x)\] and \[g(x)\]. Therefore we have
\[f(x) = 2{x^3} + a{x^2} + 3x - 5\] and \[g(x) = {x^3} + {x^2} - 4x + a\]
In the question they have mentioned that these two polynomials will have the same remainder when it is divided by \[x - 2\]
Therefore the \[x - 2\] is one of the factors of the polynomials \[f(x) = 2{x^3} + a{x^2} + 3x - 5\] and \[g(x) = {x^3} + {x^2} - 4x + a\]
So we will equate the value of x as 2 in both of the polynomials
Therefore now we have
\[ \Rightarrow f(x) = 2{x^3} + a{x^2} + 3x - 5\]
We consider the \[f(2)\] as \[{R_1}\](remainder 1)
On substituting the value of x as 2 in \[f(x) = 2{x^3} + a{x^2} + 3x - 5\], we have
\[ \Rightarrow f(2) = 2{(2)^3} + a{(2)^2} + 3(2) - 5\]
On simplifying we have
\[ \Rightarrow {R_1} = 2(8) + a(4) + 6 - 5\]
\[ \Rightarrow {R_1} = 16 + 4a + 6 - 5\]
On further simplification we have
\[ \Rightarrow {R_1} = 17 + 4a\]
Now we will consider another polynomial
\[ \Rightarrow g(x) = {x^3} + {x^2} - 4x + a\]
We consider the \[g(2)\] as \[{R_2}\](remainder 2)
On substituting the value of x as 2 in \[g(x) = {x^3} + {x^2} - 4x + a\], we have
\[ \Rightarrow g(2) = {(2)^3} + {(2)^2} - 4(2) + a\]
On simplifying we have
\[ \Rightarrow {R_2} = 8 + 4 - 4(2) + a\]
\[ \Rightarrow {R_2} = 8 + 4 - 8 + a\]
On further simplification we have
\[ \Rightarrow {R_2} = 4 + a\]
Since both remainders are the same we can equate both of them. So we have
\[ \Rightarrow {R_1} = {R_2}\]
Substituting the values we have
\[ \Rightarrow 17 + 4a = 4 + a\]
Take a to LHS and 17 to the RHS, so we have
\[ \Rightarrow 4a - a = 4 - 17\]
On simplifying we have
\[ \Rightarrow 3a = - 13\]
On dividing the above inequality by 3 we have
\[ \Rightarrow a = \dfrac{{ - 13}}{3}\]
Hence we have determined the value of a
So, the correct answer is “\[ a = \dfrac{{ - 13}}{3}\]”.

Note: While solving this question we have to know about the polynomial concept, the two polynomials will be equal then they have a solution that has zero. We must know about the concept of simple arithmetic operations which are needed to solve the question and while transferring the terms we should take care of the sign of the term.