Question

When $f\left( x \right) = {x^4} - 2{x^3} + 3{x^2} - ax - b$ is divided by $\left( {x + 1} \right)$ and $\left( {x - 1} \right)$, we get the remainders as $19$ and $5$ respectively. Find the remainder when $f\left( x \right)$ is divided by $\left( {x - 3} \right)$.

Answer 1

Hint:In the question, we are provided with a function which is divided by two divisor polynomials. The remainder in both cases is given to us in the problem itself. We first use the remainder theorem to find the remainders in both the cases and hence find the values of a and b. This question requires us to have the knowledge of basic and simple algebraic rules and operations such as substitution, addition, multiplication, subtraction and many more like these. A thorough understanding of functions, division algorithms and its applications will be of great significance. Then, we use the remainder theorem again to find remainder when the original polynomial is divided by another divisor polynomial

Complete step by step answer:
In the given question, we are given the function $f\left( x \right) = {x^4} - 2{x^3} + 3{x^2} - ax - b$. Now, the polynomial function is divided by the divisor polynomial $\left( {x + 1} \right)$.Now, we know that we can evaluate the remainder using the remainder theorem by substituting the value of the variable in dividend. First equating the divisor ${D_1}\left( x \right) = \left( {x + 1} \right)$ as zero, we get the value of x as,
$ \Rightarrow \left( {x + 1} \right) = 0$
$ \Rightarrow x = - 1$
So, we substitute the value of x as $\left( { - 1} \right)$ in the original dividend function to get the remainder using the remainder theorem.
So, $f\left( { - 1} \right) = {\left( { - 1} \right)^4} - 2{\left( { - 1} \right)^3} + 3{\left( { - 1} \right)^2} - a\left( { - 1} \right) - b$
So, we compute the powers of $\left( { - 1} \right)$.
$ \Rightarrow f\left( { - 1} \right) = 1 - 2\left( { - 1} \right) + 3\left( 1 \right) + a - b$
Opening the brackets and simplifying the expression,
$ \Rightarrow f\left( { - 1} \right) = 1 + 2 + 3 + a - b$
$ \Rightarrow f\left( { - 1} \right) = a - b + 6$
We are also given the remainder as $19$. So, we get,
$ \Rightarrow a - b + 6 = 19$
Shifting constant terms to right side of the equation, we get,
$ \Rightarrow a - b = 13 - - - - \left( 1 \right)$

Now, the polynomial function is divided by the divisor polynomial $\left( {x - 1} \right)$.Now, we know that we can evaluate the remainder using the remainder theorem by substituting the value of the variable in the dividend. First equating the divisor ${D_2}\left( x \right) = \left( {x + 1} \right)$ as zero, we get the value of x as,
$ \Rightarrow \left( {x - 1} \right) = 0$
$ \Rightarrow x = 1$
So, we substitute the value of x as $1$ in the original dividend function to get the remainder using the remainder theorem.
So, $f\left( 1 \right) = {\left( 1 \right)^4} - 2{\left( 1 \right)^3} + 3{\left( 1 \right)^2} - a\left( 1 \right) - b$
So, we compute the powers of $\left( 1 \right)$.
$ \Rightarrow f\left( 1 \right) = \left( 1 \right) - 2\left( 1 \right) + 3\left( 1 \right) - a - b$
Opening the brackets and simplifying the expression,
$ \Rightarrow f\left( 1 \right) = 2 - a - b$
We are also given the remainder as $5$. So, we get,
$ \Rightarrow 2 - a - b = 5$
Shifting constant terms to left side and terms consisting x to the right side of the equation, we get,
$ \Rightarrow a + b = - 3 - - - - \left( 2 \right)$

Now, we have to solve the equation $\left( 1 \right)$ and $\left( 2 \right)$ to find the values of a and b. Hence, adding both the equations, we get,
\[ \Rightarrow \left( {a - b} \right) + \left( {a + b} \right) = 13 + \left( { - 3} \right)\]
Cancelling the like terms with opposite signs, we get,
\[ \Rightarrow 2a = 10\]
Dividing both the sides of the equation by $5$, we get,
\[ \Rightarrow a = 5\]
So, value of a is $5$. Putting the value of a in the equation $\left( 2 \right)$, we get,
$ \Rightarrow \left( 5 \right) + b = - 3$
$ \Rightarrow b = - 8$
So, value of b is $\left( { - 8} \right)$.
Therefore, we have the polynomial as $f\left( x \right) = {x^4} - 2{x^3} + 3{x^2} - 5x + 8$.

Now, the polynomial function is divided by the divisor $\left( {x - 3} \right)$ and we have to find the value of the remainder. First equating the divisor ${D_3}\left( x \right) = \left( {x - 3} \right)$ as zero, we get the value of x as,
$ \Rightarrow \left( {x - 3} \right) = 0$
$ \Rightarrow x = 3$
So, we substitute the value of x as $\left( 3 \right)$ in the original dividend function to get the remainder using the remainder theorem.
So, $f\left( 3 \right) = {\left( 3 \right)^4} - 2{\left( 3 \right)^3} + 3{\left( 3 \right)^2} - 5\left( 3 \right) + 8$
So, we compute the powers of $\left( 3 \right)$.
$ \Rightarrow f\left( 3 \right) = 81 - 2\left( {27} \right) + 3\left( 9 \right) - 15 + 8$
Opening the brackets and simplifying the expression,
$ \Rightarrow f\left( 3 \right) = 81 - 54 + 27 - 15 + 8$
Doing the calculations, we get,
$ \Rightarrow f\left( 3 \right) = 27 + 20$
$ \therefore f\left( 3 \right) = 47$

So, we get the remainder as $47$ when the given polynomial is divided by $\left( {x - 3} \right)$.

Note:Remainder theorem is an approach of Euclidean division of polynomials.Remainder theorem requires just a simple change of variable in the function so as to find the remainder of the division procedure. Substitution of a variable involves putting a certain value in place of the variable. That specified value may be a certain number or even any other variable. We must take care of the calculations in order to be sure of the final answer.