Question

The polynomials $p\left( x \right) = {x^4} - 2{x^3} + 3{x^2} - ax + b$ when divided by $\left( {x - 1} \right)$ and $\left( {x + 1} \right)$ leaves the remainder 5 and 19 respectively. Find the values of $a$ and $b$ . Hence, find the remainder when $p\left( x \right)$ is divided by $\left( {x - 2} \right)$

Answer 1

Hint: We have been given a polynomial function, in which there are two unknown constants. It is also given that when this function is divided by factor, it leaves some remainder. We have to find the values of the unknown constant in the given polynomial. To find the value of the unknown constant, first, we equate the given factor to zero and find the value of $x$ . After that, we substitute this value of $x$ in the given polynomial and equate it to the given remainder. A similar process is obtained for another factor. Thus we obtained two-equations.
Now we solve these equations using any method of solving linear equations in two variables such as substitution method, elimination method, cross multiplication method etc.

Complete step-by-step solution:
Step1: We have been given the polynomial function $p\left( x \right) = {x^4} - 2{x^3} + 3{x^2} - ax + b$. It is given that when we divide the above polynomial function by $\left( {x - 1} \right)$ then we get a remainder of $5$.
Equating the value of $x - 1 = 0$ , we get $x = 1$ .
Substituting $x = 1$ in the given polynomial and equating it to $5$ , we get
$
p\left( 1 \right) = {\left( 1 \right)^4} - 2{\left( 1 \right)^3} + 3{\left( 1 \right)^2} - a\left( 1 \right) + b = 5 \\
   \Rightarrow 1 - 2 + 3 - a + b = 5
 $
Simplifying above equation, we get
$ \Rightarrow - a + b = 3$…………..(1)
Step 2: It is also given that when we divide the above polynomial function by $\left( {x + 1} \right)$ then we get a remainder of $19$.
Equating the value of $x + 1 = 0$ , we get $x = - 1$ .
Substituting $x = - 1$ in the given polynomial and equating it to $19$ , we get
$
  p\left( { - 1} \right) = {\left( { - 1} \right)^4} - 2{\left( { - 1} \right)^3} + 3{\left( { - 1} \right)^2} - a\left( { - 1} \right) + b = 19 \\
   \Rightarrow 1 - 2\left( { - 1} \right) + 3\left( 1 \right) + a + b = 19
 $
Simplifying above equation, we get
$ \Rightarrow a + b = 13$…………..(2)
Step 3: Now, when we add both the equations, we get
$
  \left( { - a + b} \right) + \left( {a + b} \right) = 3 + 13 \\
   \Rightarrow 2b = 16 \\
   \Rightarrow b = 8
 $
And when we subtract both the equation ,we get
$
  \left( { - a + b} \right) - \left( {a + b} \right) = 3 - 13 \\
   \Rightarrow - 2a = - 10 \\
   \Rightarrow a = 5
 $
Step 4: Now substituting the values of $a$ and $b$ in the given equation, we get the polynomial as
$p\left( x \right) = {x^4} - 2{x^3} + 3{x^2} - 5x + 8$
Step 5: Now to find the remainder, when the above polynomial is divided by $x - 2$, we equate the factor $x - 2 = 0$ which will give us the value of $x = 2$ .
Substituting $x = 2$ in given polynomial, we get
$
  p\left( 2 \right) = {\left( 2 \right)^4} - 2{\left( 2 \right)^3} + 3{\left( 2 \right)^2} - 5\left( 2 \right) + 8 \\
   \Rightarrow p\left( 2 \right) = 16 - 2 \times 8 + 3 \times 4 - 10 + 8 \\
   \Rightarrow p\left( 2 \right) = 16 - 16 + 12 - 10 + 8 \\
   \Rightarrow p\left( 2 \right) = 10 $
Which means that, when the polynomial $p\left( x \right) = {x^4} - 2{x^3} + 3{x^2} - ax + b$ is divided by $x - 2$ , it leaves the remainder of $10$.

Note: Take care of the positive and negative signs when evaluating the power of a negative number. Odd power of a negative number results in a negative number while even the power of a negative number results in an appositive number.