remainder theorem

Question

Given that x^5+ax^3+bx^2-3=(x^2-1)Q(x)-x-2, where Q(x) is a polynomial. State the degree of Q(x) and find the value of a and b. Find also the remainder when Q(x) is divide by x+2.

Answer 1

x⁵ + ax³ + bx² - 3 = (x² - 1)Q(x) - x - 2

 

If x = 1, then 1 + a + b - 3 = (0)Q(1) -1 -2

 

               So, a + b - 2 = -3   Therefore, a + b = -1

 

If x = -1, then -1 -a + b - 3 = (0)Q(-1) + 1 - 2

 

               So, -a + b - 4 = -1  Therefore, -a + b = 3

 

We have the system of equations:  a + b = -1

                                                -a  + b = 3

 

Adding the equations, we get 2b = 2   So, b = 1 and a = -2

 

Since the degree of the original polynomial is 5, the degree of Q(x) must be 3 so that when Q(x) is multiplied by x² - 1, we get a 5th degree polynomial.  

 

Substituting the obtained values of a and b into the given equation, we have:

 

x⁵ -2x³ + x² - 3 = (x² - 1)Q(x) - x - 2  (**)

 

By the Remainder theorem, the remainder when Q(x) is divided by

x + 2 (which is equivalent to x - (-2)) is Q(-2).

 

So, plugging -2 into equation (**), we have:

 

   -32 + 16 + 4 - 3 = 3Q(-2) + 2 - 2

 

                    -15 = 3Q(-2)

 

                     Q(-2) = -5