When a polynomial f(x) is divided by x-3 and x+6 , the respective remainders are 7 and 22. What is the remainder when f(x) is divided by (x-3)(x+6)
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Explanation:
=When a polynomial f(x) is divided by x-3 and x+6, the respective remainders are 7 and 22. What is the remainder when f(x) is divided by (x-3) (x+6)?
=Let’s look at a more general problem.
We know that the remainders when f(x) is divided by x−a and x−b are r and s respectively. What’s the remainder when f(x) is divided by (x−a)(x−b) ?
=Let’s assume first that a≠b and write px+q the sought remainder. This means that, for some polynomial g(x) ,
f(x)=(x−a)(x−b)g(x)+px+q(*)
=The hypotheses are equivalent to f(a)=r and f(b)=s , so we can evaluate (*) at a and b respectively, getting {pa+q=rpb+q=s
=This solves easily, by subtracting the second equation from the first, so p(a−b)=r−s and
p=r−sa−b
•Next
q=r−pa=r−r−sa−ba=as−bra−b
Hence the remainder is
r−sa−bx+as−bra−b
You can now substitute a=3 , b=−6 , r=7 and s=22 , getting
−53x+12
=Of course, if a=b the problem is underdetermined.
When ever a polynomial of degree N is divided by another polynomial of degree < N, the remainder will always be a polynomial ONE degree less than degree of denominator.
Remainder Theorem states that if a function f(x) is divided by (x-a), then f(a) is the remainder.
Taking cognizance of above two facts, we know the remainder when f(x) is divided by (x-3)(x+6) will be linear polynomial of degree ONE.
Let the remainder be represented by Ax + B
If f(x) is divided by x-3, remainder is 7
=> 3A + B = 7
If f(x) is divided by x-(-6), remainder is 22
=> -6A + B = 22
Solving the two equations, we get A = -15/9 & B = 12
So final remainder is -15x/9 + 12
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