SOLUTION: the remainders when f(x)=x³+ax²+bx+c is divided by (x-1),(x+2) and (x-2) are respectively 2,-1 and 15, find the quotient and remainder when f(x) is divided by (x+1).
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-> SOLUTION: the remainders when f(x)=x³+ax²+bx+c is divided by (x-1),(x+2) and (x-2) are respectively 2,-1 and 15, find the quotient and remainder when f(x) is divided by (x+1).
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Question 1111313: the remainders when f(x)=x³+ax²+bx+c is divided by (x-1),(x+2) and (x-2) are respectively 2,-1 and 15, find the quotient and remainder when f(x) is divided by (x+1). Answer by ikleyn(52788) (Show Source):
I will use the Remainder theorem. I will make all necessary explanations and references, but will not go in details.
Based on the Remainder theorem, from the given part you have these equations
f(1) = 2, or 1^3 + a*1^2 + b*1 + c = 2 (1)
f(-2) = -1, or (-2)^3 + a*(-2)^2 + b*(-2) + c = -1 (2)
f(2) = 15, or 2^3 + a*2^2 + b*2 + c = 15 (3)
Simplifying
1 + a + b + c = 2 (1')
-8 + 4a - 2b + c = -1 (2')
8 + 4a + 2b + c = 15 (3')
Simplifying one more time
a + b + c = 1 (1'')
4a - 2b + c = 7 (2'')
4a + 2b + c = 7 (3'')
Subtract (2'') from (3''). You will get 4b = 0 ====> b = 0.
Now substitute this value of b into eqs (1'') and (2''). You will get
a + c = 1 (4)
4a + c = 7 (5)
--------------------------------------- Subtract (4) from (5)
3a = 6 ====> a = 2
Then from (4) c = 1 - 2 = -1
Thus I restored the 3-rd degree polynomial. It is
f(x) = .
The rest is pure mechanical work:
f(x) = (x+1)*(x^2 + x -1).
Answer. The quotient under the question is (x^2 + x - 1). The remainder is 0.
-------------------- Theorem (the remainder theorem)
1. The remainder of division the polynomial by the binomial is equal to the value of the polynomial.
2. The binomial divides the polynomial if and only if the value of is the root of the polynomial , i.e. .
3. The binomial factors the polynomial if and only if the value of is the root of the polynomial , i.e. .
