SOLUTION: A polynomial has a remainder of -6 and 4 when divided by (x+1) and (x-1) respectively. Find the remainder when the polynomial is divided by x^2-1.
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Question 1075892: A polynomial has a remainder of -6 and 4 when divided by (x+1) and (x-1) respectively. Find the remainder when the polynomial is divided by x^2-1. Answer by ikleyn(52814) (Show Source):
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A polynomial has a remainder of -6 and 4 when divided by (x+1) and (x-1) respectively.
Find the remainder when the polynomial is divided by x^2-1.
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The fact that "A polynomial has a remainder of -6 when divided by (x+1)" means P(-1) = -6. (The "Remainder theorem")
The fact that "A polynomial has a remainder of 4 when divided by (x-1)" means P(1) = 4. (The same theorem)
Now, looking for the remainder when the polynomial is divided by x^2-1, we can write for this remainder
P(x) = Q(x)*(x^2-1) + ax + b (*) ( after division P(x) by (x^2-1) )
If you put x= -1 in this equation (*), you will get
P(-1) = -6 = a*(-1) + b.
If you put x= 1 in this equation (*), you will get
P(1) = 4 = a*1 + b.
Thus you have these two equations
-a + b = -6,
a + b = 4.
Solve it by any method you know to get b = -1, a = 5.
Answer. The remainder under the question is (5x -1).
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. .
