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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).
Solved.
On the "Remainder theorem" see the lesson
- Divisibility of polynomial f(x) by binomial x-a
in this site.
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. .
Also, you have this free of charge online textbook in ALGEBRA-I in this site
- ALGEBRA-I - YOUR ONLINE TEXTBOOK.
The referred lessons are the part of this online textbook under the topic
"Divisibility of polynomial f(x) by binomial (x-a). The Remainder theorem".