Question 1200513: When the polynomial F(x) of degree 3 is divided by x-1,the remainder is -1.When F(x) is divided by x+1,the remainder is -5.If F(x) is divided by (x-1)(x+1),the remainder is ax+b,where a and b are constants .
Find the value of a and b.
Answer by ikleyn(52797)   (Show Source):
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When the polynomial F(x) of degree 3 is divided by x-1, the remainder is -1.
When F(x) is divided by x+1, the remainder is -5.
If F(x) is divided by (x-1)(x+1), the remainder is ax+b, where a and b are constants .
Find the value of a and b.
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Step by step solution.
(a) When the polynomial F(x) of degree 3 is divided by x-1, the remainder is -1.
According to the Remainder theorem, it means that F(1) = -1. (1)
(b) When F(x) is divided by x+1, the remainder is -5.
According to the Remainder theorem, it means that F(-1) = -5. (2)
(c) If F(x) is divided by (x-1)(x+1), the remainder is ax+b, where a and b are constants.
It means that
F(x) = Q(x)*(x-1)*(x+1) + (ax+b). (3)
In this equation (3), put x= 1. You will get
F(1) = Q(x)*(1-1)*(1+1) + (a*1+b), or
F(1) = a + b.
But, according to (1), F(1) = -1. Hence
a + b = -1. (4)
Next, in equation (3), put x= -1. You will get
F(-1) = Q(-1)*(1-(-1))*(1+(-1)) + (a*(-1)+b), or
F(-1) = -a + b.
But, according to (2), F(-1) = -5. Hence
-a + b = -5. (5)
(d) Thus we have this system of two equations for "a" and "b"
a + b = -1 (4)
-a + b = -5 (5)
To solve it, add equations (4) and (5). You will get
2b = -1 + (-5) = -6 ===> b = -6/2 = -3.
Then from (4)
a = -1 - b = -1 -1 - (-3) = -1 + 3 = 2.
ANSWER. a = 2; b = -1.
Solved.
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On the Remainder theorem and many associated solved problems, see the lessons
- Divisibility of polynomial f(x) by binomial (x-a) and the Remainder theorem
- Typical problems on the Remainder theorem
- Advanced problems on the Remainder theorem
in this site.
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