let assume a polynomial of degree more than 2 when divided by (x -2) leaves the remainder of 1 and when divide by (x - 3) leaves the remainder of 3 find what would be the remainder if it is divided by (x-3)(x-2)
Answers
x-2=1
(remainder theorem)
=> x= 3
Similarly,
x-3=3
=> x=3-3=0
Answer:
The remainder is 2x - 3.
Step-by-step explanation:
Call the polynomial f(x).
When divided by x-a, the remainder left is the same as the value of f(a).
[ To see this, when dividing by x - a we get f(x) = (x-a)(some polynomial) + (remainder), and putting x = a, this says f(a) = remainder. ]
Since f(x) leaves remainder 1 when divided by x-2, we know f(2) = 1.
Since f(x) leaves remainder 3 when divided by x-3, we know f(3) = 3.
When dividing by (x-3)(x-2), we get
f(x) = (x-3)(x-2)(some polynomial) + Px + Q
where Px+Q is the remainder.
Putting x = 2, we get 2P + Q = f(2) = 1.
Putting x = 3, we get 3P + Q = f(3) = 3.
Solving these two equations for P and Q, we have P = 2 and Q = -3.
The remainder is 2x - 3.