Remainder theorem: If a polynomial P(x) of degree n ≥ 1 is divided by x - b, where b is a constant, then the remainder is P(b).
Example #1
In the lesson about polynomial long division, we ended with the following result:
(x^{2} - 5x + 1) ÷ (x + 3) = x + -8 with a remainder of 25
You can also write (x^{2} - 5x + 1) = (x + 3)(x + -8) + 25
Observation
(x + 3)(x + -8) + 25 = x^{2} + -8x + 3x + -24 + 25
(x + 3)(x + -8) + 25 = x^{2} + -5x + 1
Using (x^{2} - 5x + 1) = (x + 3)(x + -8) + 25
Dividend = (x^{2} - 5x + 1)
divisor = (x + 3)
quotient = (x + -8)
Remainder = 25
Dividend = divisor x quotient + remainder
When we divide (x^{2} - 5x + 1) by (x + 3), we get a remainder of 25.
Using P(x) = (x^{2} - 5x + 1), calculate P(-3).
P(-3) = (-3)^{2} - 5(-3) + 1
P(-3) = 9 - -15 + 1
P(-3) = 9 + 15 + 1
P(-3) = 25
As you can see, if a = -3, P(-3) is equal to the remainder (25) when (x^{2} - 5x + 1) is divided by (x + 3)
Example #2
In the lesson about polynomial long division, we ended also with the following result:
(x^{2} + 3x - 10) ÷ (x - 2) = x + 5
(x^{2} + 3x - 10) = (x - 2)(x + 5)
(x^{2} + 3x - 10) = (x - 2)(x + 5) + 0
P(2) = 2^{2} + 3(2) - 10
P(2) = 4 + 6 - 10
P(2) = 10 - 10
P(2) = 0
Again, you can see that if a = 3, P(2) is equal to the remainder (0) when (x^{2} - 5x + 1) is divided by (x - 2)
Sep 30, 22 04:45 PM