Send the link below via email or IMCopy
Present to your audienceStart remote presentation
- Invited audience members will follow you as you navigate and present
- People invited to a presentation do not need a Prezi account
- This link expires 10 minutes after you close the presentation
- A maximum of 30 users can follow your presentation
- Learn more about this feature in our knowledge base article
Do you really want to delete this prezi?
Neither you, nor the coeditors you shared it with will be able to recover it again.
Make your likes visible on Facebook?
You can change this under Settings & Account at any time.
3.3 Remainder Theorem
Transcript of 3.3 Remainder Theorem
dividend = divisor X quotient + remainder Please take down the notes that appear in such green boxes and write them in your 3.3 notes.
Of course, you can take down other notes which you feel are important for your learning.
We shall start now. x + 2x + 2x + 3 = (x+1)Q(x) + R 2 3 linear divisor
(degree 1) quotient in terms of x constant remainder (degree 0) *degree of remainder < degree of divisor *How do we know the remainder is a constant? Let P(x) = x + 2x + 2x + 3,
then P(x) = (x+1)Q(x) + R 2 3 When x = -1, we observe that the product of the divisor and quotient becomes zero. P(-1) = (-1 + 1)Q(x) + R
= 0 + R
= R P(-1) = (-1) + 2(-1) + 2(-1) + 3
= -1 + 2 - 2 + 3
= 2 R = 2 .. . 3 2 What about the value of the polynomial when x = -1?
Try substituting it into the polynomial. Check the remainder by doing long division.
You should still get a remainder of 2! The first method is much faster and less tedious as compared to long division. In addition, there is no need to compute the value of the quotient.
Now, we know when P(x) is divided by (x+1), the remainder is P(-1).
This leads us to the Remainder Theorem. states that: When a polynomial P(x) is divided by a linear divisor ax+b, the remainder is P(- ). b / a P(x) = (ax+b)Q(x) + R To prove the theorem formally, let's try evaluating P(- ). b a / When x = - ,
P(- ) = (-b + b)Q(- ) + R
= (0)Q(- ) + R
= R (proven) b a / b a / b a / b a / Note: we are unconcerned about the value of Q(- ). b a / Example 1
Let P(x) = 4x - 5x + 1.
Find the remainder when P(x) is divided by
Please spend 5 minutes on this before checking the solutions. Answers:
(c) -1 When divided by x-2,
remainder = P(2)
= 4(2) - 5(2) + 1
= 23 3 When divided by x+3,
remainder = P(-3)
= 4(-3) - 5(-3) + 1
= -92 3 When divided by 2x-1,
remainder = P(1/2)
= 4(1/2) - 5(1/2) + 1
= -1 3 Example 2
(a) When 4x - kx + 7 is divided by x-3, the remainder is -2. Find the value of the constant k.
(b) When x + 3x - px + 4 is divided by x-2, the remainder is p. Find the value of the constant p. 2 2 3 You can attempt both questions now, or you may look at the solution for (a) first before trying (b). Answer:
(a) 15 When divided by x-3,
remainder = 4(3) - k(3) + 7 = -2
3k = 45
k = 15 2 Answer:
(b) 8 When divided by x-2,
remainder = 2 + 3(2) - p(2) + 4 = p
24 - 2p = p
3p = 24
p = 8 2 3 The example you have just completed requires you to apply the theorem directly to find the The example you have just completed requires you to apply the theorem directly to find the Example 3
(a) Let P(x) = ax + bx + 2. When P(x) is divided by x-2, the remainder is 1. When P(x) is divided by x-1, the remainder is 2. Find the value of a and of b. Because there are two unknowns a and b, we need to form equations to evaluate them. Answers:
a = -1/2
b = 1/2 When divided by x-2,
remainder = P(2)
= 4a + 2b + 2
= 1 --- (1)
When divided by x-1,
remainder = P(1)
= a + b + 2
= 2 --- (2) (2): a = -b --- (3)
Sub (3) into (1):
-4b + 2b + 2 = 1
2b = 1
b = 1/2
Sub b = 1/2 into (3):
a = -1/2 Example 3
(b) Let Q(x) = 8x + ax + bx - 9. When Q(x) is divided by x+2, the remainder is -95. When Q(x) is divided by 2x-3, the remainder is 3. Find the value of a and of b. 2 Try this example on your own. Only the answers will be provided. Answers:
a = -6
b = -1 The example you have just completed requires you to apply the theorem to form find It's time for more practice!
Ex 3.3 (p46) 2, 4, 6, 7, 9, 10
(complete by Friday 1 April 2011) (this is not a joke) It's really time to (and start on your homework!) 3 2 3