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# 3.3 Remainder Theorem

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Tweet## Rafael Choy

on 1 April 2011#### Transcript of 3.3 Remainder Theorem

3.3 Remainder Theorem First of all, if you are not a student from class 3-5 of Woodlands Ring Secondary School.. Get out of here! Developing The Theorem Remainder Theorem Finding the Remainder Finding Unknowns Forming Equations to Find Unknowns This is the overview of what we are going to learn today. (goes clockwise) Before we state the theorem, let us look at how and why such a theorem was developed. Previously, when we divide polynomials, the long division method allows us to determine what the quotient and remainder are. However, long division is not an easy method to master and it is relatively prone to careless mistakes. At the same time, we need not find the value of the quotient if we are only required to find the remainder. We are interested in finding the remainder in a faster and simpler way. We will develop the theorem based on an example. Divide x + 2x + 2x + 3 by x + 1. 2 3 We express this in the form:

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

(a) x-2

(b) x+3

(c) 2x-1

Please spend 5 minutes on this before checking the solutions. Answers:

(a) 23

(b) -92

(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

Full transcriptdividend = 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

(a) x-2

(b) x+3

(c) 2x-1

Please spend 5 minutes on this before checking the solutions. Answers:

(a) 23

(b) -92

(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