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Math 123 - Section 5.6 - Dividing Polynomials by Binomials - Part I - Page 1

Section 5.6 Dividing Polynomials by Binomials - Part I

I. Polynomial Long Division A. B. Remember long division from 5th grade:

650 23

There are a few differences though. 1. Write all polynomials in descending order. 2. If any exponents are missing, insert a &quot;0&quot; as a placeholder. Polynomial long division follows a similar type of reasoning. as regular long division. 1. Divide the first term inside by the first term outside. 2. Multiply the denominator by that answer. 3. Subtract that from the numerator. 4. Bring down the next term. 5. Divide the new first term inside by the first term outside. 6. Multiply the denominator by that answer. 7. Subtract that from the numerator. 8. Bring down the next term. 9. Continue in this fashion until you have used up all the terms. 10. Any remainder gets written over what you divided by. Examples ­ Perform the polynomial long division. x 2 + 10x + 25 1. x+5 We begin by noticing that all the exponents are present in the numerator from 2 down to zero (remember, x0 = 1 and 1 times 25 is 25, so we can write 25 as 25x0). So we will set up the problem in the following way:

x + 5 x2 + 10x + 25

C.

D.

We next take the first term inside (x2) and divide it by the first term outside (x) to get x. Now we will multiply both terms on the outside by x to get:

x + 10x + 25 + 5x

x + 5 x2 x2

Now we need to subtract, remember that we are subtracting both x2 and 5x. So really we have (x2 + 10x) - (x2 + 5x):

x x + 5 x

2 2

+ 10x + 25 + 5x) 5x

(x

We now need to bring down the next term, +25, and then continue the process. The first term on the inside is now 5x, the first term on the

Math 123 - Section 5.6 - Dividing Polynomials by Binomials - Part I - Page 2

outside is still x, so dividing these, we get +5. Multiply both of the terms on the outside by this number, then subtract to get:

x + + 10x + + 5x) 5x (5x + 25 + 25) 5 25

x + 5 -

x2 (x 2

0 Since our remainder is 0, we do not need to add a fraction on to the end of the quotient.

Answer: x + 5 2. You try one:

a2 + 9a + 20 a+4

Answer: a + 5 3.

x 2 + 6x + 9 x-3 We see that all the exponents are present in the numerator from 2 to 0, so we don't need to add any &quot;place holder&quot; terms. So we can divide the first term of the numerator (x2) by the first term of the denominator (x) to get x, then multiply both terms of the denominator by x, then subtract. This will look like: x x - 3 x2

2

+ -

6x 3x) 9x

+ 9

- (x

Remember that when we subtract, we need to change the signs on both terms in the parentheses! Now we bring down the +9, then divide the first term on the inside (9x) by the first term on the outside (x), to get +9. Multiplying this times both terms on the outside and then subtracting, we get:

x x - 3 x

2 2

+ + + -

9 9 9

+ -

6x 3x) 9x (9x

(x

27) 36 Since we have a remainder, we need to add a fraction to the end of the quotient by putting the remainder in the numerator and what we divided by (called the divisor) in the denominator. 36 Answer : x + 9 + x-3

Math 123 - Section 5.6 - Dividing Polynomials by Binomials - Part I - Page 3

4.

Now you try one:

5 a +1

a2 - a + 3 a +1

Answer : a - 2 +

5.

-8y + y 2 9 y-3 On this one, we notice that the numerator is not in descending order. When we set this one up, we will have to rectify that, and then the process is still the same. We divide the first term inside (y2) by the first term outside (y) to get y, and then multiply both terms outside by that, then subtract. This will look like:

y y 3 y

2 2

8y 3y) 5y

9

(y

Again, remember that we need to change the sign on both terms in the parentheses when we subtract. Next, bring down the 9. The first term inside is 5y, the first term outside is still y, when we divide them we get 5. Multiplying both terms outside by this, then subtracting, we get:

y y 3 y2 8y 3y) 5y 9 15) 24 (y 2 5 9

( 5y +

Since we have a remainder, we need to add a fraction to the end of the quotient by putting the remainder in the numerator and what we divided by in the denominator.

y5+

24 y3

6.

Now you try one:

5y +10 + y 2 y+2

y+3+

4 y+2

Math 123 - Section 5.6 - Dividing Polynomials by Binomials - Part I - Page 4

7.

x3 3x 2 5x 3 x 1

We see that all the exponents are present in the numerator from 3 to 0, so we don't need to add any &quot;place holder&quot; terms. So we can divide the first term of the numerator (x3) by the first term of the denominator (x) to get x2, then multiply both terms of the denominator by x, then subtract. This will look like:

x2 3x 2 x) 2x 2

x

1

x3

3

5x

3

(x

Now we bring down the 5x. Dividing the first thing inside (2x2) by the first thing outside (x), we get 2x. Multiply everything outside by 2x and subtract to get:

x2 3x 2 x) 2x 2 (2 x 2 2x 5x

x 1 x (x3

3

3

5x 2x ) 3x

Now we bring down the 3. Dividing the first thing inside (3x) by the first thing outside (x), we get 3. Multiply everything outside by 3 and subtract to get:

x2 3x 2 x2 ) 2x 2 (2 x 2 2x 5x 3 3

x 1 x (x3

3

5x 2x ) 3x 3 (3 x 3) 0

Since the remainder is 0, we can now write our answer.

x2 + 2x + 3

Math 123 - Section 5.6 - Dividing Polynomials by Binomials - Part I - Page 5

8.

2b 2 9b 5 2b 1 We see that all the exponents are present in the numerator from 2 to 0, so we don't need to add any &quot;place holder&quot; terms. So we can divide the first term of the numerator (2b2) by the first term of the denominator (2b) to get b, then multiply both terms of the denominator by b, then subtract. This will look like:

b 9b b)

2b 1

2b2

2

5

(2b

10b

Now we bring down the 5. Dividing the first thing inside (10b) by the first thing outside (2b), we get 5. Now multiply everything outside by 5 and subtract to get:

b 2b 1 2b

2 2

5 5 5

9b b) 10b

(2b

( 10b 5) 0

Since the remainder is 0, we can now write our answer.

9.

b5

4a 2 4a 3 2a 1

Now you try one:

10.

2a + 3

On #71, remember that the area of a rectangle is length times width. So the length is equal to the area divided by the width.

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