Holt Algebra 1 12-6 Dividing Polynomials 12-6 Dividing Polynomials Holt Algebra 1 Warm Up Warm Up Lesson Presentation Lesson Presentation Lesson Quiz Lesson

Holt Algebra 1

12-6 Dividing Polynomials12-6 Dividing Polynomials

Holt Algebra 1

Warm UpWarm Up

Lesson PresentationLesson Presentation

Lesson QuizLesson Quiz

Holt Algebra 1

12-6 Dividing Polynomials

Warm Up

Divide.

1. m2n ÷ mn4 2. 2x3y2 ÷ 6xy

3. (3a + 6a2) ÷ 3a2b

Factor each expression.

4. 5x2 + 16x + 12

5. 16p2 – 72p + 81

Holt Algebra 1


Divide a polynomial by a monomial or binomial.

Objective

Holt Algebra 1


To divide a polynomial by a monomial, you can first write the division as a rational expression. Then divide each term in the polynomial by the monomial.

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Example 1: Dividing a Polynomial by a Monomial

Divide (5x3 – 20x2 + 30x) ÷ 5x

x2 – 4x + 6

Write as a rational expression.

Divide each term in the polynomial by the monomial 5x.

Divide out common factors.

Simplify.

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Check It Out! Example 1a

Divide.

(8p3 – 4p2 + 12p) ÷ (–4p2)


Divide each term in the polynomial by the monomial –4p2.


Simplify.

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Check It Out! Example 1b

Divide.

(6x3 + 2x – 15) ÷ 6x


Divide each term in the polynomial by the monomial 6x.


Simplify.

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Division of a polynomial by a binomial is similar to division of whole numbers.

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Example 2A: Dividing a Polynomial by a Binomial

Divide.

x + 5

Factor the numerator.


Simplify.

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Example 2B: Dividing a Polynomial by a Binomial

Divide.

Factor both the numerator and denominator.


Simplify.

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Put each term of the numerator over the denominator only when the denominator is a monomial. If the denominator is a polynomial, try to factor first.

Helpful Hint

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Divide.

k + 5



Simplify.

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Divide.

b – 7



Simplify.

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Check It Out! Example 2c

Divide.

s + 6



Simplify.

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Recall how you used long division to divide whole numbers as shown at right. You can also use long division to divide polynomials. An example is shown below.

) x2 + 3x + 2x + 1

x2 + 2xx + 2x + 2

0

x + 2

(x2 + 3x + 2) ÷ (x + 2)

Divisor Quotient

Dividend

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Holt Algebra 1


Example 3A: Polynomial Long Division

Divide using long division.

(x2 +10x + 21) ÷ (x + 3)

x2 + 10x + 21)Step 1 x + 3Write in long division form

with expressions in standard form.

Divide the first term of the dividend by the first term of the divisor to get the first term of the quotient.

x2 + 10x + 21)Step 2 x + 3x

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Example 3A Continued


(x2 +10x + 21) ÷ (x + 3)

Multiply the first term of the quotient by the binomial divisor. Place the product under the dividend, aligning like terms.

x2 + 10x + 21)Step 3 x + 3x

x2 + 3x

x2 + 10x + 21)Step 4 x + 3–(x2 + 3x)

x

0 + 7x

Subtract the product from the dividend.

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x2 + 10x + 21)Step 5 x + 3–(x2 + 3x)

x

+ 21

Bring down the next term in the dividend.

Repeat Steps 2-5 as necessary.

x2 + 10x + 21)Step 6 x + 3–(x2 + 3x)

x + 7

7x + 21–(7x + 21)

0The remainder is 0.

7x

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Check: Multiply the answer and the divisor.

(x + 3)(x + 7)

x2 + 3x + 7x + 21

x2 + 10x + 21

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Example 3B: Polynomial Long Division


x2 – 2x – 8 )x – 4 Write in long division form.

–(x2 – 4x)2x

x2 – 2x – 8)x – 4

–(2x – 8)

0

x2 ÷ x = xMultiply x (x – 4 ). Subtract.

Bring down the 8. 2x ÷ x =2.

Multiply 2(x – 4). Subtract.The remainder is 0.

x+ 2

– 8

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Example 3B Continued


(x + 2)(x – 4)

x2 + 2x – 4x – 8

x2 – 2x + 8

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(2y2 – 5y – 3) ÷ (y – 3)

2y2 – 5y – 3 )Step 1 y – 3Write in long division form

with expressions in standard form.

Divide the first term of the dividend by the first term of the divisor to get the first term of the quotient.

2y2 – 5y – 3)Step 2 y – 32y

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Check It Out! Example 3a Continued


(2y2 – 5y – 3) ÷ (y – 3)

Multiply the first term of the quotient by the binomial divisor. Place the product under the dividend, aligning like terms.

Subtract the product from the dividend.

2y2 – 5y – 3)Step 3 y – 32y

2y2 – 6y

–(2y2 – 6y)0 + y

2y2 – 5y – 3)Step 4 y – 3

2y

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)Step 5 y – 3

2y

– 3

Bring down the next term in the dividend.

Repeat Steps 2–5 as necessary.

2y

2y2 – 5y – 3 )Step 6 y – 3–(2y2 – 6y)

y – 3 –(y – 3)

0The remainder is 0.

2y2 – 5y – 3–(2y2 – 6y)

y

+ 1

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(y – 3)(2y + 1)

2y2 + y – 6y – 3

2y2 – 5y – 3


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(a2 – 8a + 12) ÷ (a – 6)

a2 – 8a + 12 )a – 6 Write in long division form.

–(a2 – 6a)–2a

a a2 – 8a + 12)a – 6

–(–2a + 12)

0

a2 ÷ a = a

Multiply a (a – 6 ). Subtract.

Bring down the 12. –2a ÷ a = –2.

Multiply –2(a – 6). Subtract.

The remainder is 0.

– 2

+ 12

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Check It Out! Example 3b Continued


(a – 6)(a – 2)

a2 – 2a – 6a + 12

a2 – 8a + 12

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Sometimes the divisor is not a factor of the dividend, so the remainder is not 0. Then the remainder can be written as a rational expression.

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Example 4: Long Division with a Remainder

Divide (3x2 + 19x + 26) ÷ (x + 5)

3x2 + 19x + 26 )x + 5 Write in long division form.

3x2 + 19x + 26 )x + 53x

–(3x2 + 15x)4x

3x2 ÷ x = 3x.Multiply 3x(x + 5). Subtract.

Bring down the 26. 4x ÷ x = 4.

Multiply 4(x + 5). Subtract.–(4x + 20)

6 The remainder is 6.

Write the remainder as a rational expression using the divisor as the denominator.

+ 4

+ 26

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Example 4 Continued

Divide (3x2 + 19x + 26) ÷ (x + 5)

Write the quotient with the remainder.

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Divide.

3m2 + 4m – 2 )m + 3 Write in long division form.

3m2 + 4m – 2 )m + 33m

–(3m2 + 9m)

3m2 ÷ m = 3m.Multiply 3m(m + 3). Subtract.

Bring down the –2. –5m ÷ m = –5 .

Multiply –5(m + 3). Subtract.–5m

The remainder is 13.13

–(–5m – 15)

– 5

– 2

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Divide.

Write the remainder as a rational expression using the divisor as the denominator.

Holt Algebra 1


Check It Out! Example 4b Divide.

y2 + 3y + 2 )y – 3 Write in long division form.

–(y2 – 3y)

y2 ÷ y = y.Multiply y(y – 3). Subtract.

Bring down the 2. 6y ÷ y = 6.

y y2 + 3y + 2 )y – 3

Multiply 6(y – 3). Subtract.

The remainder is 20.20

6y –(6y –18)

Write the quotient with the remainder.

+ 6

+ 2

y + 6 +

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Sometimes you need to write a placeholder for a term using a zero coefficient. This is best seen if you write the polynomials in standard form.

Holt Algebra 1


Example 5: Dividing Polynomials That Have a Zero Coefficient

Divide (x3 – 7 – 4x) ÷ (x – 3).

(x3 – 4x – 7) ÷ (x – 3) Write in standard format.

x3 + 0x2 – 4x – 7 )x – 3Write in long division form.

Use 0x2 as a placeholder for the x2 term.

Holt Algebra 1


Example 5: Dividing Polynomials That Have a Zero Coefficient

Divide (x3 – 7 – 4x) ÷ (x – 3).

x3 + 0x2 – 4x – 7 )x – 3 x3 ÷ x = x2

Multiply x2(x – 3). Subtract.

(x3 – 4x – 7) ÷ (x – 3) Write the polynomials in standard form.

Write in long division form. Use 0x2 as a placeholder for the x2 term. x2

x3 + 0x2 – 4x – 7 )x – 3

–(x3 – 3x2)

3x2 – 4x Bring down –4x.

Holt Algebra 1


Example 5 Continued

x3 + 0x2 – 4x – 7 )x – 3 3x3 ÷ x = 3xMultiply x2(x – 3). Subtract.

x2

–(x3 – 3x2)

3x2 – 4x Bring down – 4x.–(3x2 – 9x)

5x–(5x – 15)

8

Bring down – 7.

Multiply 3x(x – 3). Subtract.

The remainder is 8.

+ 3x

– 7 Multiply 5(x – 3). Subtract.

+ 5

(x3 – 4x – 7) ÷ (x – 3) =

Holt Algebra 1


Recall from Chapter 7 that a polynomial in one variable is written in standard form when the degrees of the terms go from greatest to least.

Remember!

Holt Algebra 1


Divide (1 – 4x2 + x3) ÷ (x – 2).


(x3 – 4x2 + 1) ÷ (x – 2)

x3 – 4x2 + 0x + 1x – 2)

Write in standard format.Write in long division form.

Use 0x as a placeholder for the x term.

x3 – 4x2 + 0x + 1x – 2)x2 x3 ÷ x = x2

–(–2x2 + 4x)

– 4x –(–4x + 8)

–7

Bring down 0x. – 2x2 ÷ x = –2x.

Multiply –2x(x – 2). Subtract.Bring down 1.Multiply –4(x – 2). Subtract.

–(x3 – 2x2) – 2x2

Multiply x2(x – 2). Subtract.

– 2x

+ 0x

+ 1

– 4

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Divide (1 – 4x2 + x3) ÷ (x – 2).


(1 – 4x2 + x3) ÷ (x – 2) =

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Divide (4p – 1 + 2p3) ÷ (p + 1).


(2p3 + 4p – 1) ÷ (p + 1)

2p3 + 0p2 + 4p – 1p + 1)

Write in standard format.

Write in long division form. Use 0p2 as a placeholder for the p2 term.

2p3 – 0p2 + 4p – 1p + 1)2p2

p3 ÷ p = p2

–(–2p2 – 2p)

6p –(6p + 6)

–7

Bring down 4p. – 2p2 ÷ p = –2p.

Multiply –2p(p + 1). Subtract.Bring down –1.Multiply 6(p + 1). Subtract.

–(2p3 + 2p2) – 2p2

Multiply 2p2(p + 1). Subtract.

– 2p

+ 4p

–1

+ 6

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Check It Out! Example 5b Continued

(2p3 + 4p – 1) ÷ (p + 1) =

Holt Algebra 1


Lesson Quiz: Part I

Add or Subtract. Simplify your answer.

1.

3.

2.

(12x2 – 4x2 + 20x) ÷ 4x) 3x2 – x + 5

x – 2

4. x + 3

2x + 3

Holt Algebra 1


Lesson Quiz: Part II


5.

6. (8x2 + 2x3 + 7) (x + 3)

(x2 + 4x + 7) (x + 1)

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Holt Algebra 1 12-6 Dividing Polynomials 12-6 Dividing Polynomials Holt Algebra 1 Warm Up Warm Up Lesson Presentation Lesson Presentation Lesson Quiz Lesson