topic 8.2.2

1

Topic 8.2.2Topic 8.2.2

Dividing RationalExpressions

Dividing RationalExpressions

2

Lesson

1.1.1

California Standard:13.0 Students add, subtract, multiply, and divide rational expressions and functions. Students solve both computationally and conceptually challenging problems by using these techniques.

What it means for you:You’ll divide rational expressions by factoring and cancelling.

Dividing Rational ExpressionsDividing Rational ExpressionsTopic

8.2.2

Key words:• rational• reciprocal• common factor

3

Lesson

1.1.1

Dividing by rational expressions is a lot like multiplying — you just have to do an extra step first.

Topic

8.2.2

That extra step is finding the reciprocal.

Dividing Rational ExpressionsDividing Rational Expressions

4

Lesson

1.1.1

Dividing is the Same as Multiplying by the Reciprocal

Topic

8.2.2

Given any nonzero expressions m, c, b, and v:

m

c÷ = • =

b

v

v

b

m

c

mv

cb

That is, to divide by , multiply by the reciprocal of .m

c

b

v

b

v


5

Lesson

1.1.1


Topic

8.2.2

You can extend this concept to the division of any rational expression.

The question you’re trying to answer is…

“How many times does go into 10?”

…or “How many halves are in 10?”

1

2

Suppose you pick a number such as 10 and divide by . 1

2


6

10 ÷ = 20 10 × 2 = 20

10 ÷ = 30 10 × 3 = 30

10 ÷ = 40 10 × 4 = 40

10 ÷ = 10n 10 × n = 10n

12

12

12

12

Lesson

1.1.1


Topic

8.2.2

Division Equivalent to

So, 10 divided by a fraction is equivalent to 10 multiplied by the reciprocal of that fraction.


7

Lesson

1.1.1


Topic

8.2.2

Dividing anything by a rational expression is the same as multiplying by the reciprocal of that expression.

So you can always rewrite an expression a ÷ b in the

form a • = (where b is any nonzero expression).1

b

a

b

As always, you should cancel any common factors in your answer to give a simplified fraction.


8

Example 1

Solution follows…

Topic

8.2.2

Solution

Rewrite the division as multiplication by the reciprocal of the divisor.

Simplify ÷ (k + 5).k2 – 25

2k

÷ (k + 5) can be written as:k2 – 25

2k÷

k2 – 25

2k

k + 5

1

=k2 – 25

2k

1

k + 5•

Factor as much as you can: =(k – 5)(k + 5)

2k

1

k + 5•

Solution continues…


9

Example 1

Topic

8.2.2

Solution (continued)

Check your answer. Multiply your answer by (k + 5):

Simplify ÷ (k + 5).k2 – 25

2k

Cancel any common factors between the numerators and denominators.

=(k – 5)(k + 5)

2k

1

k + 5•

1

1

=k – 5

2k

=k – 5

2k

(k – 5)(k + 5)

2k

k2 – 25

2k•

k + 5

1=


10


Example 2

Topic

8.2.2

Solution

Rewrite the division as multiplication by the reciprocal of the divisor.

Solution follows…

Factor all numerators and denominators.

Simplify ÷ .m2 – 4

m2 – 3m + 2

2m

m – 1

m2 – 4

m2 – 3m + 2

m – 1

2m•=

(m + 2)(m – 2)

(m – 2)(m – 1)

m – 1

2m•=

m + 2

2m=

1

1

1

1


11

Divide and simplify each expression.

1. 2.

3. 4.

5.

Lesson

1.1.1

Guided Practice

Topic

8.2.2

Solution follows…

1

b2c2d2

bdc3

bcd2

abc÷

a2 – 9

a2 + a – 6

a + 3

a – 2÷

a(b – 2)

b + 1

2a

(b + 1)(b – 1)÷

a2 + 3a + 2

a2 – a – 6

a2 – 1

a2 – 4a + 3÷

x2 – 5x – 6

x2 + 3x – 10

x2 – 4x – 5

x2 – 25÷

abc

d3

b2 – 3b + 2

2

a – 3

a + 3

x – 6

x – 2


12

Lesson

1.1.1

You Can Divide Long Strings of Expressions At Once

Topic

8.2.2

Just like multiplication, you can divide any number of rational expressions at once, but it makes a big difference which order you do things in.

If there are no parentheses, you always work through the calculation from left to right, so that:

a

b÷ ÷

c

d

e

f

a

b= • ÷

d

c

e

f

a

b= • •

d

c

f

e


13

Rewrite each division as a multiplication by the reciprocal of the divisor.

Example 3

Topic

8.2.2

Solution

Solution follows…

Simplifyx2 + 5x + 6

x2 + 3x

x2 + x – 2

2x2 + 2x÷

x2 + 2x + 1

x – 1÷ .

=x2 + 5x + 6

x2 + 3x

2x2 + 2x

x2 + x – 2•

x2 + 2x + 1

x – 1÷

=x2 + 5x + 6

x2 + 3x

2x2 + 2x

x2 + x – 2•

x – 1

x2 + 2x + 1•



14

Example 3

Topic

8.2.2




Simplifyx2 + 5x + 6

x2 + 3x

x2 + x – 2

2x2 + 2x÷

x2 + 2x + 1

x – 1÷ .

=x2 + 5x + 6

x2 + 3x

2x2 + 2x

x2 + x – 2•

x – 1

x2 + 2x + 1•

=(x + 2)(x + 3)

x(x + 3)• •

2x(x + 1)

(x + 2)(x – 1)

x – 1

(x + 1)(x + 1)=

2

x + 11

1 1

11

1 1

1

1

1


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Lesson

1.1.1

You Can Divide Long Strings of Expressions At Once

Topic

8.2.2

Parentheses override this order of operations, so you need to simplify any expressions in parentheses first:

a

b= ÷

c • f

d • e

a

b= •

d • e

c • f

a

b÷ ÷

c

d

e

f


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6.

7.

8.

Lesson

1.1.1

Guided Practice

Topic

8.2.2

Solution follows…

k2 – 1

2k2 – 14k

k2 + 5k – 6

k2 – 9k + 14÷

–k2 + 3k – 2

2k2 – 10k÷

x2 – 4x – 12

2x2 – 3x – 2

–x2 + 2x + 8

3x3 + 3x2 – 18x÷

6x3 – 36x2

–2x2 + 7x + 4÷

(x – 2)

(x + 3)

(x – 2)(x + 4)

(x + 3)÷ ÷

(x + 4)

(x + 1)

x + 3

2x

1

x + 1

k2 – 4k – 5

k2 + 5k – 6–


17

Lesson

1.1.1

You Can Multiply and Divide at the Same Time

Topic

8.2.2

Say you have an expression like this to simplify:

Again, you work from left to right, and anywhere you get a division, multiply by the reciprocal, so:

a

b÷ ×

c

d

e

f

a

b÷ ×

c

d

e

f

a

b× ×

d

c

e

f=


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Rewrite any divisions as multiplications by reciprocals.

Example 4

Topic

8.2.2

Solution

Solution follows…

Simplifyp2 + pq – 2q2

p2 – 2pq – 3q2

p2 + q2

pq + 2q2×

p2 – 2pq + q2

p2 – 3pq÷ .

=p2 + pq – 2q2

p2 – 2pq – 3q2

pq + 2q2

p2 + q2×

p2 – 2pq + q2

p2 – 3pq×


=p

q=

(p + 2q)(p – q)

(p – 3q)(p + q)× ×

(p – q)(p + q)

q(p + 2q)

p(p – 3q)

(p – q)(p – q)

Cancel any common factors.

1

1

1

1

1

11

11

1


19

Example 5

Topic

8.2.2

Solution

Solution follows…

Justify your work.

Show that2a2 – 7a + 3

a2 + 4a – 21÷

2a – 2

a + 1=

2a2t + at – t

2a2t + 12at – 14t

The question asks you to justify your work, so make sure you can justify all your steps.

Start with left-hand side

Definition of division

2a2 – 7a + 3

a2 + 4a – 21÷

2a2t + at – t

2a2t + 12at – 14t

=2a2 – 7a + 3

a2 + 4a – 21×

2a2t + 12at – 14t

2a2t + at – t



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=2a2 – 7a + 3

a2 + 4a – 21×

2a2t + 12at – 14t

2a2t + at – t

Example 5

Topic

8.2.2


Justify your work.


a2 + 4a – 21÷

2a – 2

a + 1=

2a2t + at – t

2a2t + 12at – 14t

=(2a – 1)(a – 3)

(a + 7)(a – 3)×

2t(a2 + 6a – 7)

t(2a2 + a – 1)

=(2a – 1)(a – 3)

(a + 7)(a – 3)×

2t(a + 7)(a – 1)

t(2a – 1)(a + 1)

Distributive property

Distributive property

Equation carried forward



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=(2a – 1)(a – 3)

(a + 7)(a – 3)×

2t(a + 7)(a – 1)

t(2a – 1)(a + 1)

=(2a – 1)

(a + 1)×

t(2a – 1)(a + 7)(a – 3)

t(2a – 1)(a + 1)(a – 3)

Example 5

Topic

8.2.2


Justify your work.


a2 + 4a – 21÷

2a – 2

a + 1=

2a2t + at – t

2a2t + 12at – 14t

=(2a – 1)

(a + 1)=

2a – 2

a + 1Inverse and identity properties, and distributive property

Commutative and associative properties of multiplication

Equation carried forward


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Simplify these rational expressions.

9.

10.

11.

12.

Lesson

1.1.1

Guided Practice

Topic

8.2.2

Solution follows…

t2 – 1

t2 + 2t – 3

t + 1

t2 + 4t + 3÷

t – 1

1×

a2 + a – 12

a2 + a – 2

a2 + 5a + 4

a2 + 2a + 1÷

a2 + 2a – 3

a2 – 2a – 3×

x2 + 5x – 14

x2 – 4x – 21

x2 + 6x – 7

x2 – 6x – 7÷

x2 + 2x – 3

x2 – 5x + 6×

a2 – 1

a2 – 4

a2 – 2a – 3

a2 – 3a – 10÷

a2 – 5a + 6

–a2 + 2a + 15×

a + 3

a + 2

x + 1

x – 3

t2 – 1

a – 1

a + 3–


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Independent Practice

Solution follows…

Topic

8.2.2


1. 2.

3. 4.

5. 6.

b2 + b – 2

k2 – m2

2k2 + km – m2

2k2 + 2m

2k2 + 3km – 2m2÷

t2 + 2t – 3

t2 + 4t + 3

3t – 3

t2 – t – 2÷

–x3 – 3x2 – 2x

x2 – 2x – 3

x2 – x – 6

x3 – 2x2 – 3x÷

b3 – 4b

b3 + b

b2 – b – 2

b4 – 1÷

x2 – 6x + 8

x2 – 4

–x3 + x

–2x2 + 4x + 16÷

y2 – y – 2

y2 + 3y – 4

–y + 2

y2 – 3y + 2÷

k2 + km – 2m2

2k + 2m

2x2 – 16x + 32

x3 – x

t – 2

3

x3 + x2

x – 3–

y2 – y – 2

y + 4–


24


Solution follows…

Topic

8.2.2


7. 8.

9. 10.

a3 – 4a

–a2 + 2a

a2 + a – 2

a2 – a – 2÷

b2 – 1

b2 – 2b – 3

b2 – 2b + 1

b2 – 4b + 3÷

(m – v)2

m2 – v2

m2 – 3mv + 2v2

(m – 2v)2÷

x2 – 3x + 2

x2 + x – 2

–x + 2

x2 – 3x – 10÷

a2 – a – 2

a – 1–

1

m – 2v

m + v

–x + 5


25


Solution follows…

Topic

8.2.2


11.

12.

13.

2x2 – 5x – 12

4x2 + 8x + 3

x2 – 16

2x2 + 7x + 3÷

x2 – 9

x2 + 2x – 8÷

y + 5

y2 – 4y – 5

y2 + 4y – 5

y + 1÷

1

y2 – 6y + 5÷

t2 – t – 6

t2 + 6t + 9÷ (t2 – 4) ÷

t + 2

t + 3

1

x – 2

x – 3

1

(t + 2)2


26


Solution follows…

Topic

8.2.2

Simplify these rational expressions.

14.

15.

16.

17.

k2 – 5k + 6

k2 + 2k – 8

–2k2 – 6k – 4

k2 – 2k – 3•

k2 + 3k + 2

k2 + 5k + 4÷

–2v2 + 4vw

3v2 – 4vw + w2×

v2 – w2

–2vw + 3w2÷

v3 – vw2

6v2 + 4vw2 – 2w3

–2v2 + 5vw – 3w2

–4v2 + 4vw + 8w2×

m2 + 2mn + n2

m2n – 3mn2÷ ÷ ÷

4m2 + 5mn + n2

2m2 – 5mn – 3n2

2m2 + 3mn + n2

–m2 + 3mn – 2n2

–m2 + 3mn – 2n2

–2m2n + 5mn2

m2 + 2mn + n2

m2n – 3mn2÷ ÷ ÷

4m2 + 5mn + n2

2m2 – 5mn – 3n2

2m2 + 3mn + n2

–m2 + 3mn – 2n2

–m2 + 3mn – 2n2

–2m2n + 5mn2

2m – 5n

4m + n–

–2

1

(m + n)2(2m + n)2(–2m + 5n)

(4m + n)(m – 2n)2(–m + n)2


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Topic

8.2.2

Round UpRound Up

It’s really important that you can justify your work step by step, because division of rational expressions can involve lots of calculations that look quite similar.


topic 8.2.2

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