Question 65419
<pre><font size = 5><b>Rational and Radical Expressions 

1.  Simplify the rational expression:

x² – 6x – 7
———————————                   
  x² – 1

Factor top and bottom:

 (x - 7)(x + 1)
————————————————                   
 (x - 1)(x + 1)

Cancel the (x + 1)'s

           1
 (x - 7)<s>(x + 1)</s>
————————————————                   
 (x - 1)<s>(x + 1)</s>
           1

and all that's left is

 x - 7
———————                   
 x - 1 
            


2.  Divide:

 x² – 3x + 2      x² – 4
————————————— ÷ —————————— 
   8x – 8        5x + 10

Invert the second fraction and change ÷ to ·

 x² – 3x + 2      5x + 10
————————————— · —————————— 
   8x – 8          x² - 4

Factor all numerators and denominators:

 (x - 2)(x - 1)        5(x + 2)
———————————————— · ———————————————— 
   8(x – 1)         (x - 2)(x + 2)

Indicate the multiplication of all factors
in the numerator and denominator all as
one fraction:

  5(x - 2)(x - 1)(x + 2)
—————————————————————————— 
  8(x – 1)(x - 2)(x + 2)

Cancel the (x - 2)'s

      1        
  5<s>(x - 2)</s>(x - 1)(x + 2)
—————————————————————————— 
  8(x – 1)<s>(x - 2)</s>(x + 2)
             1

      1      1  
  5<s>(x - 2)(x - 1)</s>(x + 2)
—————————————————————————— 
  8<s>(x – 1)(x - 2)</s>(x + 2)
      1      1

Cancel the (x + 2)'s

      1      1      1 
  5<s>(x - 2)(x - 1)(x + 2)</s>
—————————————————————————— 
  8<s>(x – 1)(x - 2)(x + 2)</s>
      1      1      1

all that's left is

       5
      ———
       8


3.  Simplify:

 __________
&#8730;100x<sup>2</sup>y<sup>16</sup>z<sup>8</sup> 

Write the 10 as 10<sup>2</sup>

To take the square root of an even power,
divide the exponent by 2:
 __________
&#8730;10<sup>2</sup>x<sup>2</sup>2y<sup>16</sup>z<sup>8</sup>

10<sup>2÷2</sup>x<sup>2÷2</sup>y<sup>16÷2</sup>z<sup>8÷2</sup>

10<sup>1</sup>x<sup>1</sup>y<sup>8</sup>z<sup>4</sup>

Erase the two 1 exponents

10xy<sup>8</sup>z<sup>4</sup>

4.  Perform the indicated operations:
 __    __    __
<font face = "symbol">Ö</font>72 + <font face = "symbol">Ö</font>32 – <font face = "symbol">Ö</font>18

72 = 8·9 = 4·2·3·3 = 2·2·2·3·3
32 = 4·8 = 2·2·4·2 = 2·2·2·2·2
18 = 2·9 = 2·3·3
 _________    _________    _____
<font face = "symbol">Ö</font>2·2·2·3·3 + <font face = "symbol">Ö</font>2·2·2·2·2 - <font face = "symbol">Ö</font>2·3·3

Group like factor into pairs:
 _____________    _____________    _______
<font face = "symbol">Ö</font>(2·2)·2·(3·3) + <font face = "symbol">Ö</font>(2·2)·(2·2)·2 - <font face = "symbol">Ö</font>2·(3·3)   

Each pair of like factors comes out in front
of the square root radical as a single factor.
That is, in the first radical the (2·2) and
the (3·3) come out in front of the radical as
as single factors multiplied 2·3, and the
unpaired 2 stays under the radical.
            _       _     _
        2·3<font face = "symbol">Ö</font>2 + 2·2<font face = "symbol">Ö</font>2 - 3<font face = "symbol">Ö</font>2
            _     _     _
          6<font face = "symbol">Ö</font>2 + 4<font face = "symbol">Ö</font>2 - 3<font face = "symbol">Ö</font>2
            _
Factor out <font face = "symbol">Ö</font>2
            _
           <font face = "symbol">Ö</font>2(6 + 4 - 3)
            _ 
           <font face = "symbol">Ö</font>2(7)
              _
            7<font face = "symbol">Ö</font>2


5.  Multiply:
  _    _    _    _
(<font face = "symbol">Ö</font>3 + <font face = "symbol">Ö</font>5)(2<font face = "symbol">Ö</font>3 – <font face = "symbol">Ö</font>5)

Use FOIL
  _    _      _   _      _    _      _   _
(<font face = "symbol">Ö</font>3)(2<font face = "symbol">Ö</font>3) – (<font face = "symbol">Ö</font>3)(<font face = "symbol">Ö</font>5) + (<font face = "symbol">Ö</font>5)(2<font face = "symbol">Ö</font>3) – (<font face = "symbol">Ö</font>5)(<font face = "symbol">Ö</font>5)

Multply under the radicals:
  ___    ___    ___    ___
2<font face = "symbol">Ö</font>3·3 - <font face = "symbol">Ö</font>3·5 + 2<font face = "symbol">Ö</font>5·3 - <font face = "symbol">Ö</font>5·5

Pair like factors under the 1st and 4th radicals
Multiply factors under the middle two radicals
which have no pairs of like factors:
  _____    __    __    _____   
2<font face = "symbol">Ö</font>(3·3) - <font face = "symbol">Ö</font>15 + 2<font face = "symbol">Ö</font>15 - <font face = "symbol">Ö</font>(5·5) 

The pair of 3's comes out eliminating the radical
in the first term. The middle two terms combine,
and the pair of 5's comes out eliminating the
radical in the 4th term:
       __
2·3 + <font face = "symbol">Ö</font>15 - 5
      __
 6 + <font face = "symbol">Ö</font>15 - 5
        __
   1 + <font face = "symbol">Ö</font>15


6.  Rationalize the denominator:
         3
    ———<u>—</u>————<u>—</u>——  
      <font face = "symbol">Ö</font>6 – <font face = "symbol">Ö</font>3

Form the two term conjugate of the denominator.
1. Its first term is_the same as the first term of  
   the denominator <font face = "symbol">Ö</font>6
2. Its second term is the second term of t<u>h</u>e   _
   denominator with its <u>s</u>ign <u>c</u>hange<u>d</u>, or_<font face = "symbol">Ö</font>6 + <font face = "symbol">Ö</font>3
3. So the conjugate of <font face = "symbol">Ö</font>6 - <font face = "symbol">Ö</font>3 is <font face = "symbol">Ö</font>6 + <font face = "symbol">Ö</font>3
4. Put the conjugate over itself
       _    _
      <font face = "symbol">Ö</font>6 + <font face = "symbol">Ö</font>3
    ———<u>—</u>————<u>—</u>——  
      <font face = "symbol">Ö</font>6 + <font face = "symbol">Ö</font>3

which just equals 1, so we can then multiply the
fraction to be rationalized by this fraction without
changing its value, since multiplication by 1 does
no change the value of an expression

                     _    _
         3          <font face = "symbol">Ö</font>6 + <font face = "symbol">Ö</font>3
    ———<u>—</u>————<u>—</u>—— · ———<u>—</u>————<u>—</u>——
      <font face = "symbol">Ö</font>6 – <font face = "symbol">Ö</font>3       <font face = "symbol">Ö</font>6 + <font face = "symbol">Ö</font>3

Indicate the multiplication of numerators and 
denominators, all as one fraction:
            _    _
         3(<font face = "symbol">Ö</font>6 + <font face = "symbol">Ö</font>3)
     ———<u>—</u>————<u>—</u>———<u>—</u>————<u>—</u>——
      (<font face = "symbol">Ö</font>6 – <font face = "symbol">Ö</font>3)(<font face = "symbol">Ö</font>6 + <font face = "symbol">Ö</font>3)

FOIL out the bottom:

               _    _
            3(<font face = "symbol">Ö</font>6 + <font face = "symbol">Ö</font>3)
     ——<u>—</u>—<u>—</u>————<u>—</u>—<u>—</u>————<u>—</u>—<u>—</u>————<u>—</u>—<u>—</u>—
      <font face = "symbol">Ö</font>6<font face = "symbol">Ö</font>6 + <font face = "symbol">Ö</font>3<font face = "symbol">Ö</font>6 - <font face = "symbol">Ö</font>6<font face = "symbol">Ö</font>3 - <font face = "symbol">Ö</font>3<font face = "symbol">Ö</font>3

               _    _
            3(<font face = "symbol">Ö</font>6 + <font face = "symbol">Ö</font>3)
     ————————<u>——</u>————<u>——</u>———————
      <font face = "symbol">Ö</font>36 + <font face = "symbol">Ö</font>18 - <font face = "symbol">Ö</font>18 - <font face = "symbol">Ö</font>9

               _    _
            3(<font face = "symbol">Ö</font>6 + <font face = "symbol">Ö</font>3)
     ————————<u>——</u>————<u>——</u>———————
        6 + <font face = "symbol">Ö</font>18 - <font face = "symbol">Ö</font>18 - 3

The middle terms cancel

               _    _
            3(<font face = "symbol">Ö</font>6 + <font face = "symbol">Ö</font>3)
           ————————————
               6 - 3

             
               _    _
            3(<font face = "symbol">Ö</font>6 + <font face = "symbol">Ö</font>3)
           ————————————
                 3

Cancel the 3's
                  
            1  _    _
            <s>3</s>(<font face = "symbol">Ö</font>6 + <font face = "symbol">Ö</font>3)
           ————————————
                 <s>3</s>
                 1 
               _    _
              <font face = "symbol">Ö</font>6 + <font face = "symbol">Ö</font>3

Edwin</pre>