S k i l l
28 MULTIPLYING AND DIVIDING
|
a) | · = | b) 2 · 3 = 6 | ||
c) | · = | = 6 | d) (2)2 = 4 · 5 = 20 |
e) | = |
Problem 2. Multiply, then simplify:
Example 1. Multiply ( + )( − ).
Solution. The student should recognize the form those factors will produce:
( + )( − ) | = | ()2 − ()2 |
= | 6 − 2 | |
= | 4. |
Problem 3. Multiply.
a) ( + )( − ) = 5 − 3 = 2
b) (2 + )(2 − ) = 4 · 3 − 6 = 12 − 6 = 6
c) (1 + )(1 − ) = 1 − (x + 1) = 1 − x − 1 = −x
d) ( + )( − ) = a − b
Problem 4. (x − 1 − )(x − 1 + )
a) What form does that produce?
The difference of two squares. x − 1 is "a." is "b.">
b) Multiply out.
(x − 1 − )(x − 1 + ) | = | (x − 1)2 − 2 | ||
= | x2 − 2x + 1 − 2, | on squaring the binomial, | ||
= | x2 − 2x − 1 |
Problem 5. Multiply out.
(x + 3 + )(x + 3 − ) | = | (x + 3)2 − 3 |
= | x2 + 6x + 9 − 3 | |
= | x2 + 6x + 6 |
Dividing radicals
For example,
= | = |
Problem 6. Simplify the following.
a) | = | b) | 8 |
= | 3 4 |
c) | = | a | = | a ·a | = | a2 |
Conjugate pairs
The conjugate of a + is a − . They are a conjugate pair.
Example 2. Multiply 6 − with its conjugate.
Solution. The product of a conjugate pair --
(6 − )(6 + )
-- is the difference of two squares. Therefore,
(6 − )(6 + ) = 36 − 2 = 34.
When we multiply a conjugate pair, the radical vanishes and we obtain a rational number.
Problem 7. Multiply each number with its conjugate.
a) x + = x2 − y
b) 2 − (2 − )(2 + ) = 4 − 3 = 1
c) + | You should be able to write the product immediately: 6 − 2 = 4. |
d) 4 − 16 − 5 = 11
Example 3. Rationalize the denominator:
1 |
Solution. Multiply both the denominator and the numerator by the conjugate of the denominator; that is, multiply them by 3 − .
1 |
= | 9 − 2 |
= | 7 |
The numerator becomes 3 − . The denominator becomes the difference of the two squares.
Example 4. | = | 3 − 4 |
= | −1 |
|
= | −(3 − 2) | ||||
= | 2 − 3. |
Problem 8. Write out the steps that show the following.
a) | 1 |
= ½() |
= | 5 − 3 |
= | 2 |
||
= | ½( −) | ||||
The definition of division |
b) | 2 3 + |
= ½(3 − ) |
2 3 + |
= | 9 − 5 |
= | 4 |
= | ½(3 − ) |
c) | _7_ 3 + |
= | 6 |
_7_ 3 + |
= | 9 · 5 − 3 |
= | 42 |
= | 6 |
d) | − 1 |
= | 3 + 2 |
= | 2 − 1 |
= | 2 + 2 + 1, | Perfect square trinomial | |||
= | 3 + 2 |
e) | 1 + |
= | x |
1 + |
= | 1 − (x + 1) |
||||
= | 1 − x − 1 |
, | Perfect square trinomial | |||
= | −x |
|||||
= | x |
on changing all the signs. |
Example 5. Simplify |
Solution. | = | on adding those fractions, | |||
= | on taking the reciprocal, | ||||
= | 6 − 5 |
on multiplying by the conjugate, | |||
= | 6 − 5 | on multiplying out. |
Problem 9. Simplify |
= | on adding those fractions, | ||||
= | on taking the reciprocal, | ||||
= | 3 − 2 |
on multiplying by the conjugate, | |||
= | 3 + 2 | on multiplying out. |
Problem 10. Here is a problem that comes up in calculus. Write out the steps that show:
= − | ____1____ |
In this case, you will have to rationalize the numerator.
= | 1 h |
· | ||
= | 1 h |
· | _____x − (x + h)_____ | |
= | 1 h |
· | ____x − x − h_____ | |
= | 1 h |
· | _______−h_______ | |
= | − | _______ 1_______ |
Next Lesson: Rational exponents
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