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28 MULTIPLYING AND DIVIDING
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| a) | b) 2 |
|||
| c) | = 6 | d) (2 |
||
| e) |
Problem 2. Multiply, then simplify:
![]()
![]()
Example 1. Multiply (
+
)(
−
).
Solution. The student should recognize the form those factors will produce:
| ( |
= | ( |
| = | 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)2 − 2 | ||
| = | x2 − 2x + 1 − 2, | on squaring the binomial, | ||
| = | x2 − 2x − 1 | |||
Problem 5. Multiply out.
| (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 |
||||
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) | = | 3 + 2 |
| = | 2 − 1 |
= | 2 + 2 |
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 |
on multiplying out. | |||
| Problem 9. Simplify | ![]() |
![]() |
= | ![]() |
on adding those fractions, | ||
| = | on taking the reciprocal, | ||||
| = | 3 − 2 |
on multiplying by the conjugate, | |||
| = | 3 |
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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