28 MULTIPLYING AND DIVIDING
|
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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