Chapter Three — Basic Algebraic Expressions
The Humongous Book of Algebra Problems
45
Multiply the resulting decimal (2.39) by 10
n
, where n is the number of digits you
moved the decimal to the left.
23,900,000 = 2.39 × 10
7
3.20 Express the number 958,000,000,000,000,000,000 using scientific notation.
Assume the decimal point is located at the end of the number, to the right
of the 18th zero. Move it 20 digits to the left so that only the nonzero digit 9
remains left of the decimal point. Multiply the resulting decimal (9.58) by 10
n
,
where n is the number of digits the decimal point moved left.
958,000,000,000,000,000,000 = 9.58 × 10
20
3.21 Express the number 0.000049 using scientific notation.
To write small numbers using scientific notation, count the number of times
you must move the decimal point to the right until exactly one nonzero digit
appears left of the decimal point. In this case, moving the decimal five digits to
the right produces the decimal 4.9. Multiply that decimal by 10
–n
, where n is the
number of digits the decimal moved right.
0.000049 = 4.9 × 10
–5
3.22 Express the number –0.00000000412 in scientific notation.
Apply the technique described in Problem 3.21—scientific notation does not
require different procedures for positive and negative numbers. Move the
decimal point nine places to the right so that a single nonzero digit appears left
of the decimal point.
–0.00000000412 = –4.12 × 10
–9
Distributive Property
Multiply one thing by a bunch of things in parentheses
3.23 Simplify the expression: –2(4x + 3y).
To multiply the entire parenthetical quantity (4x + 3y) by –2, apply the
distributive property: a(b + c) = ab + ac.
–2(4x + 3y) = (–2)(4x) + (–2)(3y)
Multiply the coefficient of each term by –2.
The “×”
multiplication symbol
is used in scientic
notation, so that the
multiplication dot
and the decimal
point don’t get
confused.
When you
move the
decimal point
RIGHT, the 10 in
scientic notation
has a negative
exponent. When you
move the decimal point
LEFT, 10 has a positive
exponent. Remember
that, because in math,
right usually means
positive and left
usually means
negative.
In other
words, every term
in the parentheses
gets multiplied by the
number outside the
parentheses, one at
a time.
Multiply
the numbers
together and
then stick the
variable on at
the end.
Chapter Three — Basic Algebraic Expressions
The Humongous Book of Algebra Problems
46
3.24 Simplify the expression: 8x(7x
2
– 3).
Multiply each term of the quantity 7x
2
– 3 by 8x.
8x(7x
2
– 3) = (8x)(7x
2
) + (8x)(–3)
To calculate the product of 8x and 7x
2
, multiply the coefficients and then
apply the technique described in Problem 3.8 to multiply the x-terms. Both
exponential terms have the same base, so the product is the common base
raised to the sum of the individual powers.
3.25 Simplify the expression: –y
3
(–12x
2
+ y
6
).
Apply the distributive property, multiplying each term of the expression
–12x
2
+ y
6
by –y
3
. Writing the coefficients of –y
3
and y
6
explicitly might facilitate
the calculation of the products.
Because y
3
and y
6
have the same base, calculate their product by adding the
powers. Note that x
2
and y
3
have different bases, so those powers should not be
summed.
3.26 Simplify the expression: x(9x
3
– 6x
2
+ 2x + 1).
Distribute x to each term in the parenthetical quantity. Express the implicit
exponents explicitly.
If a term
doesn’t have a
coefcient—the
variable doesn’t
have a number in
front of it—then you
should assume the
coefcient is 1. (Just
like the power of x is
1 if there’s no power
written: x = x
1
.) That
means you can
write –y
3
as –1y
3
and y
6
as 1y
6
.
You don’t HAVE
to do this. It’s sort
of annoying to write x
as x
1
, so when you can
get the problems right
without those 1s, I’d
stop writing them.
Chapter Three — Basic Algebraic Expressions
The Humongous Book of Algebra Problems
47
3.27 Simplify the expression: 5xy(x
2
– 4xy + 25y
2
).
Distribute 5xy to each term in the parenthetical quantity.
5xy(x
2
– 4xy + 25y
2
) = (5x
1
y
1
)(x
2
) + (5x
1
y
1
)(–4x
1
y
1
) + (5x
1
y
1
)(25y
2
)
Multiply the coefficients of each term, followed by the variables.
3.28 Simplify the expression: 6x(2x – 3y) – y(11x – 5y).
Distribute 6x to the terms of the expression (2x – 3y) and distribute –y to the
terms of the expression (11x – 5y).
Because the terms –18xy and –11xy contain the same variable expression (xy),
they are like terms. Combine like terms by combining their coefficients and
multiplying the result by the common variable expression: –18xy – 11xy = (–18
– 11)xy = –29xy.
12x
2
– 18xy – 11xy + 5y
2
= 12x
2
– 29xy + 5y
2
3.29 Apply the distributive property to the expression and eliminate negative
exponents: 10x
3
y
2
z
–1
(4x
2
y + 8xy
–5
z
2
– 3y
12
z
–4
. Do not combine the resulting
fractions.
Apply the distributive property.
Move factors with negative exponents across the fraction bar into the
denominator, a process explained in Problem 3.18.
Instead
of distributing
–y, you could
distribute a positive y
to get –(11xy – 5y
2
) and
THEN distribute the
negative sign. Either
way, you’ll end up with
–11xy + 5y
2
.
This explanation
of like terms is not
very satisfying, but if
you’re confused, don’t
worry. The whole concept
of terms (including like
terms) is explained
much more in depth in
Problems 11.9–11.16.
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