A. The Products
of Monomials and/or Polynomials
B. Quadratic Equations
C. Factoring
D. Division of Algebraic Expressions
A. The Products
of Monomials and/or Polynomials
To multiply a binomial (a polynomial
with two terms like, x + 3) by a binomial use FOIL:
First: Mulitply the first terms of each binomial.
Outside: Multiply the outside terms (the outer extreme
terms of the expression).
Inside: Multiply the inside terms (the inner extreme
terms of the expression).
Last: Multiply the last terms of each binomial.
Then add them making sure to
combine like terms.
Example 1
(x + 3)(x - 5) =
Solution
F: x × x = x
O: -5 × x = -5x
I : 3 × x = 3x
L: +3 × -5 = -15
Therefore, the answer is the sum of F +
O + I + L:
x -
5x + 3x - 15
then combine (-5x and +3x)
x - 2x - 15
Example 2
(2a + b)(a - 2b) = 2a -4ab + ab - 2b
=2a - 3ab - 2b
B. Quadratic
Equations
A quadratic equation may be written
in the standard form:
ax + bx + c = 0 where
a, b, and c are constants.
If either b or c is zero, the
equation is relatively easy to solve. If neither b nor c is zero,
we will consider only those equations in which the quadratic
expression on the left side can be factored.
Try this sample using the techniques from the section above:
x -
2x - 15
Here we do "reverse foil."
FIRST:
The first number x is easy. It must
be the product of two numbers, which in this case are most likely
x and x.
(x )(x )
LAST:
The last number -15 is trickier:
The second number in each binomial must multiply to be 15, so
we are looking for two factors:(15, 1) and (3,5). One of the
two numbers must be negative.
The middle number is -2x. The middle number is the combination
of the OUTSIDE, INSIDE steps. Note that among the factors,
3 and 5 can sum to -2x, here is how:
(x + 3)(x - 5)
OUTSIDE = -5x
INSIDE = 3x
Total= -2x
Thus,
(x + 3)(x - 5) must be the binomial solution because FOIL will
produce
x - 2x - 15.
Strategy: Quadratic equations usually have two
answers. For example X = 25 has two answers, -5 and +5. The test writers
often generate trick questions by exploiting the fact that many
students forget that these equations have two answers. Data sufficiency
questions will often turn on this issue.
Example 3
Solve x
+ 4x = 0
Solution
The left side of this equation can be factored so that we can
write it as
x(x + 4) = 0.
This equation is satisfied if either factor is zero. Consequently
we write,
x = 0
x + 4 = 0.
The solution is x = 0, x = -4.
Example 4
Solve x
- 5x + 4 = 0.
Solution
This quadratic equation has all terms present, so we expect that
the left side can be factored.
(x - 4)(x - 1) = 0
We set each factor equal to zero and solve for x:
x - 4 = 0
x - 1 = 0
The solution is x = 4, 1.
Example 5
Solve 2x - 3 = (4 / x) + x.
Solution
First, put the equation in standard form by multiplying both
sides by x.
This equation is factorable so we write
2x
- 3x = 4 + x
x - 3x - 4 = 0
(x - 4)(x + 1) = 0
Set each factor equal to zero and solve for x:
x - 4 = 0; x = 4
x + 1 = 0; x = -1
The solution is x = 4, -1.
To multiply a monomial by a monomial,
multiply the numerical coefficients and then follow the laws
of exponents with the same base. For example,
(2rs
)(-3rs)= 2(-3)(r x r)(s x s) =
-6r
s
To multiply a polynomial by a monomial, multiply each term of
the polynomial by the monomial. This is illustrated by
4x(3x- 2xy
+ y) = 12x -
8xy + 4xy
To multiply two polynomials, multiply one of them by each term
of the other and then combine like terms. The following illustrates
the process:
(2x - y)(2 + x - 3y) = 2x(2 +
x - 3y) - y(2 + x - 3y)
=
4x + 2x
- 6xy - 2y -xy + 3y
=
4x - 2y + 2x
- 7xy + 3y
C. Factoring
Before
we proceed with the rules of division, we need to introduce the
notion of factoring. If an algebraic expression can be written
as the product of two other algebraic expressions, the two other
expressions are called factors. Sometimes there may be more than
two factors. Factoring an expression will often allow us to simplify
a problem, so it is important when solving many algebra problems.
There are several algebraic expressions
that occur frequently. You should be able to recognize their
factors immediately.
Five common algebraic
factoring types:
1. ax + ay = a(x+y)
2. x
- y = (x + y)(x -
y)
3. x + 2xy + y
= (x + y)
4. x - 2xy + y
= (x-y)
5. x + (a + b)x + ab = (x + a)(x +
b)
Example 6
Factor 3ab -
15ab
Solution
Note that 3ab is common to both terms. Factor
it out and obtain
3ab - 15ab = 3ab(b
- 5a).
The two factors are 3ab
and (b - 5a).
Example 7
Factor 4x - 9.
Solution
This is the difference of two perfect squares: (2x)
and 3. Using factoring expression No.2
(see above), we see that
4x -
9 = (2x - 3)(2x + 3)
NOTE: No. 2 is the most common
expression that you will need to factor, so be sure to recognize
it quickly.
Example 8
Factor y + 6y + 9.
Solution
The number 9 is 3,
so we try type No.3 and observe that:
y +
6y + 9 = (y + 3).
Note that this can be factored
using type No.5. What two numbers add together to give 6 and
multiply together to give 9? The answer is 3 and 3 so that
y + 6y + 9 = (y + 3)(y + 3).
Example 9
Factor r
- 8r + 16.
Solution
The number 16 is 4 so we try type
No.4 and observe that
r -
8r + 16 = (r - 4).
This also can be factored using
type No.5. What two numbers add together to give -8 and multiply
together to give 16? The answer is - 4 and - 4 so that
r -
8r + 16 = (r - 4)(r - 4).
D. Division of
Algebraic Expressions
There
are two common divisions that you may be asked to perform. The
first involves dividing an algebraic expression by a monomial,
such as
(3x + 6ax + 15xa)/3x
We first
recognize the numerator as a type No.1 expression and factor
out 3x. The factor 3x then cancels out with the denominator and
we have
(3x + 6ax + 15xa)/3x
= [3x(1 + 2ax + 5a)]/3x = 1 + 2ax
+ 5a
The second type of division involves
divisions such as
(5x - 15x + 10) / (5x -10)
We
write the denominator as 5(x -2) and the numerator as 5(x - 3x + 2). The numerator is a type No.5
expression: What two numbers added together give - 3 and multiplied
together give 2? They are -2 and -1. Hence, we can write
[5(x -
3x + 2)] / [5(x - 2)] = [(x - 2) (x -1)]/[x - 2] = x - 1.
III. Manipulating
Complex Expressions
A. Multiplying
and Dividing Algebraic Fractions
B. Addition of Algebraic Fractions
C. Equations
D. Equivalent Equations
E. Linear Equations
F. Inequalities
G. Simultaneous
Equations
A. Multiplying and Dividing Algebraic Fractions
When
multiplying algebraic fractions, we search for common factors:
factors that are common to both numerators and denominators.
These factors are canceled, and the resulting fraction simplified,
if possible. When dividing algebraic fractions, we invert the
divisor (the fraction we're dividing by) and then multiply, searching
for common factors.
Example 1
Perform the division.
3a - 4 divided by 3ab - 4b
2 - 3a 4
- 9a
Solution
First, invert the divisor
and express the division as the multiplication.
3a - 4 x 4 - 9a
2 - 3a 3ab - 4b
Next, factor any factorable expressions.
4 - 9a =
(2 - 3a)(2 + 3a)
3ab - 4b = b(3a - 4)
The given division is now written
and the common factors (in red) canceled.
3a -
4 x (2 - 3a)(2 + 3a) = 2 + 3a
2
- 3a b(3a - 4) b
B. Addition of
Algebraic Fractions
When adding (or subtracting)
algebraic fractions, we proceed as in arithmetic:
- Find the lowest common denominator
(LCD) of the fractions.
- Write each fraction using the
LCD of the fractions.
- Add (or subtract) the numerators.
The denominator will be the LCD. Simplify the resulting fraction.
Example 2
Express as a single fraction:
2x + 1 - x - 3
2a 3a
Solution
The lowest common denominator is 6a. Each fraction is written
with its denominator as 6a. This gives
2x + 1 - x - 3 =
3(2x+1) - 2(x-3)
2a 3a
6a
6a
=
6x + 3 - 2x - 6
6a
6a
=
6x + 3 - 2x + 6
6a
=
4x + 9
6a
C. Equations
An
equation is a mathematical statement that two algebraic expressions
are equal. The letters in the equation are the unknowns. An equation
may be thought of as asking the question, "What numerical
value of the unknown satisfies the equation?" (Satisfies
means to make both sides equal.) The value of the unknown is
called the solution of the equation, or root of the equation.
We solve an equation by finding the numerical value of all of
its roots.
There
are equations that do not have real roots. For example , the
equation x = -4 does not have a real root;
the square of any real number, whether positive or negative (e.g.,
+ 2 or -2) is positive. Hence, there is no real root to x = -4.
D. Equivalent Equations
Two equations
that have the same roots are said to be equivalent. To solve
an equation, we often put the equation in a form that is more
easily solved with one or both of the following operations:
- Add or subtract the same term
to or from each side of an equation.
- Multiply or divide each side
of an equation by the same number or algebraic expression.
E. Linear
Equations (linear equations do not contain squares
or square roots)
To
solve a linear equation that contains fractions, first remove
all fractions by multiplying both sides by the lowest common
denominator of all the fractions.
Example 3
x/3 + x/2 = 5
Solution
Multiply both sides by the LCD which is 6: (3 x 2)
2x + 3x = 30
5x = 30
x = 6
Then
move all terms containing the unknown to one side and all terms
that do not contain an unknown to the other side. Factor out
the unknown from all terms that contain the unknown; finally,
divide each side by the coefficient of the unknown.
Example 4
Solve a(a + t) = b
- bt for t
Solution
To solve for t, first
remove the parentheses:
a +
at = b - bt
Move "- bt" to the
left side and "a" to the right
side:
bt + at = b - a
We can factor t out of the two terms on the left side; we can
also factor the right side to obtain t(b + a) = (b + a)(b - a).
Divide both sides by (b + a) and find the solution to be
t = b - a.
F.
Inequalities
An
inequality is simply a comparison of two quantities or expressions.
The following symbols with their meanings are used:
> is greater than
< is less than
>= is greater than or equal
to
<= is less than or equal to
If we place all numbers on a
number line with negative numbers to the left of zero and positive
numbers to the right of zero, then if A is to the right of B
we state that A > B; if A is to the left of B then A <
B. Consequently, we conclude that -5 > -7 and -3 < 2.
-7 -5 -3 0 2
We also form inequalities using
algebraic symbols. The inequality 3x + 2 > x - 6 is solved
just like an algebraic equation is solved. We subtract 2 from
each side of the inequality so that 3x > x - 8. Then subtract
x from each side so that 2x > - 8. Divide by 2 and obtain
x > - 4. This is the solution. Any number greater than - 4
satisfies the inequality.
There are several rules that
we must follow when manipulating inequalities:
- The same number or algebraic
expression can be added or subtracted from each side of an inequality.
- The same positive number (or
positive algebraic expression) can multiply or divide each side
of an inequality.
- Both sides of the same type
of inequality can be added and the inequality remains. (If x
< y and w < z, then x + w < y + z.)
- If a negative number (or negative
algebraic expression) multiplies or divides each side of an inequality,
the inequality sign must be reversed. (Be sure to remember this;
it often leads to errors!)
Example 5
Solve the inequality
2x - 2 > x - 5.
Solution
The inequality is treated in much the same manner as an algebraic
equation. Add 2 to each side:
2x > x - 3.
Subtract x from each side and
the solution is x > -3
NOTE: Retain the same inequality
symbol throughout the solution unless an operation reverses the
symbol.
Example 6
Solve the inequality 3r + 5 > 6r - 7.
Solution
First, subtract 5 from each side:
3r
> 6r - 12
Next, subtract 6r from each side:
= 1 / 3
