GRE

II. Complex Expressions with Exponents

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) = -6rs

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

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