SAT Prep Black Book: The Most Effective SAT Strategies Ever Published (27 page)

15
x
2
+ 41
x
+ 28

So (5
x
+ 7)(3
x
+ 4) = 15
x
2
+ 41
x
+ 28

If this seems a little complicated now, don’t worry about it. You’ll get it with practice—and there isn’t that much of it on the SAT anyway.

On the SAT, factoring an algebraic expression involves breaking the expression down into two other expressions that could be multiplied by each other to give the original expression.

Example:

If we have an algebraic expression like (8
x
+ 4)
,
we can break that down into the factors 4 and (2
x
+ 1), because 4(2
x
+ 1) = 8x + 4.

On the SAT, there are three types of factoring situations you’ll need to recognize:

o
recognizing common factors

o
doing “FOIL” in reverse

o
recognizing a difference of squares

Recognizing common factors involves noticing that every term in a given expression has a common factor, as we did in the last example.

Example:

In the expression (21
xy
+ 7
x
), both of the terms in the expression have a common factor of 7
x
, so we can factor the expression like this: 7
x
(3
y
+ 1).

Factoring polynomials basically involves doing the “FOIL” process in reverse. Trust me, it’s not as hard as it looks. It just takes a little practice.

Example:

9x2 – 21x + 12 = (3x – 3)(3x
– 4)

5x2 – 3x
– 2 = (5
x
+ 2)(
x
– 1)

When we factor the difference of two squares, there’s a shortcut we can use—the difference of two squares can be factored as the product of the sum of the square roots of the two squares times the difference of the square roots of the two squares. Let’s see an example.

Example:

9
x
2
– 4 = (3
x
+ 2)(3
x
– 2)

An exponent of a number is what we get when we multiply the number by itself a certain number of times.

Example:

x
*
x
*
x
=
x
3
is an example of an exponential expression. The 3 in this example is the exponent, and the
x
is called the “base.”

Exponents can be positive or negative.

When an exponent is positive, we multiply the base by itself as many times as the exponent indicates, just like we did in the above example.

When an exponent is negative, we treat it just like a positive exponent EXCEPT that we take the reciprocal of the final amount (take another look at the discussion of reciprocals on page).

Example:

x
5
= x * x * x * x * x

x
-5
= 1/(
x
5
)

We can multiply exponent expressions by each other when the bases are identical. To do that, we just add the exponents:

Example:

(x
6
)(x
4
) = (x * x * x * x * x * x) (x * x * x * x) =
x * x * x* x * x * x * x * x * x * x = x
10

(
x
7
)(
x
- 4
) =
x
3

We can also divide exponent expressions when they have the same base. For that we just subtract the exponents:

Example:

(
x
8
)/(
x
2
) =
x
6

Finally, we can raise exponential expressions to other exponents by multiplying the first exponent by the second one:

Example:

(
x
4
)
5
=
x
20

Note that raising any number to an exponent of zero gives you the number 1.

Example:

y
0
= 1

On the SAT, an equation is a statement that involves an algebraic expression and an equals sign.

Example:

5
x
= 20 is an equation, because it involves the algebraic expression 5
x
and an equals sign.

Solving an equation means figuring out how much the variable in the equation is worth. We can solve equations just like you learned in algebra class—by multiply, dividing, adding, or subtracting both sides of the equation by the same amounts until we’re left with a value for the variable.

Example:

  
5
x
= 20

5
x
/5 = 20/5

     x = 4

On the SAT, we can often use equations to answer a question
even when we can’t solve the equation for each variable individually
.

Example:

We might be told that (
a
+
b
)/10 = 15. How can we figure out the value of
a
+
b
? In school, you might try to figure out
a
first, and then
b
, and then add them together. But we don’t have enough information to do that. So what can we do? Well, we just solve for the entire amount
a
+
b
. In this situation, we can do that by multiplying both sides by 10, so
a
+
b
= 150. In this case, even though we can never know the individual values of
a
and
b
, we can know the sum
a
+
b
.  

On the SAT, we can also solve equations “in terms of” one particular variable. To do this, we just isolate the target variable on one side of the equation.

Example:

What if we have to solve this expression in terms of
n
?

     
4
n
+ 7
y
= 2
a

     
4
n      
  = 2
a
– 7
y

      n           = (2a – 7y)/4

               n = (2a – 7y)/4

Sometimes you’ll have a “system” of equations. A system of equations contains two or more equations with the same variables.

Example:

This is a system of equations:

  x
+
y
= 5

2
x
–
y
= 7

The easiest way to solve a system of equations is to solve one equation in terms of one variable, like we just did before. Then we substitute in the second equation and solve.

Example:

First, we’ll isolate the
y
in the first equation, giving us that equation in terms of
y
:
y =
5 –
x
. Now that we know
y
is the same thing as 5 –
x
, we just plug in 5 –
x
where
y
appears in the second equation:

2
x
– (5 –
x
) = 7

2
x
– 5 + x   = 7

3
x
– 5         = 7

3
x
               = 12

                x = 4

Now that we know
x
is 4, we just plug that back into the first equation, and we’ll be able to solve for
y
:

4 +
y
= 5

y
= 1

On the SAT, inequalities are statements that show a particular amount may be greater than or less than a second amount. They use these symbols:

The symbol < means “less than.”

The symbol > means “greater than.”

The symbol
<
means “less than or equal to.”

The symbol
>
means “greater than or equal to.”

You solve an inequality the same way you solve an equation, with one difference: when you multiply by -1 to solve for a variable, you have to switch the direction of the inequality symbol.

Example:

               
-
x
/4 = 10                            
-x
/4
<
10

                 -
x
    = 10(4)                       -
x
   
<
10(4)

                 -
x
    = 40                            -
x
   
<
40

                
x
     = -40                          
x
   
>
-40

A quadratic equation is an equation that involves three terms:

o
one term is a variable expression raised to the power of 2.

o
one term is a variable expression not raised to any power.

o
one term is a regular number with no variable.

Example:

x
2
+ 3
x
= -2 is a quadratic equation because it involves a term with
x
squared, a term with
x
, and a regular number.
 

There is only one way to solve quadratic equations on the SAT, and that is by factoring. (See the discussion of factoring above).

To solve a quadratic equation by factoring, we have to make one side of the equation equal to zero, and then factor the other side of the equation (the quadratic part).

Example: 

              x
2
+3
x
            = -2

             
x
2
+3
x
+ 2      = 0

             
(
x
+ 1)(
x
+ 2) = 0

Now that we know (
x
+ 1)(
x
+ 2) = 0, what else do we know?  We know that one of those two factors has to equal zero—either
x
+ 1 = 0 or
x
+ 2 = 0. How do we know this?  Remember that the only way to multiply two numbers and get zero is if one of the numbers is zero. So if we can multiply
x
+ 1 by
x
+ 2 and get zero, then either
x
+ 1 is zero or
x
+ 2 is zero.

Once we’ve factored, we solve for the variable by creating two small sub-equations in which each factor is set equal to zero.

                            x
+ 1 = 0   or             
x
+ 2 = 0

             
              x
       = -1  or  
x
       = -2                           

So in the equation
x
2
+ 3
x
= -2 ,
x
can equal either -1 or -2.

Quadratic equations can have multiple solutions, as we’ve just seen.