REVIEW OF SIMPLE FACTORING

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REVIEW OF SIMPLE FACTORING
You are expected to be able to factor algebraic
expressions. You will also need to be able to
solve algebraic equations by factoring. When
appropriate, you may use the quadratic formula
rather than by factoring. However the quadratic
formula only works for problems of the form ax 2
bx c 0.
The basic types of expressions you will be
factoring are of the type
Greatest Common Factor G C F
Difference of two squares a 2 b 2 (a
b)(a b)
Trinomial x 2 bx c
Trinomial ax 2 bx c
2
Greatest Common Factor G C F
The GCF must be found in all of the terms, not in
a majority of the terms. If it is not found in
every term then it does not exist or in other
words the GCF is a 1.
EXAMPLE
1. 6x 3 9x 5 y 2 12x 2 y 4
The only common term is 3x 2
3x 2 (2x 3x 3 y 2 4y 4 )
The only common factor is 1. Therefore we say
the expression is prime.
2. 4x 2 8 x 5y
The first step in factoring should always be to
factor the GCF. The remainder of the problem may
or may not be factorable.
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Difference of two squares a 2 b 2 (a b)(a
b)
Number Square Root 1
1 4 2 9
3 16 4
25 5
etc.

• Recognition
• Two terms always separated by a minus sign
• All numbers are perfect squares
• All exponents are divisible by two

• Solve
• The two parentheses will always be alike except
  one will be the other . ( )( )
• In order of the problem write the square root for
  the number and one half of the variables
  exponent.
• Write the same thing in the second parenthesis.
• The order of the signs is not important.

Example 1. 9x 2 25y 4 (3x 5y 2)(3x
5y 2)
2. 28a 2 175c 2 7(4a 2 25c 2) 7(2a
5c)(2a 5c)
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Trinomial x 2 bx c

• I call this a plain trinomial because the
  coefficient of x 2 is a one.
• Reading from left to right answer the following
  questions
• What two numbers multiply to give c
• and at the same time (if ) add to give b or (if
  ) have a difference of b?
• Do not worry about the sign of b yet. Put the
  signs in the answer last not first.

What two numbers have a product of 20 and atthe
same time add to give 12?
Example x 2 12 x 20
The numbers are 10 and 2 because 10times 2 is 20
and 10 plus 2 is 12.
( x 10)( x 2)
However we need 12 so it must have been 10
and 2.
( x 10)( x 2)
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What two numbers have a product of 33 and atthe
same time have a difference of 8?
1. a 2 8 a 33
The numbers are 3 and 11 because3 times 11 is 33
and 11 minus 3 is 8.
(a 3 )(a 11)
It must be 11 and 3 to be 8
(a 3 )(a 11)
What two numbers have a product of 84 and atthe
same time have a sum of 19?
2. c 2 19c 84
184 84 but 84 1 ? 19242 84 but 2 42 ?
19328 84 but 3 28 ? 19421 84 but 4 21
? 19614 84 but 6 14 ? 19712 84 and 7
12 19
It must have been 7 and 12 to be 19
(c 7 )(c 12)
If the problem had been c 2 19c 84,7 and 12
would still be the numbers used except it would
be a 7 and 12.
(c 7 )(c 12)
(c 7 )(c 12)
If you do not know what the numbers are, learn to
find them in a systematic way. Stop all of the
guessing! Try
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Trinomial ax 2 bx c
The value of a adds to the difficulty of
factoring the trinomial. If we use a concept
that is taught in factoring what is called group
factoring, the problem is simplified but only if
we are consistently systematic about the
procedure and not trying to guess the answer as
many students try to do.
120
8x 2 14 x 15
2. Now find two numbers that multiply to give
120 and at the same time have a difference of
14.
3. Write
8x 2 15
Works6 20
Do not work 1 120 2 60 3 40 4
30 5 24
We need 14x. So 20x 6x is 14x
4. Write
8x 2 20 x 6x 15
620 12020 6 14
5. Split the problem in half.
Write this in step 4
8x 2 20 x 6x 15
The underlined parentheses must be exactly alike.
If not, a mistake was made.
4x(2x 5)
3(2x 5)
6. Factor each half

(2x 5)

1. Write the common factor in the first parenthesis

(4x 3)
Final answer
8. and the leading coefficients in the second.
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