Download Unit 7 - Factoring

Unit 7 - Factoring
Unit 7.1 – Greatest Common Factors; Factoring by Grouping
To factor a polynomial is to write the polynomial as the product of two or more simpler
polynomials.
The product of 3x and 5x  2 is 15 x 2  6 x , and 15 x 2  6 x can be factored as the product
3x5x  2 .
Both multiplication and factoring use the distributive property, but in different directions.
Factoring “undoes” or reverses multiplying.
The first step in factoring a polynomial is to find the greatest common factor (GCF).
8 x  12  42 x   43
 42 x  3
56m  35q
12  24 x
Examples: 9 x  18
What about when the GCF includes a variable?
9 x 2  12 x 3  3x 2 3  3x 2 4 x 
notice the GCF includes a numeric part and a variable part. I used
 3x 2 3  4 x 
the variable with the smallest exponent as the variable part.
Examples: 16 y 4  8 y 3 14 x 2  9 x 3  6 x 4 5m 4 x 3  15m5 x 6  20m 4 x 6
Our first examples all used monomials as the GCF. This isn’t always the case. Let’s use
binomials as the GCF.
x  5x  6  x  52x  5 the GCF here is x  5
x  5x  6  x  52 x  5  x  5x  6  2 x  5
 x  5x  6  2 x  5
 x  53x  11
Examples: x  2x  3  x  2x  6
 y  1 y  3   y  1 y  4
k 2 x  5 y   m 2 x  5 y 
Factoring when the coefficient of the term of greatest degree is negative.
 x 3  3x 2  5 x can be factored in two ways:


 x 3  3x 2  5 x  x  x 2  x3x   x 5


or
 
 x 3  3x 2  5 x   x x 2   x  3x    x 5


 x  x 2  3x  5
  x x 2  3x  5
Either answer is correct, but we tend to favor the solution where the common factor has a
positive coefficient.
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Unit 7 - Factoring
Factoring by grouping
What if the terms of a polynomial have no common factors? In this case we may be able to use a
process called factoring by grouping.
Example: the terms of ax  ay  bx  by have no common factor, but we can group the terms into
“smaller polynomials” that do have common factors.
ax  ay  bx  by can be grouped as  ax  ay   bx  by  .
We can now factor each group a  x  y   b  x  y  which does have a common factor of  x  y  .
So the final factored form is  x  y  a  b  .
Steps in factoring by grouping
1. Group terms – Collect the terms into groups so that each group has a common factor.
2. Factor within the groups – Factor out the common factor in each group.
3. Factor the entire polynomial – If each group now has a common factor, factor it out. If not,
try a different grouping.
4. ALWAYS CHECK THE FACTORED FORM BY MULTIPLYING.
Factoring Examples:
xy  2 y  4 x  8   xy  2 y    4 x  8 
 y  x  2   4  x  2 
 y  x  2  4  x  2
  x  2  y  4 
2 xy  3 y  2 x  3   2 xy  3 y    2 x  3
 y  2 x  3   1 2 x  3 
  2 x  3 y  1
note here that the factor 1 was introduced to the second
grouping. Why is this possible?
p 2 q 2  10  2q 2  5 p 2  p 2 q 2  2q 2  5 p 2  10

 
 q  p  2  5  p
  p  2  q  5 
 p 2 q 2  2q 2  5 p 2  10
2
2
2
2
2


2
Note the need to rearrange the terms prior to grouping and then factoring. This is because as
written neither the first two nor the last two terms have a common factor.
Unit 7.2 – Factoring Trinomials
Factor trinomials when the coefficient of the squared term is 1.
Multiplying and factoring are operations that “undo” each other. Factoring trinomials involves
using FOIL “backwards”.
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Unit 7 - Factoring
I.E. since
 x  3 x  5  x 2  5 x  3x  15
 x 2  2 x  15
we can say that  x  3 x  5 is the factored form of x 2  2 x  15 .
Factoring x 2  bx  c
1. Find pairs whose product is c – Find all pairs of integers whose product is the third term of
the trinomial, c.
2. Find pairs whose sum is b – Choose the pair from Step 1 whose sum is the coefficient of the
middle term, b.
Note: if there are no such pairs in Step 2, the polynomial cannot be factored. A polynomial that
cannot be factored with integer coefficients is a prime polynomial.
Examples: x 2  9 x  20 x 2  7 x  10 x 2  8x  6 (when doing these examples be sure to first
list all the pairs of integers that equal the third term and then go through them checking to see if
they add to the second term.)
What about p 2  6ap  16a 2 ?
In this case find expressions whose product is 16a 2 and whose sum is 6a .
Example: m 2  2mn  8n 2
Factoring a trinomial with a common factor.
16 y 3  32 y 2  48 y  16 y  y 2  2 y  3 first factor out the common factor, 16y. Then factor the
remaining trinomial, if possible.
** When factoring, always look for a common factor first. And don’t forget to include the
common factor as part of the answer.
Example: 5m4  5m3  100m 2
Factoring a trinomial when the coefficient of the squared term is not 1.
Using Grouping:
Recall the steps of factoring trinomials when the coefficient of the squared term is 1 ( x 2  bx  c ):
1. Find pairs whose product is c – Find all pairs of integers whose product is the third term of
the trinomial, c.
2. Find pairs whose sum is b – Choose the pair from Step 1 whose sum is the coefficient of the
middle term, b.
The trinomial x 2  bx  c is really ax 2  bx  c with a = 1. So in Step 1 we are really finding
pairs whose product is ac. We can use this generalization and the grouping method of factoring
to factor trinomials where the coefficient of the squared term is not 1.
Example: 3x 2  7 x  2
First identify values of a, b, and c in the trinomial. In this case a  3 b  7 c  2 . The product
ac is 6 and b is 7, so we must find integers having a product of 6 and a sum of 7. The only
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Unit 7 - Factoring
possibilities here are 1 and 6. So we rewrite the trinomial thus: 3x 2  7 x  2  3x 2  1x  6 x  2 .
We can now take this trinomial and factor using the grouping method.
3x 2  x  6 x  2   3x 2  x    6 x  2 
 x  3x  1  2  3x  1
  3x  1 x  2 
That’s one method. We can also use the guess and check with FOIL method. We’ll use the same
trinomial.
3x2  7 x  2   ___ x  ___  ___ x  ___  . The goal is to fill in the blanks with the correct
numbers. All + signs are used because the original trinomial has all + signs.
Now the first two expressions have a product of 3x 2 so they must be 3x and x.
3x2  7 x  2  3x  ___  x  ___  .
The product of the last two terms has to be 2, so the numbers must be 2 and 1. And here there is
a choice, where to put each number. Plug ‘em in and use FOIL to check if they’re correct.
Let’s try this one first:  3x  2 x  1 and use FOIL to check. Using FOIL gives us
3x 2  3x  2 x  2  3x 2  5 x  2 but this gives us the incorrect middle term.
So we’ll try  3x  1 x  2 . Using FOIL gives us 3x 2  6 x  x  2  3x 2  7 x  2 which is what
we’re looking for.
The Box method.
This is a variation of the grouping method. You have to find the same values as that method.
Example: 3x 2  7 x  2 Again, we have to find integers whose product is the same as the product
ac (in this case 6) and whose sum is b (in this case 7). The only ones that work are 1 and 6. Now
we set up the box.
Set up a box with 2 columns and 2 rows.
In the top left cell, place the first term from the original
trinomial. In the bottom right cell, place the last term.
2
3x
2
2
3x
6x
1x
2
3x
X 3x2
+2 6x
(3x+1)(x+2)
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+1
1x
2
Now take the factors you just came up with (in this case 1
and 6) and place them in the remaining cells, complete
with signs and variables.
Now factor the rows moving to the left and the columns
moving up. Keep the sign of the closest number as you
factor. (this means keep the sign of the top cells for the
column factors and the left cells for the row factors)
Now just read your answer from the left side and top of
the Box!
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Unit 7 - Factoring
Try several using the Box method: 10 x 2  17 x  3 16 y 2  34 y  15 8x 2  13x  5
The Box method even works with a trinomial in two variables.
18m2  19mx  12 x 2 In this case the product ac = -216 and b = -19.
-27 * 8 = -216 and -27 + 8 = -19, so we’ll use those values in the box:
2m
-3x
2
9m 18m
-27mx
+4x 8mx
-12x2
Try a couple: 12 x 2  5xy  2 y 2 6m2  7mn  5n2 8m2  18mx  5x 2
When factoring a trinomial that has a negative number on the squared term, it can be helpful to
factor out the negative before doing anything else. It’s not mandatory; it just sometimes makes it
easier.
Unit 7.3 – Special Factoring
Remember the special products back when we were doing multiplication of polynomials.
The product of the sum and difference of two terms ( x  y x  y  ) equals a difference of
squares ( x 2  y 2 ). You can use this pattern when factoring.
Example: Factor (completely) this polynomial 4 x 2  64
1. Remember you should always factor any common factor first.
a. 4 x 2  16
2. x 2  16 is a difference of squares x 2  4 2 and factors as
3. x  4x  4
4. So the complete factoring is 4x  4x  4 (don’t forget the common factor!)



Examples:
2 x 2  18  2 x 2  9




 
9 x 2  16 y 2  3x   4 y 
2
 2x  3x  3
2
 3x  4 y 3x  4 y 
x  32  49 y 2  x  32  7 y 2
 x  3  7 y x  3  7 y 
Perfect Square Trinomial
2
x 2  2 xy  y 2   x  y 
x 2  2 xy  y 2   x  y 
Both the first and last terms must be perfect squares. In the factored form, twice the product of
the first and last terms must give the middle term of the trinomial.
Example:
2
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Unit 7 - Factoring
144 x 2  120 x  25 is this a perfect square trinomial and if so, what is the factored form?
2
144 x 2  12 x  and 25  5 2 . Since the sign on the middle term is negative, if this is a perfect
square trinomial its factored form must be 12 x  5 . Calculate twice the product of the two
terms to find out. 212 x 5  120 x so this is a perfect square trinomial and the factored form
2
is 12 x  5 .
2
Example:
4m 2  20mn  49n 2 If this is a perfect square trinomial, its factored form would have to be
2m  7n 2 . But 22m7n  20mn , so this cannot be a perfect square trinomial.
Factoring a Difference of Cubes
x3  y 3   x  y   x 2  xy  y 2 
Example:
m3  8  m3  23

  m  2  m
  m  2  m 2  2m  2 2
2
 2m  4
27 x3  8 y 3   3x    2 y 
3


3

  3x  2 y   3x    3x  2 y    2 y 
2
2

  3x  2 y   9 x 2  6 xy  4 y 2 
Factoring a Sum of Cubes
x3  y 3   x  y   x 2  xy  y 2 
Example:
x3  27  x3  33

  x  3  x
  x  3 x 2  3x  32
2
 3x  9


27 x3  125   3 x   53
3
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Unit 7 - Factoring
Special Types of Factoring
Difference of Squares x2  y 2   x  y  x  y 
x 2  2 xy  y 2   x  y 
2
Perfect Square Trinomial
x 2  2 xy  y 2   x  y 
2

  x  y x

 xy  y 
Difference of Cubes
x3  y 3   x  y  x 2  xy  y 2
Sum of Cubes
x3  y 3
2
2
Unit 7.4 – Solving Equations by Factoring
Think back on the many equations we have solved so far in this class. They have all been linear
equations, or equations of degree 1. Solving equations of degree greater than 1 requires different
methods. One of the methods is factoring. Solving equations by factoring relies on the ZeroFactor Property.
Zero-Factor Property – If two numbers have a product of zero, then at least one of the numbers
must be zero. In other words, if ab = 0 then either a = 0 or b = 0.
Example:  x  6 2 x  3  0 By the zero factor property this can only be true if x  6  0 or
2x  3  0 . So solve BOTH equations.
x60
2x  3  0
3
3

x  6
2 x  3 the solutions are x  6 or x  or 6, 
2
2

3
x
2
Quadratic Equation – An equation that can be written in the form ax 2  bx  c where a  0 ,
is a quadratic equation. This is the standard form. (In the example above the product
 x  6 2x  3  2x2  9x 18 and is a quadratic equation.
Solving a Quadratic Equation by Factoring
1. Write in standard form – Rewrite the equation if necessary so that one side is 0.
2. Factor the polynomial.
3. Use the zero-factor property – Set each variable factor equal to 0.
4. Find the solution(s) – Solve each equation from Step 3.
5. Check – Check each solution in the original equation.
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Unit 7 - Factoring
Example: Solve 2 x 2  3 x  2
2 x 2  3x  2
1.
2 x 2  3x  2  0
2.  x  2 2 x 1  0
3. x  2  0 or 2x 1  0
1
x
4. x  2
2
Examples:
4 x2  4 x 1
5 x 2  25 x  0
 x  6 x  2  8  x
 x3  x 2  6 x  0 (multiply by -1 first, then factor out the common x)
A piece of sheet metal is in the shape of a parallelogram. The longer sides are each 8 m longer
than the distance between them. The area is 48 m2. Find the length of the longer sides and the
distance between them.
If a small rocket is launched vertically with an initial velocity of 128 ft per second, then its
height in feet after t seconds is a function defined by h  t   16t 2  128t , if air resistance is
ignored. After how many seconds will the rocket be 220 ft above the ground?
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