Factoring Polynomials
Grouping, Trinomials, Binomials,
GCF ,Quadratic form & Solving
Equations
Student will be able to Factor
by Grouping terms

When polynomials contain four terms, it
is sometimes easier to group like terms in
order to factor.
 Your goal is to create a common factor.
 You can also move terms around in the
polynomial to create a common factor.
 Practice makes you better in recognizing
common factors.
Factoring Four Term
Polynomials
Do now: find the GCf of the first two terms and the last two terms:
3x 12x  6x  24
3
2
2
3x

and 6
Group together and Factor each one
separately:
(3x 12x )  (6x  24)
3
2
3x (x  4)  6(x  4)
2
They share a common factor of (x-4)
Write 2 factors:

Write the common factor once and put
the outside terms together:
(3x 3  12x 2 )  (6x  24)
3x 2 (x  4)  6(x  4)
(3x 2  6)(x  4)
Factor by Grouping
Example 1:
FACTOR: 3xy - 21y + 5x – 35
 Factor the first two terms:
3xy – 21y
 Factor the last two terms:
+ 5x - 35 =

Factor by Grouping
Example 1:
FACTOR: 3xy - 21y + 5x – 35
 Factor the first two terms:
3xy - 21y = 3y (x – 7)
 Factor the last two terms:
+ 5x - 35 = 5 (x – 7)
 The green parentheses are the same so
it’s the common factor

Factor by Grouping
Example 1:
FACTOR: 3xy - 21y + 5x – 35
 Factor the first two terms:
3xy - 21y = 3y (x – 7)
 Factor the last two terms:
+ 5x - 35 = 5 (x – 7)
 The green parentheses are the same so
it’s the common factor
Now you have a common factor
(x - 7) (3y + 5)

Factor by Grouping
Example 2:
FACTOR: 6mx – 4m + 3rx – 2r
 Factor the first two terms:
6mx – 4m =
 Factor the last two terms:
+ 3rx – 2r =

Factor by Grouping
Example 2:
FACTOR: 6mx – 4m + 3rx – 2r
 Factor the first two terms:
6mx – 4m = 2m (3x - 2)
 Factor the last two terms:
+ 3rx – 2r = r (3x - 2)
 The green parentheses are the same so
it’s the common factor
Now you have a common factor
(3x - 2) (2m + r)

Factor by Grouping
Example 3:

FACTOR: y3– 5y2 -4y +20
Factor by Grouping
Example 3:
FACTOR: y3– 5y2 - 4y +20
 Factor the first two terms:
y3– 5y2 = y2 (y - 5)
 Factor the last two terms:
- 4y +20 = -4 (y – 5)
 The green parentheses are the same!
y2 (y - 5) and -4 (y - 5)
 Now you have the difference of two squares!
look at red ( ):
(y - 5) (y2 - 4) :
answer: (y - 5) (y - 2) (y + 2)

See worksheet “Factor by
grouping”

Try first 4 problems.
Using Factor by Grouping to solve a polynomial
function:
From the last example, suppose it was an equation…..
y3– 5y2 - 4y +20 = 0

(y - 5) (y - 2) (y + 2) = 0
y=5 y = 2 y=-2
So the solution set is { 5,2,-2}
Factor first, then set factors = 0
3x 12x  6x  24  0
3
2
(3x 2  6)(x  4)  0
3x 2  6  0 X-4=0
solve
3x 12x  6x  24  0
3
2
(3x 2  6)(x  4)  0
3x 2  6  0
3x  6
X-4=0
2
x2  2
X=4
x  2
{4, 2, 2}
Hand this one in:

Solve for all roots:
3x3 - 4x2 -27x +36 = 0
Factoring Trinomials
Factoring Trinominals
1. When trinomials have a degree of “2”,
they are known as quadratics.
2. We learned earlier to use the last term’s
factors to factor trinomials that had a
“1” in front of the squared term.
x2 + 12x + 35
So… 7 and 5 or 35 and 1
Factoring Trinominals
1. When trinomials have a degree of “2”,
they are known as quadratics.
2. We learned earlier to use the last term’s
factors to factor trinomials that had a
“1” in front of the squared term.
x2 + 12x + 35
(x + 7)(x + 5)
Because 7 + 5 = 12!
More Factoring Trinomials
3. When there is a coefficient larger than
“1” in front of the squared term, we can
use a method we will call, the “am” add,
multiply method to find the factors.
3. Always remember to look for a GCF
before you do ANY other factoring.
More Factoring Trinomials
5. Let’s try this example
3x2 + 13x + 4
(3x )(x )
Write the factors of the last term…1,4 2,2
Multiply using foil until you get the middle
term of the trinomial. If so, you’re
done!
More Factoring Trinomials
3x2 + 13x + 4
(3x + 1 )(x + 4 )
3x2 + 12x + 1x + 4
= 3x2 + 13x + 4 ✓
Difference of Squares
Difference of Squares

When factoring using a difference of
squares, look for the following three
things:
– only 2 terms
– minus sign between them
– both terms must be perfect squares
– No common factors
 If all of the above are true, write two
( ), one with a + sign and one with a – sign
: ( + ) ( - ).
Try These, (if possible)
 1.
 2.
 3.
 4.
 5.
 6.
a2 – 16
x2 – 25
4y2 – 16
9y2 – 25
3r2 – 81
2a2 + 16
answers:
 1.
 2.
 3.
 4.
 5.
 6.
a2 – 16 (a + 4) (a – 4)
x2 – 25 (x + 5) (x – 5)
4y2 – 9 (2y + 3) (2y – 3)
9y2 – 25 (3y + 5) (3y – 5)
3r2 – 81 *3 is not a square!
a2 + 16 Not a difference!
Perfect Square Trinomials
Perfect Square Trinomials

When factoring using perfect square
trinomials, look for the following three
things:
– 3 terms
– last term must be positive
– first and last terms must be perfect
squares
 If all three of the above are true, write
one ( )2 using the sign of the middle
term.
Try These
a2 – 8a + 16
 2. x2 + 10x + 25
 3. 4y2 + 16y + 16
 4. 9y2 + 30y + 25
 1.
Factoring Completely
Factoring Completely

Now that we’ve learned all the types of
factoring, we need to remember to use them all.

Whenever it says to factor, you must break
down the expression into the smallest possible
factors.

Let’s review all the ways to factor.
Types of Factoring
1. Look for GCF first.
2. Count the number of terms:
a) 4 terms – factor by grouping
b) 3 terms 1. look for perfect square trinomial
2. if not, try “am” method
c) 2 terms look for difference of squares
 If any ( ) still has an exponent of 2 or
more, see if you can factor again.
These may take 2 steps!
1. 3r2 – 18r + 27
 2. 2a2 + 8a - 8

Answers:
1. 3r2 – 30r + 27

3(r2 - 10r + 9)

3(r – 9) (r – 1)

2. 2a2 + 8a – 8
 2(a2 + 4a – 4)

Solving Equations by Factoring
Completely
Do Now:
1)Factor completely
2)Solve for x
1) x  4 x
5
2
2) 3x  18x  27  0
2
Steps to Solve Equations by Factoring
Completely
set each factor = 0 and solve for the unknown.
x3 + 12x2 = 0 1. Factor GCF
x2 (x + 12) = 0
Steps to Solve Equations by Factoring
Completely
set each factor = 0 and solve for the unknown.
x3 + 12x2 = 0 1. Factor GCF
x2 (x + 12) = 0 2. (set each factor = 0, & solve)
x2 = 0
x + 12 = 0
x=0
x = -12
You now have 2 answers, x = 0 and x = -12
Factor completely:
x  x 12x  0
3
2
Factor completely:
x  x 12x  0
3
2
x(x  x 12)  0
x(x  4)(x  3)  0
2
X=
0 , 4 , 3
Solving higher degree functions
Quadratic form:
 ax4 + bx2 + c = 0
 Example: x4 +2x2 -24 = 0
 Factor: (x2
)(x2 )=0

Solving higher degree functions
Quadratic form:
 ax4 + bx2 + c = 0
 Example: x4 +2x2 -24 = 0
 Factor: (x2 +6 )(x2 – 4 ) = 0

x2 +6=0
x2 – 4 =0

x2 =-6
x2 = 4

x = 2, -2
x  i 6


Try this one:

X4 – 13x2 +36 = 0
Factor first:

X4 – 13x2 +36 = 0
(x2 – 9)(x2 – 4)=0
Solutions:

X4 – 13x2 +36 = 0
(x2 – 9)(x2 – 4)
X2-9=0 x2-4=0
X=3,-3
x= 2,-2
This one can be verified on the calculator.
X = 2,-2,3,-3
Hand this one in

Quadratic form:
x  6x  27  0
4
2
Ans: x 4  6x 2  27  0

X=
(x  9)(x  3)  0
2
2
x 9  0 x 3 0
2
2
x  9 , x  3
2
2
x  3i,3i , x  3, 3