#### Transcript Objective The student will be able to: factor quadratic trinomials. Trial and Error Method SOL: A.2c Designed by Skip Tyler, Varina High School.

```Objective
The student will be able to:
Trial and Error Method
SOL: A.2c
Designed by Skip Tyler, Varina High School
Factoring Chart
which method of factoring to use.
Type
Number of Terms
1. GCF
2 or more
2. Diff. Of Squares 2
3. Trinomials
3
Review: (y + 2)(y + 4)
Multiply using FOIL or using the
Box Method.
Box Method: y + 4
y y2 +4y
+ 2 +2y +8
Combine like terms.
2
FOIL: y + 4y + 2y + 8
y2 + 6y + 8
1) Factor.
2
y
+ 6y + 8
Put the first and last terms into the box
as shown.
y2
+8
What are the factors of y2?
y and y
1) Factor.
2
y
+ 6y + 8
Place the factors outside the box as shown.
y
y
y2
+8
What are the factors of + 8?
+1 and +8, -1 and -8
+2 and +4, -2 and -4
1) Factor.
2
y
+ 6y + 8
Which box has a sum of + 6y?
y
+1
y
+2
y
2
y
+ 8 + 8y
+y
+8
y
2
y
+ 4 + 4y
+ 2y
+8
The second box works. Write the numbers
on the outside of box for your solution.
1) Factor.
2
y
+ 6y + 8
(y + 2)(y + 4)
factors.
1) When the last term is positive, the factors
will have the same sign as the middle term.
2) When the last term is negative, the factors
will have different signs.
2) Factor.
2
x
- 2x - 63
Put the first and last terms into the box
as shown.
x2
- 63
What are the factors of x2?
x and x
2) Factor.
2
x
- 2x - 63
Place the factors outside the box as shown.
x
x
2
x
- 63
What are the factors of - 63?
Remember the signs will be different!
2) Factor. x2 - 2x - 63
Use trial and error to find the correct
combination!
x
-3
x
-7
x
x2
-3x
+ 21 +21x - 63
x
x2
+ 9 +9x
-7x
- 63
Do any of these combinations work?
The second one has the wrong sign!
2) Factor. x2 - 2x - 63
Change the signs of the factors!
x
+7
x
-9
2
x
+7x
-9x
- 63
(x + 7)(x - 9)
3) Factor.
2
5x
- 17x + 14
Put the first and last terms into the box
as shown.
5x2
+ 14
What are the factors of 5x2?
5x and x
3) Factor. 5x2 - 17x + 14
5x
x
5x2
+ 14
What are the factors of + 14?
Since the last term is positive, the signs of the
factors are the same! Since the middle term
is negative, the factors must be negative!
3) Factor. 5x2 - 17x + 14
When the coefficient is not 1, you must try
both combinations!
5x
-2
5x
-7
x
-7
5x2
- 2x
x
-35x + 14 - 2
5x2
- 7x
-10x + 14
Do any of these combinations work?
3) Factor. 5x2 - 17x + 14
(5x - 7)(x - 2)
It is not the easiest of things to do,
but the more problems you do, the
easier it gets! Trust me!
2
2x
4) Factor
+ 9x + 10
(x + 2)(2x + 5)
5) Factor.
2
6y
- 13y - 5
(2y - 5)(3y + 1)
6) 12x2 + 11x - 5
(4x + 5)(3x - 1)
7) 5x - 6 + x2
2
x + 5x - 6
(x - 1)(x + 6)
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