1. Can you list all the different types of factoring?
When do you you these?
Anytime
Two terms both being perfect squares or cubes
Three terms the outside terms need to make middle term
Four terms the first two and last two must factor in a way that a common factor is created.
Common x + 8
Difference of squares x2 – 4
Sum of cubes x3 + 27
Difference of cubes x3 – 27
Trinomial factoring x2 + 4x – 12
2x2 + 4x + 2
Grouping x3 + 3x2 – 4x – 12

2. Common Factoring Factor the following: 4x – 6 2x2 + 3x 3x3 + 6x2 – 12x

3. Difference of squares/Sum and difference of cubes Factoring(2 terms)
Factor the following:
x2 – x4 – 25x2 x4 + 4
(x+3)(x-3) (x2 -5x)(x2+5x) prime (can’t factor)
(Square root of first term + the square root of second term)(square root of first term – square root of second term)
Factor the following:
x3 – 8 8x x6 – 64x3
(x-2)(x2+2x+4) (2x+3)(4x2-6x+9) (x2-4x)(x4+4x3+14x2)
(cube root of first term +/- cube root of second term)(cube root of first term squared +/- cube root of first term times cube root second term + cube root of second term squared)
What else can be done to the last expression?
The first parenthesis has a common factor and difference of squares the second parenthesis has a common factor.

4. Trinomial factoring Factoring
Factor the following:
x2 + 5x x2 - 3x + 2 x4 - 6x2 – 27
(x+7)(x-2) (x-1)(x-2) (x2-9)(x2+3)
Factor the following:
2x2 + 5x x4 – 5x2 – 6
(2x+3)(x+1) (3x2 -2)(2x2+3)
The first term of the two factors must multiply to make the first term of the trinomial the second term of the two factors must multiply to make the third term of the trinomial and when you multiply the the first term of the first factor and the second term of the second factor and combine it with the multiplying the second term of the first factor and the first term of the second factor it must make the second term of the trinomial.

5. Common Factoring Factor the following:
4x3 – 6x2 + 2x x3 – 2x2 – 5x x4 + 6x3 – 2x2 - 4x
2x2(2x-3)+1(2x-3) x2(x-2)-5(x-2) 3x3(x+2)-2x(x+2)
Took out a common factor from the first two terms then took out the common factor from the last two terms(even if it is 1).
(2x-3)(2x2+1) (x-2)(x2-5) (x+2)(3x2-2x)
Took out the common factor(binomial). Like the last problem on slide 2.
What can you still do on problem 3?
How else could you have factored problem 3?
The second factor can take out a common x
You could have taken a common factor out of the original problem first.

6. Practice
2 - 8x x – x x2 + 8x + 2
(3x-2)3 – (2x+1)2 + (5x-6)2(2x+1)2
2(1-4x2) -(x2-2x-15) (3x2+4x+1)
2(1+2x)(1+2x) -(x+3)(x-5) (3x+1)(x+1)
(3x-2 – 3)((3x-2)2+3(3x-2)+9) (2x+1)(5(2x+1)+4(5x-6)
(3x-5)(9x2-12x+4+9x-6+9) (2x+1)(10x+5+20x-24)
(3x-5)(9x2-3x+7) (2x+1)(30x-19)