The document provides a comprehensive overview of the methods for factoring various algebraic expressions, including monomials, binomials, and trinomials, as well as special products such as the difference of squares and cubes. It emphasizes the importance of finding the greatest common factor (GCF) and offers detailed examples to illustrate the factoring process. The document serves as a resource for understanding the fundamentals of factoring in algebra, catering to both simple and complex expressions.

1. Factoring - Reviewed
Monomials, Binomials,
Trinomials
&
More…

2. A Quadratic
Activity

3. Scratching Your Head Over Factoring
Quadratics?
Then you must get back to basics
Quadratic Activity A
to Z
?

4. Back to Basics With
Quadratics
If you listen attentively
And keep eyes and hands busy
Factoring Quadratics can be as easy
as 123
+ ÷ - x

5. Linear
Circular
Parabola
• Everything’s not straight.
• The power of 1; that’s me
• Y = mX + b
• Some things are round
• The power of 2; add you
• X2 + Y2 = r2
• Other things are turned upside down!
• The power of 1 and 2; that’s me and you
• Y = a X2 + b X + c -

6. 1. Greatest
Common Factor
In order to factor completely
You must consider the GCF
Factoring out the GCF is like studying a family with children.
• Let m2w + mw2 represent the family unit, where m represents the father, and
let w represents the mother.
• What does the family unit have in common?
• You must factor to find out: mw(m + w)
• So mw represents the child, the greatest common factor in the family.
• If there is more than one child, the family could look like this: m3w2 + m2w3
• Factor out all the children: (mw)(mw)(m + w) = m2w2(m + w)

7. 2. Factoring and
Special Products
1. Sometimes Alone
2. Sometimes together
- Always work out the difference
between each other
(x + y)2 = (x + y)(x + y)
(x – y)2 = (x - y)(x – y)
(x + y)(x – y)
x2 – y2

8. Example: (2x2 - 12x + 18)
2(x2 - 6x + 9) – That took care of the GCF.
? Is the first term a perfect square?
? Is the third term a perfect square?
? Is the 3rd term ½ the middle coefficient squared?
(6/2 = 3, and 32 = 9) ?
If the answer to all three is yes, then:
(2x2 - 12x + 18) factors completely as
2(x – 3)(x – 3)
2(x – 3)2 ; and that’s a perfect square.
Special Products
Cont’d
1. Sometimes Alone - Factoring Perfect
Squares

9. Don’t recoil, just remember how to F.O.I.L.
(x2 – xy + xy - y2)
You may be asking, “Why should I care?”
‘Cause when you bring them together, the
middles disappear!
(x2 – xy + xy - y2) = x2 - y2
Let’s Teach By Example:
Example 1: 4x2 – 25
- 4 is perfect
- x2 is perfect
- 25 is perfect. Now let’s un-F.O.I.L. it.
Special Products
Cont’d
2. Sometimes Together - Factoring Binomials
(x + y)(x – y)

10. To Factor Difference of
Squares – Mix the signs
and take the square root.
√x2 = x; √4 = 2; √25 = 5
4x2 – 25 = (2x – 5)(2x + 5)
It’s not always clear that the
square is there
Example 2: -27x3 + 75x
• Think again bout factoring
completely;
• You must remember the GCF
-3x(9x2- 25)
-27x3 + 75x = -3x(3x – 5)(3x + 5)
Special Products
Cont’d

11. 1. There is no GCF
2. The factors of 6 are (1, 6), (-1, -6) and (2, 3), (-2.
-3)
3. None of the sums or differences of the pairs of
factors of 6 add up to the middle term, -3.
4. Conclusion: This trinomial cannot be factored
x2 -3x +6 is prime
1. Prime
x2 -3x +6
3. Factoring Trinomials

12. 1. There is no GCF
2. The factors of 6 are (1, 6), (-1, -6) and (2, 3), (-2,
-3)
3. The pair (-2, -3) sum to -5, the middle term.
4. Conclusion: This trinomial is factorable
5. Write x as the first term in the pair of parentheses
(x )(x )
6. Write each factors – with its sign – as the second
term of each parenthesis (x - 2)(x - 3)
The trinomial with first coefficient ‘1’ is done!
Factoring Trinomials
cont’d
x2 -5x +6
2. First coefficient is 1 - Factorable

13. 1. Split the middle term
1. 3p2 + 6p + 3p + 6 : Notice that the sum of the
split term equals the original middle term.
2. Notice that the product of the split term (6)(3) =
18 = (3)(6), the product of the first and last term.
2. Group: Create binomials with the first two terms and
the last two terms
(3p2 + 6p) + (3p + 6)
- If it’s 3 then split it
- When it’s 4 then group it
1. Trinomials – First Coefficient Not ‘1’
3p2 + 9p + 6
.
4. Factoring by Grouping

14. 3. Factor out the GCF – 3p(p + 2) + 3(p + 2)
4. Then factor out the GCF again
(p + 2)(3p + 3)
Notice that the GCF in this case is (p + 2)
Factoring by Grouping
1. Trinomials – First Coefficient Not ‘1’ Cont’d
When split in 4, group and factor.
1. (7q3 + q2)+ (14q + 2)
2. Factor Out the GCF – q2(7q + 1)+ 2(7q +1)
3. And Factor out the GCF again
(7q + 1)(q2 + 2)
2. Four Terms – Group them – 7q3 + q2
+ 14q + 2

15. 1. Look at c3 + d3 – The first and second terms are
cubes
2. Take the cube root of both terms: cube root of c3,
and d3 is c, and d respectively.
3. Then write the product of a binomial and a
trinomial
You Can’t get out of the cube without
factoring.
5.Factoring
Cubes
Factoring Cubes – Is All About Products
1. Sum of Cubes

16. 1. Sum of Cubes Cont’d
1. Write the binomial (c + d)
2. Now create a trinomial like this:
1. Square the first term of the binomial c2 – that’s a
product - and place it as the first term of a trinomial
(c2 )
2. Square the second term of the binomial, d2 - that’s a
product - and place it as the third term of the trinomial
(c2 d2)
3. Find the product of the first and second terms of the
binomial and place it as the middle term of the trinomial.
(c2 cd d2).
The Sum and
Difference of Cubes

17. 1. Sum of Cubes Cont’d
1. Now What about the signs?
1. The sign of the binomial is always the same
as the sign of the cube.
(c + d)
2. The first sign in the trinomial is always the
opposite of the sign of the binomial
(c2 – cd d2).
3. The second sign in the trinomial is always
positive (c2 – cd + d2).
The Sum and
Difference of Cubes

18. The Sum and
Difference of Cubes
2. Difference of Cubes
1. Let’s show this one by Example
2. 8q3 – 27
1. The cube root of 8 is 2
2. The cube root of q3 is q
3. The cube root of 27 is 3
3. Using the same general rules as for the sum of squares
1. Rewrite 8q3 – 27 as (2q)3 – 33.
2. 8q3 – 27 = (2q – 3)((2q)2 – (2q)(3) + 32)
(2q – 3)(4q2 – 6q + 9)