1. Factoring Polynomials 3

2. Quadratic Trinomials
A quadratic trinomial is a polynomial with the form
𝑎 𝑥 2 +𝑏𝑥+𝑐 where 𝑎, 𝑏, and 𝑐 are real numbers and 𝑎≠0
You have already learned how to factor such a polynomial when 𝑎=1
For example, 𝑥 2 −7𝑥+12 is factored by finding the factors of 12 that add to give −7
The factors of 12 are 1⋅12, 2⋅6, 3⋅4
Since −3⋅−4=12 and −3+ −4 =−7, then 𝑥 2 −7𝑥+12=(𝑥−4)(𝑥−3)

3. Quadratic Trinomials When 𝑎≠1 then a quadratic trinomial factors as
(𝑝𝑥+𝑞)(𝑟𝑥+𝑡)
where 𝑝𝑟=𝑎 and 𝑝𝑞=𝑐 in 𝑎 𝑥 2 +𝑏𝑥+𝑐
We must discover a method for factoring a quadratic trinomial
Start by multiplying 𝑝𝑥+𝑞 𝑟𝑥+𝑡
𝑝𝑥+𝑞 𝑟𝑥+𝑡 =𝑝𝑟 𝑥 2 +𝑝𝑡𝑥+𝑞𝑟𝑥+𝑞𝑡
The two middle terms are not like terms, but we can factor the common monomial 𝑥
𝑝𝑟 𝑥 2 +𝑥 𝑝𝑡+𝑞𝑟 +𝑞𝑡

4. Quadratic Trinomials 𝑝𝑟 𝑥 2 +𝑥 𝑝𝑡+𝑞𝑟 +𝑞𝑡 Compare this to 𝑎 𝑥 2 +𝑏𝑥+𝑐
We see that 𝑝𝑟 is in the same position as 𝑎, and that 𝑞𝑡 is in the same position as 𝑐
The value of 𝑏 comes from multiplying the factors of 𝑎 and 𝑐 in a particular way, then adding them
Our method must discover how to determine the value of b
We will use a method similar to the earlier one, and then use factoring by grouping

5. Quadratic Trinomials 𝑝𝑟 𝑥 2 +𝑥 𝑝𝑡+𝑞𝑟 +𝑞𝑡 𝑎 𝑥 2 +𝑏𝑥+𝑐
Notice that if we multiply 𝑎⋅𝑐 (or 𝑝𝑟⋅𝑞𝑡), we put together all of the factors of a and c into one number
We then find the factors of this new number that add together to produce b (similar to what we did in the simpler case)
The factors of 𝑝𝑟𝑞𝑡 are 𝑝𝑟⋅𝑞𝑡, 𝑝𝑞⋅𝑟𝑡, 𝑝𝑡⋅𝑞𝑟
The last is the one we want because we can then find 𝑝𝑡+𝑞𝑟

6. Quadratic Trinomials 𝑝𝑟 𝑥 2 +𝑥 𝑝𝑡+𝑞𝑟 +𝑞𝑡 𝑎 𝑥 2 +𝑏𝑥+𝑐
Now, we write 𝑝𝑟 𝑥 2 +𝑝𝑡𝑥+𝑞𝑟𝑥+𝑞𝑡 (note that we have included x)
We group the expression as 𝑝𝑟 𝑥 2 +𝑝𝑡𝑥 +(𝑞𝑟𝑥+𝑞𝑡)
In the first grouping, the common factor is 𝑝𝑥 and in the second the common factor is q
𝑝𝑥⋅𝑟𝑥+𝑝𝑥⋅𝑡 +(𝑞⋅𝑟𝑥+𝑞⋅𝑡)
This gives 𝑝𝑥 𝑟𝑥+𝑡 +𝑞(𝑟𝑥+𝑡)
Finally, factor the common binomial to get (𝑝𝑥+𝑞)(𝑟𝑥+𝑡)

7. Quadratic Trinomials
The steps in this process of factoring 𝑎 𝑥 2 +𝑏𝑥+𝑐 are
Multiply the lead coefficient a by the constant c
Find the factors of the product 𝑎𝑐 that add to give 𝑏
Write the four terms (the two middle terms will have x) and group each pair
Factor the greatest common monomial from each group
Factor the common binomial

8. Quadratic Trinomials Example: factor 8 𝑥 2 +2𝑥−3
Always check for a common monomial first! (this one has no common monomial)
Multiply 8⋅3=24
The factors of 24 are 1⋅24, 2⋅12, 3⋅8, 6⋅4
We have −6⋅4=−24 and 6⋅−4=24; but, 6−4=2 so the factors are 6 and −4
Rewrite the expression as 8 𝑥 2 +6𝑥−4𝑥−3= 8 𝑥 2 +6𝑥 + −4𝑥−3 −4𝑥−3 =2𝑥 4𝑥+3 −(4𝑥+3)
Factor the common binomial: (4𝑥+3)(2𝑥−1)

9. Quadratic Trinomials Example: factor 6 𝑥 2 +11𝑥+5
Always check for a common monomial first! (this one has no common monomial)
Multiply 6⋅5=30
The factors of 30 are 1⋅30, 2⋅15, 3⋅10, 5⋅6
We have 5⋅6=30 and 5+6=11
Rewrite the expression as 6 𝑥 2 +5𝑥+6𝑥+5= 6 𝑥 2 +5𝑥 + 6𝑥+5 = 𝑥 6𝑥+5 +(6𝑥+5)
Factor the common binomial: (6𝑥+5)(𝑥+1)

10. Quadratic Trinomials Example: factor 12 𝑥 2 +33𝑥+18
Always check for a common monomial first!: 3⋅4 𝑥 2 +3⋅11𝑥+3⋅ 6=3(4 𝑥 2 +11𝑥+6); set aside the three for now
Multiply 4⋅6=24
The factors of 24 are 1⋅24, 2⋅12, 3⋅8, 4⋅6
We have 3⋅8=24 and 3+8=11
Rewrite the expression as 4 𝑥 2 +3𝑥+8𝑥+6= 4 𝑥 2 +3𝑥 + 8𝑥+6 = 𝑥 4𝑥+3 +2(4𝑥+3)
Factor the common binomial: (4𝑥+3)(𝑥+2)
The final result is 3(4𝑥+3)(𝑥+2)

11. Quadratic Trinomials Example: factor 14 𝑥 2 −13𝑥+3
Always check for a common monomial first! (no common monomial)
Multiply 14⋅3=42
The factors of 42 are 1⋅42, 2⋅21, 3⋅14, 6⋅7
We have −6⋅−7=42 and −6+ −7 =−13
Rewrite the expression as 14 𝑥 2 −6𝑥−7𝑥+3= 14 𝑥 2 −6𝑥 + −7𝑥+3 =2𝑥 7𝑥−3 −(7𝑥−3)
Factor the common binomial: (7𝑥−3)(2𝑥−1)

12. Quadratic Trinomials Example: factor 4 𝑥 2 −6𝑥+4
Always check for a common monomial first!: all are divisible by 2; 2(2 𝑥 2 −3𝑥+2); ignore the 2 for now
Multiply 2⋅2=42
The factors of 4 are 1⋅4, 2⋅2
Although −4+1=−3, −4⋅1=−4 but we need −4
When this happens, the quadratic trinomial is not factorable
But the expression is factorable: 2(2 𝑥 2 −3𝑥+2)

13. Quadratic Trinomials Example: factor 9 𝑏 2 −30𝑏+25
Always check for a common monomial first! (no common factor)
Multiply 9⋅25=225
The factors of 225 are 1⋅225, 3⋅75, 5⋅45, 15⋅15
We have that −15⋅−15=225 and −15+ −15 =−30
Now write 9 𝑏 2 −15𝑏−15𝑏+25= 9 𝑏 2 −15𝑏 +(−15𝑏+25); the common factors are 3𝑏 3𝑏−5 −5(3𝑏−5)
Factor the common binomial: 3𝑏−5 3𝑏−5 = 3𝑏−5 2
This is a perfect square trinomial.

14. Guided Practice Factor the quadratic trinomial, if possible.
3 𝑥 2 +2𝑥−4
4 𝑥 2 −25
3 𝑥 2 +4𝑥+2
4 𝑥 2 −20𝑥+25
8 𝑥 2 −10𝑥−3

15. Guided Practice Factor the quadratic trinomial, if possible.
3 𝑥 2 +2𝑥−4=(3𝑥−2)(𝑥+2)
4 𝑥 2 −25=(2𝑥+5)(2𝑥−5)
3 𝑥 2 +4𝑥+2; not factorable
4 𝑥 2 −20𝑥+25= 2𝑥−5 2
8 𝑥 2 −10𝑥−3=(4𝑥+1)(2𝑥−3)

16. Special Forms
Some polynomials have special forms that are easier to remember than to factor by any method
You have already seen one of these; two more are added here
Difference of squares: 𝑥 2 − 𝑎 2 = 𝑥+𝑎 𝑥−𝑎
Difference of cubes: 𝑥 3 − 𝑎 3 = 𝑥−𝑎 𝑥 2 +𝑎𝑥+ 𝑎 2
Sum of cubes: 𝑥 3 + 𝑎 3 =(𝑥+𝑎)( 𝑥 2 −𝑎𝑥+ 𝑎 2 )

17. Special Forms Examples:
Factor 4 𝑥 2 −81
Rewrite these as perfect squares: 2𝑥 2 − 9 2
Then use the pattern 𝑎 2 − 𝑏 2 = 𝑎+𝑏 𝑎−𝑏
Since 𝑎=2𝑥 and 𝑏=9, then 4 𝑥 2 −81=(2𝑥+9)(2𝑥−9)
Factor 36 𝑛 2 −25
Rewrite these as perfect squares: 6𝑛 2 − 5 2
Since 𝑎=6𝑛 and 𝑏=5, then 36 𝑛 2 −25=(6𝑛+5)(6𝑛−5)
Note that these can also be factored as quadratic polynomials

18. Special Forms Examples: Factor 8 𝑥 3 −27
Rewrite these as perfect cubes: 2𝑥 3 − 3 3
Then use the pattern 𝑎 3 − 𝑏 3 = 𝑎−𝑏 𝑎 2 +𝑎𝑏+ 𝑏 2
Since 𝑎=2𝑥 and 𝑏=3, then 8 𝑥 3 −27=(2𝑥−3)(4 𝑥 2 +6𝑥+9)
Factor 𝑛 2 −216
Rewrite these as perfect squares: 𝑛 3 − 6 2
Since 𝑎=6 and 𝑏=6, then 𝑛 3 −216=(𝑛−6)( 𝑛 2 +6𝑛+36)

19. Special Forms Examples: Factor 𝑥 3 +1
Rewrite these as perfect cubes: 𝑥
Then use the pattern 𝑎 3 + 𝑏 3 = 𝑎+𝑏 𝑎 2 −𝑎𝑏+ 𝑏 2
Since 𝑎=𝑥 and 𝑏=1, then 𝑥 3 +1=(𝑥+1)( 𝑥 2 −𝑥+1)
Factor 125𝑦 3 +64
Rewrite these as perfect squares: 5𝑦
Since 𝑎=5𝑦 and 𝑏=4, then 125 𝑦 3 +64=(5𝑦+4)(25 𝑦 2 −20𝑛+16)

20. Guided Practice Factor the given polynomials. 9 𝑛 2 −49 64 𝑝 3 −343
729 𝑥 3 +1

21. Practice 7
Handout