1. Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu
Chabot Mathematics
§5.4 Factor TriNomials
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer

2. 5.3 Review § Any QUESTIONS About Any QUESTIONS About HomeWork
MTH 55
Review §
Any QUESTIONS About
§5.3 → Factoring by GCF and/or Grouping
Any QUESTIONS About HomeWork
§5.3 → HW-13

3. Factor (+1)·x2 + bx + c
Recall the FOIL method of multiplying two binomials:
F O I L
(x + 2)(x + 5) = x2 + 5x + 2x + 10
= x x

4. Factor (+1)·x2 + bx + c
To factor x2 + 7x + 10, think of FOIL: The first term, x2, is the product of the First terms of two binomial factors, so the first term in each binomial must be x.
The challenge is to find two numbers p and q such that p•q = c, and p+q = b
x2 + 7x + 10 = (x + p)(x + q)
= x2 + qx + px + pq

5. Factor (+1)·x2 + bx + c Need to find two numbers p & q such that
x2 + 7x + 10 = (x + p)(x + q)
= x2 + qx + px + pq
Thus the numbers p and q must be selected so that their
PRODUCT is 10
SUM is 7
The Factor Pairs for 10 [and their sums]
1·10 [11]; (−1)·(−10) [−11]; 2·5 [7]; (−2)·(−5) [−7];

6. Factor (+1)·x2 + bx + c (x + 2)(x + 5) or (x + 5)(x + 2).
In this case, Examination of the “c” term factor-pairs revealed the desired numbers
p = 2 & q = 5
Thus the factorization for x2 + 7x + 10
(x + 2)(x + 5) or (x + 5)(x + 2).
The FACTORING Process changed TERMS (addition) into FACTORS (pure multiplication)
Note that the FACTORING Process converted an ADDITION-chain into a pure MULTIPLICATION-Chain

7. Example  FOIL Factoring
Multiplying binomials uses the FOIL method, Factoring uses the FOIL method backwards
Product of x and x is x2. F L
Product of 5 and – is –35.
Sum of the product of outer and inner terms O I

8. Factor x2 + bx + c for Positive c
When the constant term of a trinomial is positive, look for two numbers with the same sign. The sign is that of the middle term:
x2 – 7x + 10 = (x – 2)(x – 5);
x2 + 7x + 10 = (x + 2)(x + 5);

9. Example  Factor x2 + 7x + 12 SOLUTION: Think of FOIL in reverse:
We need a constant term that has a product of 12 and a SUM of 7.
We list some pairs of numbers that multiply to 12
Pairs of Factors of 12
Sums of Factors
1, 12 13
2, 6 8
3, 4 7
1, 12
13
2, 6 8
3, 4 7

10. Example  Factor x2 + 7x + 12
Since 3  4 = 12 and = 7, the factorization of x2 + 7x + 12 is (x + 3)(x + 4).
To check we simply multiply the two binomials.
CHECK by FOIL:
(x + 3)(x + 4) = x2 + 4x + 3x + 12
= x2 + 7x + 12 

11. Example  Factor y2 – 8y + 15
SOLUTION: Since the constant term is positive and the coefficient of the middle term is negative, we look for the factorization of 15 in which both factors are negative. Their SUM must be −8.
Pairs of Factors of 15
Sums of Factors
–1, –15
–16
–3, –5 –8
Sum of −8
y2 − 8y + 15 = (y − 3)(y – 5)

12. Factor x2 + bx + c for Negative c
When the constant term of a trinomial is negative, look for two numbers whose product is negative. One must be positive and the other negative:
x2 – 4x – 21 = (x + 3)(x – 7);
x2 + 4x – 21 = (x – 3)(x + 7).
Select the two numbers so that the number with the LARGER absolute value has the SAME SIGN as b, the coefficient of the middle term

13. Example  Factor x2 – 5x – 24 x2 − 5x − 24 = (x + 3)(x – 8)
SOLUTION: The constant term must be expressed as the product of negative & positive numbers.
Since the sum of the two numbers must be negative, the negative number must have the greater absolute value.
Pairs of Factors of 24
Sums of Factors
1, 24
23
2, 12
10
3, 8 5
4, 6 2
6, 4 2
8, 3 5
x2 − 5x − 24 = (x + 3)(x – 8)

14. Example  Factor t2 – 32 + 4t t2 + 4t − 32 = (t + 8)(t − 4) SOLUTION:
Rewrite the trinomial t2 + 4t − 32.
We need one positive and one negative factor. The sum must be 4, so the positive factor must have the larger absolute value
Pairs of Factors of 32
Sums of Factors
1, 32 31
2, 16 14
4, 8 4
t2 + 4t − 32 = (t + 8)(t − 4)

15. Example  Two Variables
Factor: a2 + ab − 30b2
SOLUTION
We need the factors of a2 & 30b2 that when added equal ab.
Those factors are a, and −5b & 6b.
a2 + ab − 30b2 = (a − 5b)(a + 6b)

16. Prime Polynomials
A polynomial that canNOT be factored is considered prime.
Example: x2 − x + 7
Often factoring requires two or more steps. Remember, when told to factor, we should factor completely. This means the final factorization should contain only prime polynomials.

17. Example  Factor 2x3−24x2+72x 2x(x2 − 12x + 36)
SOLUTION Always look first for a common factor. In this case factor out 2x:
2x(x2 − 12x + 36)
Since the constant term is positive and the coefficient of the middle term is negative, we look for the factorization of 36 in which both factors are negative.
Their SUM must be −12.

18. Example  Factor 2x3–24x2+72x 2x3 – 242x – 72x = 2x(x – 6 )(x – 6)
The factorization of (x2 – 12x + 36) is (x – 6)(x – 6) or (x – 6)2
Pairs of Factors of 36
Sums of Factors
1, 36
37
2, 18
20
3, 12
15
4, 9
13
6, 6
12
The factorization of x3 – 24x2 + 72x is 2x(x – 6)2 or 2x(x – 6)(x – 6)
2x3 – 242x – 72x = 2x(x – 6 )(x – 6)

19. To Factor (+1)·x2 + bx + c Distribute out Common Factors
Find a pair of factors that have c as their product and b as their sum.
If c is positive, its factors will have the same sign as b.
If c is negative, one factor will be positive and the other will be negative. Select the factors such that the factor with the larger absolute value has the same sign as b.
CHECK by MULTIPLYING

20. Factoring When: LeadCoeff ≠ 1
Factoring Trinomials of the Type ax2 + bx + c
Factoring with FOIL
The Grouping Method

21. Factor ax2+bx+c by FOIL

22. Example  Factor 3x2 – 14x – 5
First, check for a common factor, or GCF for all Terms. There is none other than 1 or −1.
Find the First terms whose product is 3x2. The only possibilities are 3x and x:
(3x + )(x + )
Find the Last terms whose product is −5. Possibilities are (−5)(1) & (5)(−1)
Important!: Since the First terms are not identical, we must also consider the above factors in reverse order: (1)(−5), & (−1)(5).

23. Example  Factor 3x2 – 14x – 5
Knowing that the First and Last products will check, inspect the Outer and Inner products resulting from steps (2) and (3) Look for the combination in which the sum of the products is the middle term.
(3x – 5)(x + 1) = 3x2 + 3x – 5x – 5
= 3x2 – 2x – 5
(3x – 1)(x + 5) = 3x2 + 15x – x – 5
= 3x2 + 14x – 5
Wrong middle term
Wrong middle term
close

24. Example  Factor 3x2 – 14x – 5 Keep Trying the factors of −5.
(3x + 5)(x – 1) = 3x2 – 3x + 5x – 5
= 3x2 + 2x – 5
(3x + 1)(x – 5) = 3x2 – 15x + x – 5
= 3x2 – 14x – 5
Wrong middle term
CORRECT middle term!
Thus
3x2 – 14x – 5 = (3x + 1)(x – 5)

25. LdCoeff ≠ 1 Factorization Notes
Reversing the signs in the binomials reverses the sign of the middle term
Organize your work so that you can keep track of which possibilities you have checked.
Remember to include the largest common factor - if there is one - in the final factorization.
ALWAYS CHECK by multiplying

26. Example  Factor 14x + 5 – 3x2 SOLUTION:
It is an important problem-solving strategy to find a way to make problems look like problems we already know how to solve. Rewrite the equation in descending order.
14x + 5 – 3x2 = – 3x2 + 14x + 5

27. Example  Factor 14x + 5 − 3x2 Starting with −3x2 + 14x + 5
Factor out the –1:
−3x2 + 14x + 5 = −1(3x2 − 14x − 5)
= −1(3x + 1)(x − 5)
The factorization of 14x + 5 − 3x2 is −1(3x + 1)(x − 5).
or (−3x − 1)(x − 5) or (3x + 1)(−x + 5)

28. Example  2Vars: 6x2 − xy − 12y2
SOLUTION: No common factors exist, we examine the first term, 6x2. There are two possibilities:
(2x + )(3x + ) or (6x + )(x + ).
The last term −12y2, has pairs of factors:
12y, −y 6y, −2y 4y, −3y
and −12y, y −6y, 2y −4y, 3y
as well as each pairing reversed.

29. Example  2Vars: 6x2 − xy − 12y2 SOLUTION:
Some trials such as (2x – 6y)(3x + 2y) and (6x + 4y)(x – 3y), cannot be correct because (2x – 6y) and (6x + 4y) contain a common factor, 2.
Trial
(2x + 3y)(3x − 4y)
Product
6x2 − 8xy + 9xy − 12y2
= 6x2 + xy − 12y2

30. Example  2Vars: 6x2 − xy − 12y2
SOLUTION: the Trial (2x + 3y)(3x − 4y) incorrect, but only because of the sign of the middle term.
To correctly factor, simply change the signs in the binomials.
Trial
(2x − 3y)(3x + 4y)
Product
6x2 + 8xy − 9xy − 12y2
= 6x2 − xy − 12y2
The factorization: (2x − 3y)(3x + 4y)

31. Example  18m2 – 19mn – 12n2 SOLUTION There are no common factors.
Factor the first term, 18m2 and get the following possibilities: 18mm, 9m2m, and 6m3m.
Factor the last term, −12n2, which is negative. The possibilities are: (−12n)(n), (−n)(12n), (−2n)(6n), (6n)(−2n), (−4n)(3n) or (−3n)(4n)

32. Example  18m2 – 19mn – 12n2
Look for combinations of factors such that the sum of the outside and the inside products is the middle term, (−19mn).
(9m + n)(2m − 12n) = 18m2 − 106mn − 12n2
(9m − 12n)(2m + n) = 18m2 − 15mn − 12n2
(9m − 3n)(2m + 4n) = 18m2 + 30mn − 12n2
(9m + 4n)(2m – 3n) = 18m2 − 19mn − 12n2
Thus ANS → (9m + 4n)(2m − 3n)
correct middle term

33. Factoring by Substitution
Some times substituting a single varaible for a (complicated) expression reveals an easily factored PolyNomial
Example  factor p2q2 + 7pq + 6
SOLUTION
Rewrite using Product-to-Power Exponent rule → (pq)2 + 7(pq) + 6
Now engage a substitution

34. Factoring by Substitution
factor p2q2 + 7pq + 6 = (pq)2 + 7(pq) + 6
Now to engage a substitution LET
u = pq
Replace in the Expression pq with u
u2 + 7u + 6
The Expression in u is easily FOIL-factored
u2 + 7u + 6 = (u + 6)(u + 1)

35. Factoring by Substitution
Now BACK Substitute u = pq
(u)2 + 7 (u) + 6 = (u + 6)(u + 1) 
(pq)2 + 7(pq) + 6 = (pq + 6)(pq + 1)
p2q2 + 7pq + 6 = (pq + 6)(pq + 1)

36. WhiteBoard Work Problems From §5.4 Exercise Set Shaded Area Equals
30, 44, 66, 82, 92
Shaded Area Equals

37. All Done for Today
F.O.I.L. Factoring

38. Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu
Chabot Mathematics
Appendix
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer –

39. Factoring ax2+bx+c by Grouping
Factor out the largest common factor, if one exists.
Multiply the leading coefficient a and the constant c; i.e., form the a•c product
Find a pair of factors of a•c whose sum is b.
Rewrite the middle term, bx, as a sum or difference using the factors found in step (3).
Factor by grouping.
Include any common factor from step (1) and check by multiplying.

40. Example  Factor 4x2 – 5x – 6 SOLUTION
First, we note that there is no common factor (other than 1 or −1).
We multiply the leading coefficient, 4 and the constant, −6: (4)(−6) = −24.
We next look for the factorization of −24 in which the sum of the factors is the coefficient of the middle term, −5.

41. Example  Factor 4x2 – 5x – 6 Pairs of Factors of -24 Sums of Factors
1, –24
–23
–1, 24 23
2, –12
–10
–2, 12 10
3, –8 –5
–3, 8 5
4, –6 –2
–4, 6 2
We would normally stop listing pairs of factors once we have found the one we need

42. Example  Factor 4x2 – 5x – 6
Next, we express the middle term as a sum or difference using the factors found in step (3):
−5x = −8x + 3x.
5. We now factor by grouping as follows:
4x2 − 5x − 6 = [4x2 − 8x] + [3x − 6]
= 4x(x − 2) + 3(x − 2)
= (x − 2)(4x + 3)

43. Example  Factor 4x2 – 5x – 6 CHECK by FOIL:
(x − 2)(4x + 3) = 4x2 + 3x − 8x − 6
= 4x2 − 5x − 6 
The factorization of 4x2 − 5x − 6 is
(x − 2)(4x + 3).

44. Example: Factor 8x3 + 10x2 – 12x SOLUTION
Factor out the Greatest Common Factor (GCF), 2x:
8x3 + 10x2 − 12x = 2x(4x2 + 5x − 6)
2. To factor 4x2 + 5x − 6 by grouping, we multiply the leading coefficient, 4 and the constant term (−6): 4(−6) = −24.

45. Example: Factor 8x3 + 10x2 – 12x
next look for pairs of factors of −24 whose sum is 5.
Pairs of Factors of −24
Sums of Factors
3, −8 −5
−3, 8 5
We then rewrite the 5x in 4x2 + 5x − 6 using:
5x = −3x + 8x

46. Example: Factor 8x3 + 10x2 – 12x Next, factor by grouping:
4x2 + 5x − 6 = [4x2 − 3x] + [8x − 6]
= x(4x − 3) + 2(4x − 3)
= (x + 2)(4x − 3)
The factorization of the original trinomial 8x3 + 10x2 − 12x is
2x(x + 2)(4x − 3)

47. Graph y = |x|
Make T-table