3. We evaluated expressions with numerical inputs.
Factoring Formulas and Substitution
Evaluating A2 – B2 with A = 2, B = 1 as inputs, the output is 3.
4. We evaluated expressions with numerical inputs.
Factoring Formulas and Substitution
Evaluating A2 – B2 with A = 2, B = 1 as inputs, the output is 3.
We extend this procedure to using expressions as inputs and
say that we “substitute the variables with the expressions...".
5. We evaluated expressions with numerical inputs.
Factoring Formulas and Substitution
Example A. Substitute A2 – B2 with A = x + 2 and B = 2x.
Expand and simplify the result.
Evaluating A2 – B2 with A = 2, B = 1 as inputs, the output is 3.
We extend this procedure to using expressions as inputs and
say that we “substitute the variables with the expressions...".
6. We evaluated expressions with numerical inputs.
Factoring Formulas and Substitution
Example A. Substitute A2 – B2 with A = x + 2 and B = 2x.
Expand and simplify the result.
Replace A by (x + 2), and B by (2x)
Evaluating A2 – B2 with A = 2, B = 1 as inputs, the output is 3.
We extend this procedure to using expressions as inputs and
say that we “substitute the variables with the expressions...".
7. We evaluated expressions with numerical inputs.
Factoring Formulas and Substitution
Example A. Substitute A2 – B2 with A = x + 2 and B = 2x.
Expand and simplify the result.
Replace A by (x + 2), and B by (2x)
We also say that we
“plug in” (x + 2) for A
Evaluating A2 – B2 with A = 2, B = 1 as inputs, the output is 3.
We extend this procedure to using expressions as inputs and
say that we “substitute the variables with the expressions...".
8. We evaluated expressions with numerical inputs.
Factoring Formulas and Substitution
Example A. Substitute A2 – B2 with A = x + 2 and B = 2x.
Expand and simplify the result.
Replace A by (x + 2), and B by (2x)
make sure to encase the
inputs in the ( )’s
We also say that we
“plug in” (x + 2) for A
Evaluating A2 – B2 with A = 2, B = 1 as inputs, the output is 3.
We extend this procedure to using expressions as inputs and
say that we “substitute the variables with the expressions...".
9. We evaluated expressions with numerical inputs.
Factoring Formulas and Substitution
Example A. Substitute A2 – B2 with A = x + 2 and B = 2x.
Expand and simplify the result.
Replace A by (x + 2), and B by (2x)
we have (x + 2)2 – (2x)2
make sure to encase the
inputs in the ( )’s
We also say that we
“plug in” (x + 2) for A
Evaluating A2 – B2 with A = 2, B = 1 as inputs, the output is 3.
We extend this procedure to using expressions as inputs and
say that we “substitute the variables with the expressions...".
10. We evaluated expressions with numerical inputs.
Factoring Formulas and Substitution
Example A. Substitute A2 – B2 with A = x + 2 and B = 2x.
Expand and simplify the result.
Replace A by (x + 2), and B by (2x)
we have (x + 2)2 – (2x)2
= x2 + 4x + 4 – 4x2
= –3x2 + 4x + 4
make sure to encase the
inputs in the ( )’s
We also say that we
“plug in” (x + 2) for A
Evaluating A2 – B2 with A = 2, B = 1 as inputs, the output is 3.
We extend this procedure to using expressions as inputs and
say that we “substitute the variables with the expressions...".
11. We evaluated expressions with numerical inputs.
Factoring Formulas and Substitution
Example A. Substitute A2 – B2 with A = x + 2 and B = 2x.
Expand and simplify the result.
Replace A by (x + 2), and B by (2x)
we have (x + 2)2 – (2x)2
= x2 + 4x + 4 – 4x2
= –3x2 + 4x + 4
make sure to encase the
inputs in the ( )’s
Reverse the procedure.
We also say that we
“plug in” (x + 2) for A
Evaluating A2 – B2 with A = 2, B = 1 as inputs, the output is 3.
We extend this procedure to using expressions as inputs and
say that we “substitute the variables with the expressions...".
12. We evaluated expressions with numerical inputs.
Factoring Formulas and Substitution
Example A. Substitute A2 – B2 with A = x + 2 and B = 2x.
Expand and simplify the result.
Replace A by (x + 2), and B by (2x)
we have (x + 2)2 – (2x)2
= x2 + 4x + 4 – 4x2
= –3x2 + 4x + 4
make sure to encase the
inputs in the ( )’s
Reverse the procedure. Sometime the inputs can be easily
identified when the formula and the outcome are given.
We also say that we
“plug in” (x + 2) for A
Evaluating A2 – B2 with A = 2, B = 1 as inputs, the output is 3.
We extend this procedure to using expressions as inputs and
say that we “substitute the variables with the expressions...".
13. We evaluated expressions with numerical inputs.
Factoring Formulas and Substitution
Example A. Substitute A2 – B2 with A = x + 2 and B = 2x.
Expand and simplify the result.
Replace A by (x + 2), and B by (2x)
we have (x + 2)2 – (2x)2
= x2 + 4x + 4 – 4x2
= –3x2 + 4x + 4
Example B. Evaluating A2 – B2 and we obtain
a. 4x2 – 9y2, what are the A and B?
make sure to encase the
inputs in the ( )’s
Reverse the procedure. Sometime the inputs can be easily
identified when the formula and the outcome are given.
We also say that we
“plug in” (x + 2) for A
Evaluating A2 – B2 with A = 2, B = 1 as inputs, the output is 3.
We extend this procedure to using expressions as inputs and
say that we “substitute the variables with the expressions...".
14. We evaluated expressions with numerical inputs.
Factoring Formulas and Substitution
Example A. Substitute A2 – B2 with A = x + 2 and B = 2x.
Expand and simplify the result.
Replace A by (x + 2), and B by (2x)
we have (x + 2)2 – (2x)2
= x2 + 4x + 4 – 4x2
= –3x2 + 4x + 4
Example B. Evaluating A2 – B2 and we obtain
a. 4x2 – 9y2, what are the A and B?
Matching the A2 = 4x2, we’re asking ( ? )2 = 4x2, so A = (2x).
make sure to encase the
inputs in the ( )’s
Reverse the procedure. Sometime the inputs can be easily
identified when the formula and the outcome are given.
We also say that we
“plug in” (x + 2) for A
Evaluating A2 – B2 with A = 2, B = 1 as inputs, the output is 3.
We extend this procedure to using expressions as inputs and
say that we “substitute the variables with the expressions...".
15. We evaluated expressions with numerical inputs.
Factoring Formulas and Substitution
Example A. Substitute A2 – B2 with A = x + 2 and B = 2x.
Expand and simplify the result.
Replace A by (x + 2), and B by (2x)
we have (x + 2)2 – (2x)2
= x2 + 4x + 4 – 4x2
= –3x2 + 4x + 4
Example B. Evaluating A2 – B2 and we obtain
a. 4x2 – 9y2, what are the A and B?
Matching the A2 = 4x2, we’re asking ( ? )2 = 4x2, so A = (2x).
Similarly we conclude that B = (3y).
make sure to encase the
inputs in the ( )’s
Reverse the procedure. Sometime the inputs can be easily
identified when the formula and the outcome are given.
We also say that we
“plug in” (x + 2) for A
Evaluating A2 – B2 with A = 2, B = 1 as inputs, the output is 3.
We extend this procedure to using expressions as inputs and
say that we “substitute the variables with the expressions...".
16. c. x2 + 2xy + y2 – 9, what are the A and B?
Factoring Formulas and Substitution
b. 25x2y4 – 1, what are the A and B?
17. c. x2 + 2xy + y2 – 9, what are the A and B?
Factoring Formulas and Substitution
b. 25x2y4 – 1, what are the A and B?
Matching the A2 = 25x2y4 and B2 = 1, we have that
A = 5xy2 and B = 1.
18. c. x2 + 2xy + y2 – 9, what are the A and B?
Factoring Formulas and Substitution
b. 25x2y4 – 1, what are the A and B?
Matching the A2 = 25x2y4 and B2 = 1, we have that
A = 5xy2 and B = 1.
Factor x2 + 2xy + y2 = (x + y)(x + y) = (x + y)2,
19. c. x2 + 2xy + y2 – 9, what are the A and B?
Factoring Formulas and Substitution
b. 25x2y4 – 1, what are the A and B?
Matching the A2 = 25x2y4 and B2 = 1, we have that
A = 5xy2 and B = 1.
so x2 + 2xy + y2 – 9 = (x + y)2 – 32.
Factor x2 + 2xy + y2 = (x + y)(x + y) = (x + y)2,
20. c. x2 + 2xy + y2 – 9, what are the A and B?
Factoring Formulas and Substitution
b. 25x2y4 – 1, what are the A and B?
Matching the A2 = 25x2y4 and B2 = 1, we have that
A = 5xy2 and B = 1.
so x2 + 2xy + y2 – 9 = (x + y)2 – 32.
Hence A = (x + y) and B = 3.
Factor x2 + 2xy + y2 = (x + y)(x + y) = (x + y)2,
21. c. x2 + 2xy + y2 – 9, what are the A and B?
Factoring Formulas and Substitution
b. 25x2y4 – 1, what are the A and B?
Matching the A2 = 25x2y4 and B2 = 1, we have that
A = 5xy2 and B = 1.
so x2 + 2xy + y2 – 9 = (x + y)2 – 32.
Hence A = (x + y) and B = 3.
a. we obtain x3 – 125, what are the A and B?
b. we obtain 8y3 – (2 – y)3 , what are the A and B?
Factor x2 + 2xy + y2 = (x + y)(x + y) = (x + y)2,
Example C. Evaluating A3 – B3 with inputs A and B
22. c. x2 + 2xy + y2 – 9, what are the A and B?
Factoring Formulas and Substitution
b. 25x2y4 – 1, what are the A and B?
Matching the A2 = 25x2y4 and B2 = 1, we have that
A = 5xy2 and B = 1.
so x2 + 2xy + y2 – 9 = (x + y)2 – 32.
Hence A = (x + y) and B = 3.
a. we obtain x3 – 125, what are the A and B?
Matching A3 = x3, we have A = x.
Similarly we’ve B3 = 125 so B = 5.
b. we obtain 8y3 – (2 – y)3 , what are the A and B?
Factor x2 + 2xy + y2 = (x + y)(x + y) = (x + y)2,
Example C. Evaluating A3 – B3 with inputs A and B
23. c. x2 + 2xy + y2 – 9, what are the A and B?
Factoring Formulas and Substitution
b. 25x2y4 – 1, what are the A and B?
Matching the A2 = 25x2y4 and B2 = 1, we have that
A = 5xy2 and B = 1.
so x2 + 2xy + y2 – 9 = (x + y)2 – 32.
Hence A = (x + y) and B = 3.
a. we obtain x3 – 125, what are the A and B?
Matching A3 = x3, we have A = x.
Similarly we’ve B3 = 125 so B = 5.
b. we obtain 8y3 – (2 – y)3 , what are the A and B?
Matching the A3 = 8y3, we have A = 2y.
By inspection, we have B = (2 – y).
Factor x2 + 2xy + y2 = (x + y)(x + y) = (x + y)2,
Example C. Evaluating A3 – B3 with inputs A and B
24. c. x2 + 2xy + y2 – 9, what are the A and B?
Factoring Formulas and Substitution
b. 25x2y4 – 1, what are the A and B?
Matching the A2 = 25x2y4 and B2 = 1, we have that
A = 5xy2 and B = 1.
so x2 + 2xy + y2 – 9 = (x + y)2 – 32.
Hence A = (x + y) and B = 3.
a. we obtain x3 – 125, what are the A and B?
Matching A3 = x3, we have A = x.
Similarly we’ve B3 = 125 so B = 5.
b. we obtain 8y3 – (2 – y)3 , what are the A and B?
Matching the A3 = 8y3, we have A = 2y.
By inspection, we have B = (2 – y).
Recall the conjugate product formula
(A – B)(A + B) = A2 – B2.
Factor x2 + 2xy + y2 = (x + y)(x + y) = (x + y)2,
Example C. Evaluating A3 – B3 with inputs A and B
25. The reverse of the conjugate product formula is the
factoring formula for difference of squares:
A2 – B2 = (A – B) (A + B)
Factoring Formulas and Substitution
26. The reverse of the conjugate product formula is the
factoring formula for difference of squares:
A2 – B2 = (A – B) (A + B)
As suggested by the name, we use this formula when
the expression is the subtraction of two perfect-square terms.
Factoring Formulas and Substitution
27. Example D.
a. x2 – 25
The reverse of the conjugate product formula is the
factoring formula for difference of squares:
A2 – B2 = (A – B) (A + B)
As suggested by the name, we use this formula when
the expression is the subtraction of two perfect-square terms.
Factoring Formulas and Substitution
28. Example D.
a. x2 – 25
= (x)2 – (5)2
The reverse of the conjugate product formula is the
factoring formula for difference of squares:
A2 – B2 = (A – B) (A + B)
As suggested by the name, we use this formula when
the expression is the subtraction of two perfect-square terms.
Factoring Formulas and Substitution
29. Example D.
a. x2 – 25
= (x)2 – (5)2
= (x + 5)(x – 5)
The reverse of the conjugate product formula is the
factoring formula for difference of squares:
A2 – B2 = (A – B) (A + B)
As suggested by the name, we use this formula when
the expression is the subtraction of two perfect-square terms.
Factoring Formulas and Substitution
30. Example D.
a. x2 – 25
= (x)2 – (5)2
= (x + 5)(x – 5)
b. 4x2 – 9y2
The reverse of the conjugate product formula is the
factoring formula for difference of squares:
A2 – B2 = (A – B) (A + B)
As suggested by the name, we use this formula when
the expression is the subtraction of two perfect-square terms.
Factoring Formulas and Substitution
31. Example D.
a. x2 – 25
= (x)2 – (5)2
= (x + 5)(x – 5)
b. 4x2 – 9y2
= (2x)2 – (3y)2
The reverse of the conjugate product formula is the
factoring formula for difference of squares:
A2 – B2 = (A – B) (A + B)
As suggested by the name, we use this formula when
the expression is the subtraction of two perfect-square terms.
Factoring Formulas and Substitution
32. Example D.
a. x2 – 25
= (x)2 – (5)2
= (x + 5)(x – 5)
b. 4x2 – 9y2
= (2x)2 – (3y)2
= (2x – 3y)(2x + 3y)
The reverse of the conjugate product formula is the
factoring formula for difference of squares:
A2 – B2 = (A – B) (A + B)
As suggested by the name, we use this formula when
the expression is the subtraction of two perfect-square terms.
Factoring Formulas and Substitution
33. Example D.
a. x2 – 25
= (x)2 – (5)2
= (x + 5)(x – 5)
b. 4x2 – 9y2
= (2x)2 – (3y)2
= (2x – 3y)(2x + 3y)
c. 25A2B4 – 1
The reverse of the conjugate product formula is the
factoring formula for difference of squares:
A2 – B2 = (A – B) (A + B)
As suggested by the name, we use this formula when
the expression is the subtraction of two perfect-square terms.
Factoring Formulas and Substitution
34. Example D.
a. x2 – 25
= (x)2 – (5)2
= (x + 5)(x – 5)
b. 4x2 – 9y2
= (2x)2 – (3y)2
= (2x – 3y)(2x + 3y)
c. 25A2B4 – 1
= (5AB2)2 – (1)2
The reverse of the conjugate product formula is the
factoring formula for difference of squares:
A2 – B2 = (A – B) (A + B)
As suggested by the name, we use this formula when
the expression is the subtraction of two perfect-square terms.
Factoring Formulas and Substitution
35. Example D.
a. x2 – 25
= (x)2 – (5)2
= (x + 5)(x – 5)
b. 4x2 – 9y2
= (2x)2 – (3y)2
= (2x – 3y)(2x + 3y)
c. 25A2B4 – 1
= (5AB2)2 – (1)2
= (5AB2 – 1)(5AB2 + 1)
The reverse of the conjugate product formula is the
factoring formula for difference of squares:
A2 – B2 = (A – B) (A + B)
As suggested by the name, we use this formula when
the expression is the subtraction of two perfect-square terms.
Factoring Formulas and Substitution
36. d. (A + 2B)2 – 16
Factoring Formulas and Substitution
37. d. (A + 2B)2 – 16
= (A + 2B)2 – (4)2
Factoring Formulas and Substitution
38. d. (A + 2B)2 – 16
= (A + 2B)2 – (4)2
= ((A + 2B) – 4)((A + 2B) + 4)
Factoring Formulas and Substitution
39. d. (A + 2B)2 – 16
= (A + 2B)2 – (4)2
= ((A + 2B) – 4)((A + 2B) + 4)
= (A + 2B – 4)(A + 2B + 4)
Factoring Formulas and Substitution
40. d. (A + 2B)2 – 16
= (A + 2B)2 – (4)2
= ((A + 2B) – 4)((A + 2B) + 4)
= (A + 2B – 4)(A + 2B + 4)
Factoring Formulas and Substitution
41. d. (A + 2B)2 – 16
= (A + 2B)2 – (4)2
= ((A + 2B) – 4)((A + 2B) + 4)
= (A + 2B – 4)(A + 2B + 4)
Factoring Formulas and Substitution
Caution: The expressions of two squares of the same sign:
(A2 + B2) or (– A2 – B2) are prime.
42. d. (A + 2B)2 – 16
= (A + 2B)2 – (4)2
= ((A + 2B) – 4)((A + 2B) + 4)
= (A + 2B – 4)(A + 2B + 4)
Factoring Formulas and Substitution
Caution: The expressions of two squares of the same sign:
(A2 + B2) or (– A2 – B2) are prime.
For example, –9 – 4x2 is not factorable.
43. d. (A + 2B)2 – 16
= (A + 2B)2 – (4)2
= ((A + 2B) – 4)((A + 2B) + 4)
= (A + 2B – 4)(A + 2B + 4)
Sum and Difference of Cubes
Factoring Formulas and Substitution
Caution: The expressions of two squares of the same sign:
(A2 + B2) or (– A2 – B2) are prime.
For example, –9 – 4x2 is not factorable.
44. d. (A + 2B)2 – 16
= (A + 2B)2 – (4)2
= ((A + 2B) – 4)((A + 2B) + 4)
= (A + 2B – 4)(A + 2B + 4)
Sum and Difference of Cubes
Both the difference and the sum of cubes may be factored.
Factoring Formulas and Substitution
Caution: The expressions of two squares of the same sign:
(A2 + B2) or (– A2 – B2) are prime.
For example, –9 – 4x2 is not factorable.
45. d. (A + 2B)2 – 16
= (A + 2B)2 – (4)2
= ((A + 2B) – 4)((A + 2B) + 4)
= (A + 2B – 4)(A + 2B + 4)
Sum and Difference of Cubes
Both the difference and the sum of cubes may be factored.
If we try to factor A3 – B3, we might guess that
A3 – B3 = (A – B)(A2 + B2)
Factoring Formulas and Substitution
Caution: The expressions of two squares of the same sign:
(A2 + B2) or (– A2 – B2) are prime.
For example, –9 – 4x2 is not factorable.
46. d. (A + 2B)2 – 16
= (A + 2B)2 – (4)2
= ((A + 2B) – 4)((A + 2B) + 4)
= (A + 2B – 4)(A + 2B + 4)
Sum and Difference of Cubes
Both the difference and the sum of cubes may be factored.
If we try to factor A3 – B3, we might guess that
A3 – B3 = (A – B)(A2 + B2)
But this does not work because we have the inner
and outer products –BA2 and AB2 remaining in the expansion
as shown here.
Factoring Formulas and Substitution
Caution: The expressions of two squares of the same sign:
(A2 + B2) or (– A2 – B2) are prime.
For example, –9 – 4x2 is not factorable.
47. d. (A + 2B)2 – 16
= (A + 2B)2 – (4)2
= ((A + 2B) – 4)((A + 2B) + 4)
= (A + 2B – 4)(A + 2B + 4)
Sum and Difference of Cubes
Both the difference and the sum of cubes may be factored.
If we try to factor A3 – B3, we might guess that
A3 – B3 = (A – B)(A2 + B2) = A3 –BA2 + AB2 – B3
But this does not work because we have the inner
and outer products –BA2 and AB2 remaining in the expansion
as shown here.
Factoring Formulas and Substitution
Caution: The expressions of two squares of the same sign:
(A2 + B2) or (– A2 – B2) are prime.
For example, –9 – 4x2 is not factorable.
48. But if we put in the adjustment +AB, then all the cross-
products with A and B are eliminated.
(A – B)(A2 + AB + B2)
Factoring Formulas and Substitution
49. But if we put in the adjustment +AB, then all the cross-
products with A and B are eliminated. That is
(A – B)(A2 + AB + B2)
= A3 + A2B + AB2 – BA2 – B2A – B3
= A3 – B3
Factoring Formulas and Substitution
50. But if we put in the adjustment +AB, then all the cross-
products with A and B are eliminated. That is
(A – B)(A2 + AB + B2)
= A3 + A2B + AB2 – BA2 – B2A – B3
= A3 – B3
Hence A3 – B3 = (A – B)(A2 + AB + B2)
Factoring Formulas and Substitution
51. But if we put in the adjustment +AB, then all the cross-
products with A and B are eliminated. That is
(A – B)(A2 + AB + B2)
= A3 + A2B + AB2 – BA2 – B2A – B3
= A3 – B3
Hence A3 – B3 = (A – B)(A2 + AB + B2)
Through a similar argument we have that
A3 + B3 = (A + B)(A2 – AB + B2)
Factoring Formulas and Substitution
52. But if we put in the adjustment +AB, then all the cross-
products with A and B are eliminated. That is
(A – B)(A2 + AB + B2)
= A3 + A2B + AB2 – BA2 – B2A – B3
= A3 – B3
Hence A3 – B3 = (A – B)(A2 + AB + B2)
Through a similar argument we have that
A3 + B3 = (A + B)(A2 – AB + B2)
We write these two formulas together as
A3 B3 = (A B)(A2 AB + B2)+– +–+–
Factoring Formulas and Substitution
53. But if we put in the adjustment +AB, then all the cross-
products with A and B are eliminated. That is
(A – B)(A2 + AB + B2)
= A3 + A2B + AB2 – BA2 – B2A – B3
= A3 – B3
Hence A3 – B3 = (A – B)(A2 + AB + B2)
Through a similar argument we have that
A3 + B3 = (A + B)(A2 – AB + B2)
We write these two formulas together as
A3 B3 = (A B)(A2 AB + B2)+– +–+–
Note that the signs are reversed for the adjustments in
these formulas.
Factoring Formulas and Substitution
55. Example E.
a. 8x3 – 27 (Note that both 8 and 27 are cubes)
Factoring Formulas and Substitution
56. Example E.
a. 8x3 – 27 (Note that both 8 and 27 are cubes)
= (2x)3 – (3)3
Factoring Formulas and Substitution
57. Example E.
a. 8x3 – 27 (Note that both 8 and 27 are cubes)
= (2x)3 – (3)3
= (2x – 3)((2x)2 + (2x)(3) +(3)2)
Factoring Formulas and Substitution
58. Example E.
a. 8x3 – 27 (Note that both 8 and 27 are cubes)
= (2x)3 – (3)3
= (2x – 3)((2x)2 + (2x)(3) +(3)2)
= (2x – 3)(4x2 + 6x + 9)
Factoring Formulas and Substitution
59. Example E.
a. 8x3 – 27 (Note that both 8 and 27 are cubes)
= (2x)3 – (3)3
= (2x – 3)((2x)2 + (2x)(3) +(3)2)
= (2x – 3)(4x2 + 6x + 9)
b. 64A3 + 125
Factoring Formulas and Substitution
60. Example E.
a. 8x3 – 27 (Note that both 8 and 27 are cubes)
= (2x)3 – (3)3
= (2x – 3)((2x)2 + (2x)(3) +(3)2)
= (2x – 3)(4x2 + 6x + 9)
b. 64A3 + 125 (both 64 and 125 are cubes)
Factoring Formulas and Substitution
61. Example E.
a. 8x3 – 27 (Note that both 8 and 27 are cubes)
= (2x)3 – (3)3
= (2x – 3)((2x)2 + (2x)(3) +(3)2)
= (2x – 3)(4x2 + 6x + 9)
b. 64A3 + 125 (both 64 and 125 are cubes)
= (4A)3 + (5)3
Factoring Formulas and Substitution
62. Example E.
a. 8x3 – 27 (Note that both 8 and 27 are cubes)
= (2x)3 – (3)3
= (2x – 3)((2x)2 + (2x)(3) +(3)2)
= (2x – 3)(4x2 + 6x + 9)
b. 64A3 + 125 (both 64 and 125 are cubes)
= (4A)3 + (5)3
= (4A + 5)((4A)2 – (4A)(5) +(5)2)
Factoring Formulas and Substitution
63. Example E.
a. 8x3 – 27 (Note that both 8 and 27 are cubes)
= (2x)3 – (3)3
= (2x – 3)((2x)2 + (2x)(3) +(3)2)
= (2x – 3)(4x2 + 6x + 9)
b. 64A3 + 125 (both 64 and 125 are cubes)
= (4A)3 + (5)3
= (4A + 5)((4A)2 – (4A)(5) +(5)2)
= (4A + 5)(16A2 – 20A + 25)
Factoring Formulas and Substitution
64. Example E.
a. 8x3 – 27 (Note that both 8 and 27 are cubes)
= (2x)3 – (3)3
= (2x – 3)((2x)2 + (2x)(3) +(3)2)
= (2x – 3)(4x2 + 6x + 9)
b. 64A3 + 125 (both 64 and 125 are cubes)
= (4A)3 + (5)3
= (4A + 5)((4A)2 – (4A)(5) +(5)2)
= (4A + 5)(16A2 – 20A + 25)
Summary on Factoring
Factoring Formulas and Substitution
65. Example E.
a. 8x3 – 27 (Note that both 8 and 27 are cubes)
= (2x)3 – (3)3
= (2x – 3)((2x)2 + (2x)(3) +(3)2)
= (2x – 3)(4x2 + 6x + 9)
b. 64A3 + 125 (both 64 and 125 are cubes)
= (4A)3 + (5)3
= (4A + 5)((4A)2 – (4A)(5) +(5)2)
= (4A + 5)(16A2 – 20A + 25)
Summary on Factoring
I. Always factor out the GCF first.
Factoring Formulas and Substitution
66. Example E.
a. 8x3 – 27 (Note that both 8 and 27 are cubes)
= (2x)3 – (3)3
= (2x – 3)((2x)2 + (2x)(3) +(3)2)
= (2x – 3)(4x2 + 6x + 9)
b. 64A3 + 125 (both 64 and 125 are cubes)
= (4A)3 + (5)3
= (4A + 5)((4A)2 – (4A)(5) +(5)2)
= (4A + 5)(16A2 – 20A + 25)
Summary on Factoring
I. Always factor out the GCF first.
II. Make sure the leading term is positive and all terms are
arranged in order.
Factoring Formulas and Substitution
67. Example E.
a. 8x3 – 27 (Note that both 8 and 27 are cubes)
= (2x)3 – (3)3
= (2x – 3)((2x)2 + (2x)(3) +(3)2)
= (2x – 3)(4x2 + 6x + 9)
b. 64A3 + 125 (both 64 and 125 are cubes)
= (4A)3 + (5)3
= (4A + 5)((4A)2 – (4A)(5) +(5)2)
= (4A + 5)(16A2 – 20A + 25)
Summary on Factoring
I. Always factor out the GCF first.
II. Make sure the leading term is positive and all terms are
arranged in order.
III. If left with a trinomial, use reverse-FOIL or ac-method.
Factoring Formulas and Substitution
68. Example E.
a. 8x3 – 27 (Note that both 8 and 27 are cubes)
= (2x)3 – (3)3
= (2x – 3)((2x)2 + (2x)(3) +(3)2)
= (2x – 3)(4x2 + 6x + 9)
b. 64A3 + 125 (both 64 and 125 are cubes)
= (4A)3 + (5)3
= (4A + 5)((4A)2 – (4A)(5) +(5)2)
= (4A + 5)(16A2 – 20A + 25)
Summary on Factoring
I. Always factor out the GCF first.
II. Make sure the leading term is positive and all terms are
arranged in order.
III. If left with a trinomial, use reverse-FOIL or ac-method.
IV. If it is a two-term expression, try factoring by formula.
Factoring Formulas and Substitution
69. Example E.
a. 8x3 – 27 (Note that both 8 and 27 are cubes)
= (2x)3 – (3)3
= (2x – 3)((2x)2 + (2x)(3) +(3)2)
= (2x – 3)(4x2 + 6x + 9)
b. 64A3 + 125 (both 64 and 125 are cubes)
= (4A)3 + (5)3
= (4A + 5)((4A)2 – (4A)(5) +(5)2)
= (4A + 5)(16A2 – 20A + 25)
Summary on Factoring
I. Always factor out the GCF first.
II. Make sure the leading term is positive and all terms are
arranged in order.
III. If left with a trinomial, use reverse-FOIL or ac-method.
IV. If it is a two-term expression, try factoring by formula.
V. If it is a four term expression, try the grouping-method.
Factoring Formulas and Substitution
71. Example F. Factor completely.
A. 20A3 – 45A Take out the common factor
= 5A(4A2 – 9)
Summary on Factoring
72. Example F. Factor completely.
A. 20A3 – 45A Take out the common factor
= 5A(4A2 – 9) Difference of square
= 5A(2A – 3)(2A + 3)
Summary on Factoring
73. Example F. Factor completely.
A. 20A3 – 45A Take out the common factor
= 5A(4A2 – 9) Difference of square
= 5A(2A – 3)(2A + 3)
B. –20A3 + 45A2 – 10A
Summary on Factoring
74. Example F. Factor completely.
A. 20A3 – 45A Take out the common factor
= 5A(4A2 – 9) Difference of square
= 5A(2A – 3)(2A + 3)
B. –20A3 + 45A2 – 10A Take out the common factor and -1
= –5A(4A2 – 9A + 2)
Summary on Factoring
75. Example F. Factor completely.
A. 20A3 – 45A Take out the common factor
= 5A(4A2 – 9) Difference of square
= 5A(2A – 3)(2A + 3)
B. –20A3 + 45A2 – 10A Take out the common factor and -1
= –5A(4A2 – 9A + 2)
Summary on Factoring
= –5A(4A – 1)(A – 2)
76. Exercise A. Simplify with the given substitution.
Simplify your answers when possible.
Summary on Factoring
For 1- 8, Simplify A2 – B2 with the given substitution.
1. A = x, B = 1 2. A = 3, B = 2y
3. A = 2xy, B = 1 4. A = x + y, B = 2
5. A = (x – 3), B = x 6. A = 2x, B = (x – 1)
7. A = (x – 3), B = (x – 1) 8. A = (2x + 1), B = (2x – 1)
For 9-14, simplify A3 – B3 with the given substitution.
9. A = x, B = 2 10. A = 3x, B = 2
11. A = 2x, B = 3y 12. A = 1, B = 2xy
13. A = 4x, B = 5 14. A = 6x, B = 5y
15. Simplify x2 – 2x + 10 with the substitution x = (2 + h)
16. Simplify x2 + 3x – 4 with the substitution x = (–3 + h)
77. B. Identify the substitution for A and B then factor the
expression using the pattern A2 – B2 = (A – B)(A + B).
Summary on Factoring
17. x2 – 1 18. 9 – 4y2
23. 4x2y2 – 1
26. x2 + 2xy + y2 – 4
21. 25x2 – 64 22. 9 – 49y2
25. 4 – 81x2y2
24. 98x2 – 8
19. x2 – 9 20. 25 – y2
C. Identify the substitution for A and B then factor the
expression using the pattern A3 ± B3 = (A ± B)(A2 AB + B2).+
29. 8x2z – 98z
27. 50x2 – 18y2 28. 36x3 – 25x
30. x2 – y2 – 2y – 1
31. x3 – 1 32. x3 – 8
37. x3 – 125
36. 27 – 8y334. x3 + 8 35. 64 + y3
33. 27 – y3
38. 27x3 – 64 39. x3 – 1000