1. Section 4-6
Solving Polynomial Equations

2. Essential Questions
• How do you factor polynomials?
• How do you solve polynomial equations by
factoring?

3. Vocabulary
1. Prime Polynomials:
2. Quadratic Form:

4. Vocabulary
1. Prime Polynomials: Polynomials that cannot be
factored
2. Quadratic Form:

5. Vocabulary
1. Prime Polynomials: Polynomials that cannot be
factored
2. Quadratic Form: Rewriting a polynomial so that it
fits the pattern of a standard form quadratic;
occurs with the degree of the first term is twice
that of the second, with a constant third term

6. Number
of Terms
Factoring Technique General Case
Any
Greatest Common Factor
(GCF)
2 Difference of Squares
2 Sum of Cubes
2 Difference of Cubes
3 Perfect Square Trinomial
3 General Trinomials
4+ Grouping

7. Number
of Terms
Factoring Technique General Case
Any
Greatest Common Factor
(GCF)
2 Difference of Squares
2 Sum of Cubes
2 Difference of Cubes
3 Perfect Square Trinomial
3 General Trinomials
4+ Grouping
4a5
b7
+12ab = 4ab(a4
b6
+ 3)

8. Number
of Terms
Factoring Technique General Case
Any
Greatest Common Factor
(GCF)
2 Difference of Squares
2 Sum of Cubes
2 Difference of Cubes
3 Perfect Square Trinomial
3 General Trinomials
4+ Grouping
4a5
b7
+12ab = 4ab(a4
b6
+ 3)
a2
− b2
= (a + b)(a − b)

9. Number
of Terms
Factoring Technique General Case
Any
Greatest Common Factor
(GCF)
2 Difference of Squares
2 Sum of Cubes
2 Difference of Cubes
3 Perfect Square Trinomial
3 General Trinomials
4+ Grouping
4a5
b7
+12ab = 4ab(a4
b6
+ 3)
a2
− b2
= (a + b)(a − b)
a3
+ b3
= (a + b)(a2
− ab + b2
)

10. Number
of Terms
Factoring Technique General Case
Any
Greatest Common Factor
(GCF)
2 Difference of Squares
2 Sum of Cubes
2 Difference of Cubes
3 Perfect Square Trinomial
3 General Trinomials
4+ Grouping
4a5
b7
+12ab = 4ab(a4
b6
+ 3)
a2
− b2
= (a + b)(a − b)
a3
+ b3
= (a + b)(a2
− ab + b2
)
a3
− b3
= (a − b)(a2
+ ab + b2
)

11. Number
of Terms
Factoring Technique General Case
Any
Greatest Common Factor
(GCF)
2 Difference of Squares
2 Sum of Cubes
2 Difference of Cubes
3 Perfect Square Trinomial
3 General Trinomials
4+ Grouping
4a5
b7
+12ab = 4ab(a4
b6
+ 3)
a2
− b2
= (a + b)(a − b)
a3
+ b3
= (a + b)(a2
− ab + b2
)
a3
− b3
= (a − b)(a2
+ ab + b2
)
a2
+ 2ab + b2
= (a + b)2

12. Number
of Terms
Factoring Technique General Case
Any
Greatest Common Factor
(GCF)
2 Difference of Squares
2 Sum of Cubes
2 Difference of Cubes
3 Perfect Square Trinomial
3 General Trinomials
4+ Grouping
4a5
b7
+12ab = 4ab(a4
b6
+ 3)
a2
− b2
= (a + b)(a − b)
a3
+ b3
= (a + b)(a2
− ab + b2
)
a3
− b3
= (a − b)(a2
+ ab + b2
)
a2
+ 2ab + b2
= (a + b)2
a2
− 2ab + b2
= (a − b)2

13. Number
of Terms
Factoring Technique General Case
Any
Greatest Common Factor
(GCF)
2 Difference of Squares
2 Sum of Cubes
2 Difference of Cubes
3 Perfect Square Trinomial
3 General Trinomials
4+ Grouping
4a5
b7
+12ab = 4ab(a4
b6
+ 3)
a2
− b2
= (a + b)(a − b)
a3
+ b3
= (a + b)(a2
− ab + b2
)
a3
− b3
= (a − b)(a2
+ ab + b2
)
a2
+ 2ab + b2
= (a + b)2
a2
− 2ab + b2
= (a − b)2
acx 2
+ (ad + bc)x + bd
= (ax + b)(cx + d )

14. Number
of Terms
Factoring Technique General Case
Any
Greatest Common Factor
(GCF)
2 Difference of Squares
2 Sum of Cubes
2 Difference of Cubes
3 Perfect Square Trinomial
3 General Trinomials
4+ Grouping
4a5
b7
+12ab = 4ab(a4
b6
+ 3)
a2
− b2
= (a + b)(a − b)
a3
+ b3
= (a + b)(a2
− ab + b2
)
a3
− b3
= (a − b)(a2
+ ab + b2
)
a2
+ 2ab + b2
= (a + b)2
a2
− 2ab + b2
= (a − b)2
acx 2
+ (ad + bc)x + bd
= (ax + b)(cx + d )
ax + bx + ay + by
= (a + b)(x + y )

15. Example 1
Factor each polynomial. If the polynomial cannot be
factored, write prime.
a. x 3
−1000
b. 24x 5
+ 3x 2
y 3
c. x 3
+ 400

16. Example 1
Factor each polynomial. If the polynomial cannot be
factored, write prime.
a. x 3
−1000
(x −10)(x 2
+10x +100)
b. 24x 5
+ 3x 2
y 3
c. x 3
+ 400

17. Example 1
Factor each polynomial. If the polynomial cannot be
factored, write prime.
a. x 3
−1000
(x −10)(x 2
+10x +100)
b. 24x 5
+ 3x 2
y 3
3x 2
(8x 3
+ y 3
)
c. x 3
+ 400

18. Example 1
Factor each polynomial. If the polynomial cannot be
factored, write prime.
a. x 3
−1000
(x −10)(x 2
+10x +100)
b. 24x 5
+ 3x 2
y 3
3x 2
(8x 3
+ y 3
)
3x 2
(2x + y )(4x 2
− 2xy + y 2
)
c. x 3
+ 400

19. Example 1
Factor each polynomial. If the polynomial cannot be
factored, write prime.
a. x 3
−1000
(x −10)(x 2
+10x +100)
b. 24x 5
+ 3x 2
y 3
3x 2
(8x 3
+ y 3
)
3x 2
(2x + y )(4x 2
− 2xy + y 2
)
c. x 3
+ 400
Prime

20. Example 2
Factor each polynomial. If the polynomial cannot be
factored, write prime.
a. x 3
+ 5x 2
− 2x −10

21. Example 2
Factor each polynomial. If the polynomial cannot be
factored, write prime.
a. x 3
+ 5x 2
− 2x −10
(x 3
+ 5x 2
)+ (−2x −10)

22. Example 2
Factor each polynomial. If the polynomial cannot be
factored, write prime.
a. x 3
+ 5x 2
− 2x −10
(x 3
+ 5x 2
)+ (−2x −10)
x 2
(x + 5)− 2(x + 5)

23. Example 2
Factor each polynomial. If the polynomial cannot be
factored, write prime.
a. x 3
+ 5x 2
− 2x −10
(x 3
+ 5x 2
)+ (−2x −10)
x 2
(x + 5)− 2(x + 5)
(x + 5)(x 2
− 2)

24. Example 2
Factor each polynomial. If the polynomial cannot be
factored, write prime.
b. a2
+ 3ay + 2ay 2
+ 6y 3

25. Example 2
Factor each polynomial. If the polynomial cannot be
factored, write prime.
b. a2
+ 3ay + 2ay 2
+ 6y 3
(a2
+ 3ay )+ (2ay 2
+ 6y 3
)

26. Example 2
Factor each polynomial. If the polynomial cannot be
factored, write prime.
b. a2
+ 3ay + 2ay 2
+ 6y 3
(a2
+ 3ay )+ (2ay 2
+ 6y 3
)
a(a + 3y )+ 2y 2
(a + 3y )

27. Example 2
Factor each polynomial. If the polynomial cannot be
factored, write prime.
b. a2
+ 3ay + 2ay 2
+ 6y 3
(a2
+ 3ay )+ (2ay 2
+ 6y 3
)
a(a + 3y )+ 2y 2
(a + 3y )
(a + 3y )(a + 2y 2
)

28. Example 3
Factor each polynomial. If the polynomial cannot be
factored, write prime.
a. x 2
y 3
− 3xy 3
+ 2y 3
+ x 2
z3
− 3xz3
+ 2z3

29. Example 3
Factor each polynomial. If the polynomial cannot be
factored, write prime.
a. x 2
y 3
− 3xy 3
+ 2y 3
+ x 2
z3
− 3xz3
+ 2z3
(x 2
y 3
− 3xy 3
+ 2y 3
)+ (x 2
z3
− 3xz3
+ 2z3
)

30. Example 3
Factor each polynomial. If the polynomial cannot be
factored, write prime.
a. x 2
y 3
− 3xy 3
+ 2y 3
+ x 2
z3
− 3xz3
+ 2z3
(x 2
y 3
− 3xy 3
+ 2y 3
)+ (x 2
z3
− 3xz3
+ 2z3
)
y 3
(x 2
− 3x + 2)+ z3
(x 2
− 3x + 2)

31. Example 3
Factor each polynomial. If the polynomial cannot be
factored, write prime.
a. x 2
y 3
− 3xy 3
+ 2y 3
+ x 2
z3
− 3xz3
+ 2z3
(x 2
y 3
− 3xy 3
+ 2y 3
)+ (x 2
z3
− 3xz3
+ 2z3
)
y 3
(x 2
− 3x + 2)+ z3
(x 2
− 3x + 2)
(x 2
− 3x + 2)(y 3
+ z3
)

32. Example 3
Factor each polynomial. If the polynomial cannot be
factored, write prime.
a. x 2
y 3
− 3xy 3
+ 2y 3
+ x 2
z3
− 3xz3
+ 2z3
(x 2
y 3
− 3xy 3
+ 2y 3
)+ (x 2
z3
− 3xz3
+ 2z3
)
y 3
(x 2
− 3x + 2)+ z3
(x 2
− 3x + 2)
(x 2
− 3x + 2)(y 3
+ z3
)
(x − 2)(x −1)(y + z)(y 2
− yz + z2
)

33. Example 3
Factor each polynomial. If the polynomial cannot be
factored, write prime.
b. 64x 6
− y 6

34. Example 3
Factor each polynomial. If the polynomial cannot be
factored, write prime.
b. 64x 6
− y 6
(8x 3
+ y 3
)(8x 3
− y 3
)

35. Example 3
Factor each polynomial. If the polynomial cannot be
factored, write prime.
b. 64x 6
− y 6
(8x 3
+ y 3
)(8x 3
− y 3
)
(2x + y )(4x 2
− 2xy + y 2
)(2x − y )(4x 2
+ 2xy + y 2
)

36. Example 4
A small cube is cut from a larger cube.
Determine the dimensions of the cubes if the
length of the smaller cube is is one half the
length of the larger cube, and the volume of the
object is 23,625 cubic centimeters.

37. Example 4
A small cube is cut from a larger cube.
Determine the dimensions of the cubes if the
length of the smaller cube is is one half the
length of the larger cube, and the volume of the
object is 23,625 cubic centimeters.
x
x
x

38. Example 4
A small cube is cut from a larger cube.
Determine the dimensions of the cubes if the
length of the smaller cube is is one half the
length of the larger cube, and the volume of the
object is 23,625 cubic centimeters.
x 3
−
x
2
⎛
⎝⎜
⎞
⎠⎟
3
= 23625
x
x
x

39. Example 4
A small cube is cut from a larger cube.
Determine the dimensions of the cubes if the
length of the smaller cube is is one half the
length of the larger cube, and the volume of the
object is 23,625 cubic centimeters.
x 3
−
x
2
⎛
⎝⎜
⎞
⎠⎟
3
= 23625
x
x
x
x 3
−
1
8
x 3
= 23625

40. Example 4
A small cube is cut from a larger cube.
Determine the dimensions of the cubes if the
length of the smaller cube is is one half the
length of the larger cube, and the volume of the
object is 23,625 cubic centimeters.
x 3
−
x
2
⎛
⎝⎜
⎞
⎠⎟
3
= 23625
x
x
x
x 3
−
1
8
x 3
= 23625
7
8
x 3
= 23625

41. Example 4
A small cube is cut from a larger cube.
Determine the dimensions of the cubes if the
length of the smaller cube is is one half the
length of the larger cube, and the volume of the
object is 23,625 cubic centimeters.
x 3
−
x
2
⎛
⎝⎜
⎞
⎠⎟
3
= 23625
x
x
x
x 3
−
1
8
x 3
= 23625
7
8
x 3
= 23625
x 3
= 27000

42. Example 4
A small cube is cut from a larger cube.
Determine the dimensions of the cubes if the
length of the smaller cube is is one half the
length of the larger cube, and the volume of the
object is 23,625 cubic centimeters.
x 3
−
x
2
⎛
⎝⎜
⎞
⎠⎟
3
= 23625
x
x
x
x 3
−
1
8
x 3
= 23625
7
8
x 3
= 23625
x 3
= 27000
x 33
= 270003

43. Example 4
A small cube is cut from a larger cube.
Determine the dimensions of the cubes if the
length of the smaller cube is is one half the
length of the larger cube, and the volume of the
object is 23,625 cubic centimeters.
x 3
−
x
2
⎛
⎝⎜
⎞
⎠⎟
3
= 23625
x
x
x
x 3
−
1
8
x 3
= 23625
7
8
x 3
= 23625
x 3
= 27000
x 33
= 270003
x = 30

44. Example 4
A small cube is cut from a larger cube.
Determine the dimensions of the cubes if the
length of the smaller cube is is one half the
length of the larger cube, and the volume of the
object is 23,625 cubic centimeters.
x 3
−
x
2
⎛
⎝⎜
⎞
⎠⎟
3
= 23625
x
x
x
x 3
−
1
8
x 3
= 23625
7
8
x 3
= 23625
x 3
= 27000
x 33
= 270003
x = 30
The larger cube has a length of 30cm.
The smaller cube has a length of 15 cm.

45. Example 5
Write each expression in quadratic form, if possible.
a. 2x 6
− x 3
+ 9 b. x 4
+ 2x 3
−1

46. Example 5
Write each expression in quadratic form, if possible.
a. 2x 6
− x 3
+ 9
2(x 3
)2
− (x 3
)+ 9
b. x 4
+ 2x 3
−1

47. Example 5
Write each expression in quadratic form, if possible.
a. 2x 6
− x 3
+ 9
2(x 3
)2
− (x 3
)+ 9
Let u = x 3
b. x 4
+ 2x 3
−1

48. Example 5
Write each expression in quadratic form, if possible.
a. 2x 6
− x 3
+ 9
2(x 3
)2
− (x 3
)+ 9
2u2
−u + 9
Let u = x 3
b. x 4
+ 2x 3
−1

49. Example 5
Write each expression in quadratic form, if possible.
a. 2x 6
− x 3
+ 9
2(x 3
)2
− (x 3
)+ 9
2u2
−u + 9
Let u = x 3
b. x 4
+ 2x 3
−1
Not possible

50. Example 6
Solve.
x 4
− 29x 2
+100 = 0

51. Example 6
Solve.
x 4
− 29x 2
+100 = 0
Let u = x 2

52. Example 6
Solve.
x 4
− 29x 2
+100 = 0
Let u = x 2
u2
− 29u +100 = 0

53. Example 6
Solve.
x 4
− 29x 2
+100 = 0
Let u = x 2
u2
− 29u +100 = 0
(u − 25)(u − 4) = 0

54. Example 6
Solve.
x 4
− 29x 2
+100 = 0
Let u = x 2
u2
− 29u +100 = 0
(u − 25)(u − 4) = 0
(x 2
− 25)(x 2
− 4) = 0

55. Example 6
Solve.
x 4
− 29x 2
+100 = 0
Let u = x 2
u2
− 29u +100 = 0
(u − 25)(u − 4) = 0
(x 2
− 25)(x 2
− 4) = 0
(x + 5)(x − 5)(x + 2)(x − 2) = 0

56. Example 6
Solve.
x 4
− 29x 2
+100 = 0
Let u = x 2
u2
− 29u +100 = 0
(u − 25)(u − 4) = 0
(x 2
− 25)(x 2
− 4) = 0
(x + 5)(x − 5)(x + 2)(x − 2) = 0
x = −5,x = 5,x = −2,x = 2