Download presentation

Presentation is loading. Please wait.

Published byLillian McCoy Modified over 6 years ago

1
Splash Screen

2
Concept

3
Example 1 Sum and Difference of Cubes A. Factor the polynomial x 3 – 400. If the polynomial cannot be factored, write prime. Answer:The first term is a perfect cube, but the second term is not. It is a prime polynomial.

4
Example 1 Sum and Difference of Cubes B. Factor the polynomial 24x 5 + 3x 2 y 3. If the polynomial cannot be factored, write prime. 24x 5 + 3x 2 y 3 =3x 2 (8x 3 + y 3 )Factor out the GCF. 8x 3 and y 3 are both perfect cubes, so we can factor the sum of the two cubes. (8x 3 + y 3 ) =(2x) 3 + (y) 3 (2x) 3 = 8x 3 ; (y) 3 = y 3 =(2x + y)[(2x) 2 – (2x)(y) + (y) 2 ] Sum of two cubes

5
Example 1 Sum and Difference of Cubes =(2x + y)[4x 2 – 2xy + y 2 ] Simplify. 24x 5 + 3x 2 y 3 =3x 2 (2x + y)[4x 2 – 2xy + y 2 ] Replace the GCF. Answer: 3x 2 (2x + y)(4x 2 – 2xy + y 2 )

6
Example 1 A. Factor the polynomial 54x 5 + 128x 2 y 3. If the polynomial cannot be factored, write prime. A. B. C. D.prime

7
A. B. C. D.prime Example 1 B. Factor the polynomial 64x 9 + 27y 5. If the polynomial cannot be factored, write prime.

8
Concept

9
Example 2 Factoring by Grouping A. Factor the polynomial x 3 + 5x 2 – 2x – 10. If the polynomial cannot be factored, write prime. x 3 + 5x 2 – 2x – 10Original expression = (x 3 + 5x 2 ) + (–2x – 10)Group to find a GCF. = x 2 (x + 5) – 2(x + 5)Factor the GCF. = (x + 5)(x 2 – 2)Distributive Property Answer: (x + 5)(x 2 – 2)

10
Example 2 Factoring by Grouping B. Factor the polynomial a 2 + 3ay + 2ay 2 + 6y 3. If the polynomial cannot be factored, write prime. a 2 + 3ay + 2ay 2 + 6y 3 Original expression = (a 2 + 3ay) + (2ay 2 + 6y 3 )Group to find a GCF. = a(a + 3y) + 2y 2 (a + 3y)Factor the GCF. = (a + 3y)(a + 2y 2 )Distributive Property Answer: (a + 3y)(a + 2y 2 )

11
Example 2 A.(d + 2)(d 2 + 2) B.(d – 2)(d 2 – 4) C.(d + 2)(d 2 + 4) D.prime A. Factor the polynomial d 3 + 2d 2 + 4d + 8. If the polynomial cannot be factored, write prime.

12
Example 2 A.(r – 2s)(r + 4s 2 ) B.(r + 2s)(r + 4s 2 ) C.(r + s)(r – 4s 2 ) D.prime B. Factor the polynomial r 2 + 4rs 2 + 2sr + 8s 3. If the polynomial cannot be factored, write prime.

13
Example 3 Combine Cubes and Squares A. Factor the polynomial x 2 y 3 – 3xy 3 + 2y 3 + x 2 z 3 – 3xz 3 + 2z 3. If the polynomial cannot be factored, write prime. With six terms, factor by grouping first. Group to find a GCF. Factor the GCF.

14
Example 3 Combine Cubes and Squares Sum of cubes Distributive Property Factor.

15
Example 3 Combine Cubes and Squares B. Factor the polynomial 64x 6 – y 6. If the polynomial cannot be factored, write prime. This polynomial could be considered the difference of two squares or the difference of two cubes. The difference of two squares should always be done before the difference of two cubes for easier factoring. Difference of two squares

16
Example 3 Combine Cubes and Squares Sum and difference of two cubes

17
Example 3 A. Factor the polynomial r 3 w 2 + 6r 3 w + 9r 3 + w 2 y 3 + 6wy 3 + 9y 3. If the polynomial cannot be factored, write prime. A. B. C. D.prime

18
Example 3 B. Factor the polynomial 729p 6 – k 6. If the polynomial cannot be factored, write prime. A. B. C. D.prime

19
Example 6 Solve Equations in Quadratic Form Solve x 4 – 29x 2 + 100 = 0. Original equation Factor. Zero Product Property Replace u with x 2.

20
Example 6 Solve Equations in Quadratic Form Take the square root. Answer: The solutions of the equation are 5, –5, 2, and –2.

21
Example 4 Solve Polynomial Functions by Factoring GEOMETRY Determine the dimensions of the cubes below if the length of the smaller cube is one half the length of the larger cube, and the volume of the shaded figure is 23,625 cubic centimeters.

22
Example 4 Solve Polynomial Functions by Factoring Since the length of the smaller cube is half the length of the larger cube, then their lengths can be represented by x and 2x, respectively. The volume of the object equals the volume of the larger cube minus the volume of the smaller cube. Volume of object Subtract. Divide.

23
Example 4 Solve Polynomial Functions by Factoring Answer:Since 15 is the only real solution, the lengths of the cubes are 15 cm and 30 cm. Subtract 3375 from each side. Difference of cubes Zero Product Property

24
Example 6 A.2, 3 B.–2, –3 C.–2, 2, –3, 3 D.no solution Solve x 6 – 35x 3 + 216 = 0.

25
Homework: P. 346 # 3 – 33 (x3), 42- 44

26
Example 4 A.7 cm and 14 cm B.9 cm and 18 cm C.10 cm and 20 cm D.12 cm and 24 cm GEOMETRY Determine the dimensions of the cubes below if the length of the smaller cube is one half the length of the larger cube, and the volume of the shaded figure is 5103 cubic centimeters.

27
Concept

28
Example 5 Quadratic Form A. Write 2x 6 – x 3 + 9 in quadratic form, if possible. 2x 6 – x 3 + 9 = 2(x 3 ) 2 – (x 3 ) + 9 Answer: 2(x 3 ) 2 – (x 3 ) + 9

29
Example 5 Quadratic Form B. Write x 4 – 2x 3 – 1 in quadratic form, if possible. Answer: This cannot be written in quadratic form since x 4 ≠ (x 3 ) 2.

30
Example 5 A.3(2x 5 ) 2 – (2x 5 ) – 3 B.6x 5 (x 5 ) – x 5 – 3 C.6(x 5 ) 2 – 2(x 5 ) – 3 D.This cannot be written in quadratic form. A. Write 6x 10 – 2x 5 – 3 in quadratic form, if possible.

31
Example 5 A.(x 8 ) 2 – 3(x 3 ) – 11 B.(x 4 ) 2 – 3(x 3 ) – 11 C.(x 4 ) 2 – 3(x 2 ) – 11 D.This cannot be written in quadratic form. B. Write x 8 – 3x 3 – 11 in quadratic form, if possible.

32
End of the Lesson

Similar presentations

© 2021 SlidePlayer.com Inc.

All rights reserved.

To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy, including cookie policy.

Ads by Google