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Splash Screen

Five-Minute Check (over Lesson 5–4) CCSS Then/Now New Vocabulary Key Concept: Sum and Difference of Cubes Example 1: Sum and Difference of Cubes Concept Summary: Factoring Techniques Example 2: Factoring by Grouping Example 3: Combine Cubes and Squares Example 4: Real-World Example: Solve Polynomial Functions by Factoring Key Concept: Quadratic Form Example 5: Quadratic Form Example 6: Solve Equations in Quadratic Form Lesson Menu

Which is not a zero of the function f(x) = x3 – 3x2 – 10x + 24? B. –2 C. 2 D. 4 5-Minute Check 1

Which is not a zero of the function f(x) = x3 – 3x2 – 10x + 24? B. –2 C. 2 D. 4 5-Minute Check 1

Use the table of values for f(x) = x4 – 12x2 + 5 Use the table of values for f(x) = x4 – 12x2 + 5. Estimate the x-coordinates at which any relative maxima and relative minima occur. Which is not a possible relative maximum or relative minimum? A. x = –2.5 B. x = 0 C. x = 1.5 D. x = 2.5 5-Minute Check 2

Use the table of values for f(x) = x4 – 12x2 + 5 Use the table of values for f(x) = x4 – 12x2 + 5. Estimate the x-coordinates at which any relative maxima and relative minima occur. Which is not a possible relative maximum or relative minimum? A. x = –2.5 B. x = 0 C. x = 1.5 D. x = 2.5 5-Minute Check 2

Estimate the x-value at which the relative minimum of f(x) = x4 + x + 2 occurs. B. 0 C. –0.5 D. –1.5 5-Minute Check 3

Estimate the x-value at which the relative minimum of f(x) = x4 + x + 2 occurs. B. 0 C. –0.5 D. –1.5 5-Minute Check 3

For which part(s) of its domain does the function f(x) = x3 – 2x2 – 11x + 12 have negative f(x) values? A. (–∞, –3), (1, 4) B. (–4, –3), (1, 3) C. (–∞, –3), (1, ∞) D. (–∞, –2), (1, 2) 5-Minute Check 4

For which part(s) of its domain does the function f(x) = x3 – 2x2 – 11x + 12 have negative f(x) values? A. (–∞, –3), (1, 4) B. (–4, –3), (1, 3) C. (–∞, –3), (1, ∞) D. (–∞, –2), (1, 2) 5-Minute Check 4

Mathematical Practices 4 Model with mathematics. Content Standards A.CED.1 Create equations and inequalities in one variable and use them to solve problems. Mathematical Practices 4 Model with mathematics. CCSS

You solved quadratic functions by factoring. Factor polynomials. Solve polynomial equations by factoring. Then/Now

prime polynomials quadratic form Vocabulary

Concept

Sum and Difference of Cubes A. Factor the polynomial x 3 – 400. If the polynomial cannot be factored, write prime. Answer: 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. Example 1

24x 5 + 3x 2y 3 = 3x 2(8x 3 + y 3) Factor out the GCF. Sum and Difference of Cubes B. Factor the polynomial 24x 5 + 3x 2y 3. If the polynomial cannot be factored, write prime. 24x 5 + 3x 2y 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 Example 1

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

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

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

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

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

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

Concept

x 3 + 5x 2 – 2x – 10 Original expression 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 – 10 Original 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: Example 2

x 3 + 5x 2 – 2x – 10 Original expression 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 – 10 Original 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) Example 2

a 2 + 3ay + 2ay 2 + 6y 3 Original expression 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: Example 2

a 2 + 3ay + 2ay 2 + 6y 3 Original expression 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) Example 2

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

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

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

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

With six terms, factor by grouping first. Combine Cubes and Squares A. Factor the polynomial x 2y 3 – 3xy 3 + 2y 3 + x 2z 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. Example 3

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

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

Difference of two squares 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 Example 3

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

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

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

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

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

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

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. 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. Example 4

Subtract 3375 from each side. Solve Polynomial Functions by Factoring Subtract 3375 from each side. Difference of cubes Zero Product Property Answer: Example 4

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

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. A. 7 cm and 14 cm B. 9 cm and 18 cm C. 10 cm and 20 cm D. 12 cm and 24 cm Example 4

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. A. 7 cm and 14 cm B. 9 cm and 18 cm C. 10 cm and 20 cm D. 12 cm and 24 cm Example 4

Concept

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: Example 5

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 Example 5

B. Write x 4 – 2x 3 – 1 in quadratic form, if possible. Answer: Example 5

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. Example 5

A. Write 6x 10 – 2x 5 – 3 in quadratic form, if possible. A. 3(2x 5)2 – (2x 5) – 3 B. 6x5(x5) – x5 – 3 C. 6(x 5)2 – 2(x 5) – 3 D. This cannot be written in quadratic form. Example 5

A. Write 6x 10 – 2x 5 – 3 in quadratic form, if possible. A. 3(2x 5)2 – (2x 5) – 3 B. 6x5(x5) – x5 – 3 C. 6(x 5)2 – 2(x 5) – 3 D. This cannot be written in quadratic form. Example 5

B. Write x 8 – 3x 3 – 11 in quadratic form, if possible. 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. Example 5

B. Write x 8 – 3x 3 – 11 in quadratic form, if possible. 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. Example 5

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

Take the square root. Answer: Solve Equations in Quadratic Form Example 6

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

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

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

End of the Lesson