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

Over Lesson 5–4

A. –3

B. –2

C. 2

D. 4

Which is not a zero of the function f(x) = x3 – 3x2 – 10x + 24?

Over Lesson 5–4

A. x = –2.5

B. x = 0

C. x = 1.5

D. x = 2.5

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?

Over Lesson 5–4

A. 0.5

B. 0

C. –0.5

D. –1.5

Estimate the x-value at which the relative minimum of f(x) = x4 + x + 2 occurs.

Over Lesson 5–4

A. (–∞, –3), (1, 4)

B. (–4, –3), (1, 3)

C. (–∞, –3), (1, ∞)

D. (–∞, –2), (1, 2)

For which part(s) of its domain does the function f(x) = x3 – 2x2 – 11x + 12 have negative f(x) values?

Content Standards

A.CED.1 Create equations and inequalities in one variable and use them to solve problems.

Mathematical Practices

4 Model with mathematics.

You solved quadratic functions by factoring.

• Factor polynomials.

• Solve polynomial equations by factoring.

• prime polynomials

• quadratic form

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.

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

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)

A. Factor the polynomial 54x

5 + 128x

2y

3. If the polynomial cannot be factored, write prime.

A.

B.

C.

D. prime

A.

B.

C.

D. prime

B. Factor the polynomial 64x

9 + 27y

5. If the polynomial cannot be factored, write prime.

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)

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)

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.

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.

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.

Combine Cubes and Squares

Sum of cubes

Distributive Property

Factor.

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

Combine Cubes and Squares

Sum and difference of two cubes

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

B. Factor the polynomial 729p

6 – k

6. If the polynomial cannot be factored, write prime.

A.

B.

C.

D. prime

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.

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.

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

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.

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

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.

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.

A. Write 6x

10 – 2x

5 – 3 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.

B. Write x

8 – 3x

3 – 11 in quadratic form, if possible.

Solve Equations in Quadratic Form

Solve x

4 – 29x

2 + 100 = 0.

Original equation

Factor.

Zero Product Property

Replace u with x

2.

Solve Equations in Quadratic Form

Take the square root.

Answer: The solutions of the equation are 5, –5, 2, and –2.

A. 2, 3

B. –2, –3

C. –2, 2, –3, 3

D. no solution

Solve x

6 – 35x

3 + 216 = 0.