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
Which is not a zero of the function
f(x) = x3 – 3x2 – 10x + 24?
A. –3
B. –2
C. 2
D. 4
Over Lesson 5–4
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
Over Lesson 5–4
Estimate the x-value at which the relative minimum
of f(x) = x4 + x + 2 occurs.
A. 0.5
B. 0
C. –0.5
D. –1.5
Over Lesson 5–4
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)
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
B. Factor the polynomial 64x 9 + 27y 5. If the
polynomial cannot be factored, write prime.
A.
B.
C.
D. 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. 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
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
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
Distributive
Property
Sum of
cubes
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
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.
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
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. 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.
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.
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.
Solve x 6 – 35x 3 + 216 = 0.
A. 2, 3
B. –2, –3
C. –2, 2, –3, 3
D. no solution