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Amy Ferguson Moncure

Algebra 1 · Unit 7

Factor Cubics and Quartics

Factor third- and fourth-degree polynomials with GCF, grouping, difference of squares, and sum or difference of cubes.

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Key Ideas Easy Medium Hard Super Challenge
What to Remember

GCF first

A cubic or quartic often starts with a GCF, then a quadratic you already know.

Grouping

Four terms: pair them, factor each pair, then factor the common binomial.

Difference of squares, twice

x⁴ − 16 = (x² + 4)(x² − 4) = (x² + 4)(x + 2)(x − 2).

Sum or difference of cubes

x³ − 8 = (x − 2)(x² + 2x + 4). x³ + 8 = (x + 2)(x² − 2x + 4).

Level 1

Easy

GCF, then a quadratic or a difference of squares.

Example
Factor completely. 2x³ − 8x

GCF = 2x

2x(x² − 4)

2x(x + 2)(x − 2)

Answer: 2x(x + 2)(x − 2)

1
Factor completely. 3x³ + 6x² + 3x
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3x(x² + 2x + 1)

3x(x + 1)²

Answer: 3x(x + 1)²

2
Factor completely. x³ − 9x
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x(x² − 9)

x(x + 3)(x − 3)

Answer: x(x + 3)(x − 3)

3
Factor completely. 4x³ − 4x
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4x(x² − 1)

4x(x + 1)(x − 1)

Answer: 4x(x + 1)(x − 1)

4
Factor completely. x⁴ − 1
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(x² + 1)(x² − 1)

(x² + 1)(x + 1)(x − 1)

Answer: (x² + 1)(x + 1)(x − 1)

5
Factor completely. 5x² − 20
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5(x² − 4)

5(x + 2)(x − 2)

Answer: 5(x + 2)(x − 2)

Level 2

Medium

Grouping four terms, or a sum/difference of cubes.

Example
Factor by grouping. x³ + 4x² + 3x + 12

(x³ + 4x²) + (3x + 12)

x²(x + 4) + 3(x + 4)

(x² + 3)(x + 4)

Answer: (x² + 3)(x + 4)

1
Factor. x³ − 8
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(x − 2)(x² + 2x + 4)

Answer: (x − 2)(x² + 2x + 4)

2
Factor. x³ + 27
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(x + 3)(x² − 3x + 9)

Answer: (x + 3)(x² − 3x + 9)

3
Factor by grouping. 2x³ + 2x² + x + 1
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2x²(x + 1) + 1(x + 1)

(2x² + 1)(x + 1)

Answer: (2x² + 1)(x + 1)

4
Factor completely. x⁴ − 16
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(x² + 4)(x² − 4)

(x² + 4)(x + 2)(x − 2)

Answer: (x² + 4)(x + 2)(x − 2)

5
Factor completely. 2x³ − 18x
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2x(x² − 9)

2x(x + 3)(x − 3)

Answer: 2x(x + 3)(x − 3)

Level 3

Hard

More than one method in a row.

Example
Factor completely. 3x⁴ − 48

GCF = 3

3(x⁴ − 16)

3(x² + 4)(x² − 4)

3(x² + 4)(x + 2)(x − 2)

Answer: 3(x² + 4)(x + 2)(x − 2)

1
Factor completely. x³ + x² − 4x − 4
Show Answer

(x³ + x²) + (−4x − 4)

x²(x + 1) − 4(x + 1)

(x² − 4)(x + 1)

(x + 2)(x − 2)(x + 1)

Answer: (x + 2)(x − 2)(x + 1)

2
Factor completely. 2x³ + 16
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2(x³ + 8)

2(x + 2)(x² − 2x + 4)

Answer: 2(x + 2)(x² − 2x + 4)

3
Factor completely. x⁴ + 4x² + 4
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(x² + 2)²

Answer: (x² + 2)²

4
Factor completely. 4x³ − 4x² − 24x
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4x(x² − x − 6)

4x(x − 3)(x + 2)

Answer: 4x(x − 3)(x + 2)

5
Factor completely. x⁴ − 5x² + 4
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Let u = x²: u² − 5u + 4 = (u − 1)(u − 4)

(x² − 1)(x² − 4)

(x + 1)(x − 1)(x + 2)(x − 2)

Answer: (x + 1)(x − 1)(x + 2)(x − 2)

Super Challenge

Super Challenge

1
Factor x⁴ − 81 completely, and explain why x² + 9 does not factor over the reals.
Show Answer

(x² + 9)(x² − 9)

(x² + 9)(x + 3)(x − 3)

x² + 9 = 0 has no real x, so that quadratic stays.

Answer: (x² + 9)(x + 3)(x − 3)

2
A student stops at 2x(x² − 16) for 2x³ − 32x. Why is that not complete?
Show Answer

x² − 16 is still a difference of squares.

2x(x + 4)(x − 4)

Answer: 2x(x + 4)(x − 4)

3
Show grouping and a cubic formula are not both needed for x³ + 2x² − x − 2. Factor completely.
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(x³ + 2x²) + (−x − 2)

x²(x + 2) − 1(x + 2)

(x² − 1)(x + 2)

(x + 1)(x − 1)(x + 2)

Answer: (x + 1)(x − 1)(x + 2)

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