Solving Polynomial Equations Algebraically

Q: How does a difference of cubes factor?

$a^3 - b^3 = (a - b)(a^2 + ab + b^2)$. The linear factor gives a real root; the quadratic usually gives complex roots.

To solve a polynomial equation exactly, set it equal to zero, factor completely, and use the zero product property on each factor.

Special forms help: a difference of cubes like $x^3 - 27$ factors by a known pattern, giving one real root and a complex pair.

Factor completely, then split

Move everything to one side so the equation equals zero. Factor as far as possible — common factors first, then special patterns.

Set each factor to zero. Each linear factor gives a real root; irreducible quadratic factors give complex roots.

Difference and sum of cubes

$a^3 - b^3 = (a - b)(a^2 + ab + b^2)$ and $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$ . The linear factor gives the real root.

For $x^3 - 27 = 0$ , the factor $x - 3$ gives the real solution $x = 3$ ; the quadratic factor gives two complex roots.

Worked examples

Example 1: difference of cubes

Find the real solution of $x^3 - 27 = 0$ .

Factor as a difference of cubes

(x - 3)(x^2 + 3x + 9) = 0

The linear factor gives the real root

x - 3 = 0

Solve

x = 3

Answer: $x = 3$

Example 2: common factor first

Solve $x^3 - 4x = 0$ .

Factor out x

x(x^2 - 4) = 0

Factor the difference of squares

x(x-2)(x+2) = 0

Set each to zero

x = 0, 2, -2

Answer: $x = 0, 2, -2$

Example 3: sum of cubes

Find the real solution of $x^3 = -8$ .

Move everything to one side

x^3 + 8 = 0

Factor as a sum of cubes

(x + 2)(x^2 - 2x + 4) = 0

The linear factor gives the real root

x + 2 = 0

Solve

x = -2

Answer: $x = -2$

Try one yourself

Common questions

What is the first step?

Set the equation equal to zero, then factor completely before applying the zero product property.

How does a difference of cubes factor?

$a^3 - b^3 = (a - b)(a^2 + ab + b^2)$ . The linear factor gives a real root; the quadratic usually gives complex roots.

Do I always get real solutions?

Not necessarily. Irreducible quadratic factors produce complex solutions, so a cubic can have one real root and two complex ones.

Want the video version?

Allday Everyday Math has video lessons, practice, and an AI tutor for every topic, Pre-Algebra through Algebra 2.

Try it for $1

Solving Polynomial Equations Algebraically

Factor completely, then split

Difference and sum of cubes

Worked examples

Example 1: difference of cubes

Example 2: common factor first

Example 3: sum of cubes

Try one yourself

Common questions

Keep learning

Solving Polynomial Equations by Graphing

The Remainder & Factor Theorems

Roots & Zeros