### Concept (1)

When asked to simplify expressions, sometimes we come across complicated expressions that are not easily factored by traditional methods. When factoring complex expressions, one strategy that we can use is substitution. When an expression has complex terms, we can substitute a single variable, factor and then re-substitute the original term for the variable once we have completely factored the expression.

### Sample Problems (7)

Need help with "Factoring Complicated Expressions" problems? Watch expert teachers solve similar problems to develop your skills.

Factor:

x⁶ − 7x³ + 6
###### Problem 1
How to factor a quadratic in disguise.

Factor:

(3x − 1)² − (x + 3)²
###### Problem 2
How to factor the difference of squares by substitution.

Factor:

6x⁻² − x⁻³ − 2x⁻⁴
###### Problem 3
How to factor an expression with negative exponents.

Factor:

x⁶ − 64
###### Problem 4
How to factor the difference of cubes by substitution.
###### Problem 5
Using the sum or difference of cubes formulas to factor binomials
###### Problem 6
Factoring complicated difference of perfect square binomials
###### Problem 7
Factoring with negative exponents