## Intermediate Algebra for College Students (7th Edition)

$(1-3x)(1+3x)(1+9x^2)$
The given binomial can be written as: $=1^2-(9x^2)^2$ The binomial is a difference of two squares. RECALL: $a^2-b^2=(a-b)(a+b)$ Factor the difference of two squares using the formula above with $a=1$ and $b=9x^2$ to obtain: $=(1-9x^2)(1+9x^2) \\=[1^2-(3x)^2)](1+9x^2)$ The first factor is a difference of two squares. Factor using the formula above with $a=1$ and $b=3x$ to obtain: $=(1-3x)(1+3x)(1+9x^2)$