## Algebra 1: Common Core (15th Edition)

$\boxed{3x+1}$,$\boxed{2x-5}$ and $\boxed{x^2}$
First of all, you must take $x^2$: $x^2(6x^2-13x-5)$ Use Middle-term factoring to factor $6x^2-13x-5$ $6x^2 - 15x + 2x - 5$ $3x(2x-5) +1 (2x-5)$ $(3x+1)(2x-5)$ The dimensions are $\boxed{3x+1}$,$\boxed{2x-5}$ and $\boxed{x^2}$