text: # How do you factor completely 45t^4 - 90t^3 + 25t^2?
  
  Apr 18, 2016
  
  Just to clarify what the other contributor was saying...
  
  #### Explanation:
  
  Factor out the GCF.
  
  $5 {t}^{2} \left(9 {t}^{2} - 18 t + 5\right)$
  
  The expression inside the parentheses can be factored as a trinomial of the form $y = a {x}^{2} + b x + c , a \ne 1$. To do this, we must find two numbers that multiply to $a \times c$ and that add to b, and then follow the following process. Two numbers that do this are -15 and -3.
  
  $5 {t}^{2} \left(9 {t}^{2} - 3 t - 15 t + 5\right)$
  
  $5 {t}^{2} \left(3 t \left(3 t - 1\right) - 5 \left(3 t - 1\right)\right)$
  
  $5 {t}^{2} \left(3 t - 5\right) \left(3 t - 1\right)$
  
  This is completely factored.
  
  Hopefully this helps!