SOLUTION: How would I factor this? {{{12(4x-1)^3(2x+5)^9 +3(4x-1)^4(2x+5)^8}}} My attempt: 12(4x-1)^3(2x+5)^8[3(4x-1)(2x+5) 12(4x-1)^3(2x+5)^8[3(8x^2+18x-5] 36(4x-1)^3(2x+5)^8(8x^2+

Algebra ->  Expressions-with-variables -> SOLUTION: How would I factor this? {{{12(4x-1)^3(2x+5)^9 +3(4x-1)^4(2x+5)^8}}} My attempt: 12(4x-1)^3(2x+5)^8[3(4x-1)(2x+5) 12(4x-1)^3(2x+5)^8[3(8x^2+18x-5] 36(4x-1)^3(2x+5)^8(8x^2+      Log On


   

Question 909080: How would I factor this?
12%284x-1%29%5E3%282x%2B5%29%5E9+%2B3%284x-1%29%5E4%282x%2B5%29%5E8
My attempt:
12(4x-1)^3(2x+5)^8[3(4x-1)(2x+5)
12(4x-1)^3(2x+5)^8[3(8x^2+18x-5]
36(4x-1)^3(2x+5)^8(8x^2+18x-5)

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
12%284x-1%29%5E3%282x%2B5%29%5E9+%2B3%284x-1%29%5E4%282x%2B5%29%5E8

12%2A3%284x-1%29%5E3%282x%2B5%29%5E8%284%282x%2B5%29+%2B+%284x-1%29%29

36%284x-1%29%5E3%282x%2B5%29%5E8%288x%2B20+%2B+4x-1%29

36%284x-1%29%5E3%282x%2B5%29%5E8%2812x%2B19%29

36%2812x%2B19%29%284x-1%29%5E3%282x%2B5%29%5E8

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So 12%284x-1%29%5E3%282x%2B5%29%5E9+%2B3%284x-1%29%5E4%282x%2B5%29%5E8 completely factors to 36%2812x%2B19%29%284x-1%29%5E3%282x%2B5%29%5E8