# SOLUTION: I need help factoring out this problem{{{20x^3+16x^2+25x+20=0}}} I did it in two group (20x^3+16x^2) (25x+20)=0 4x^2(5x+4) 5(5x+4)=0 (5x+4) (4x^2+5)=0 5x+4=0 5x+4-4=0-4

Algebra ->  Algebra  -> Polynomials-and-rational-expressions -> SOLUTION: I need help factoring out this problem{{{20x^3+16x^2+25x+20=0}}} I did it in two group (20x^3+16x^2) (25x+20)=0 4x^2(5x+4) 5(5x+4)=0 (5x+4) (4x^2+5)=0 5x+4=0 5x+4-4=0-4       Log On

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 Question 598315: I need help factoring out this problem I did it in two group (20x^3+16x^2) (25x+20)=0 4x^2(5x+4) 5(5x+4)=0 (5x+4) (4x^2+5)=0 5x+4=0 5x+4-4=0-4 5x=-4 divide x= -4/5 I could not factor out (4x^2+5)=0 Found 2 solutions by jim_thompson5910, Alan3354:Answer by jim_thompson5910(28595)   (Show Source): You can put this solution on YOUR website!You are correct. You cannot factor . So you simply solve for x. Now if you haven't learned about complex/imaginary numbers yet, then you would ignore this (and the rest of the solution) and simply say that the only solution is However, if you have learned about complex/imaginary numbers, then... or So if you have learned about complex/imaginary numbers, then there are three solutions and they are , or Answer by Alan3354(30993)   (Show Source): You can put this solution on YOUR website!I need help factoring out this problem I did it in two group (20x^3+16x^2) (25x+20)=0 4x^2(5x+4) 5(5x+4)=0 (5x+4) (4x^2+5)=0 5x+4=0 5x+4-4=0-4 5x=-4 divide x= -4/5 I could not factor out (4x^2+5)=0 --------------- 4x^2 + 5 = 0 4x^2 = -5 x^2 = -5/4 ------ i = sqrt(-1)