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 ->  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


   

Question 598315: I need help factoring out this problem20x%5E3%2B16x%5E2%2B25x%2B20=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
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(35256) About Me  (Show Source):
You can put this solution on YOUR website!
You are correct. You cannot factor 4x%5E2%2B5. So you simply solve for x.

4x%5E2%2B5=0

4x%5E2=-5

x%5E2=-5%2F4

x=%22%22%2B-sqrt%28-5%2F4%29

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 x=-4%2F5


However, if you have learned about complex/imaginary numbers, then...

x=%22%22%2B-sqrt%28-5%2F4%29

x=%28i%2Asqrt%285%29%29%2F2 or x=-%28i%2Asqrt%285%29%29%2F2

So if you have learned about complex/imaginary numbers, then there are three solutions and they are x=-4%2F5, x=%28i%2Asqrt%285%29%29%2F2 or x=-%28i%2Asqrt%285%29%29%2F2

Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
I need help factoring out this problem20x%5E3%2B16x%5E2%2B25x%2B20=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
5x=-4
divide x= -4/5
I could not factor out (4x^2+5)=0
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4x^2 + 5 = 0
4x^2 = -5
x^2 = -5/4
x+=+0+%2B-+sqrt%285%29i%2F2
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i = sqrt(-1)