SOLUTION: Im having trouble factoring polynomials: please show the steps and explain (simply) what changes were made and why they were made between steps: a few examples of what i am h

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 Question 634771: Im having trouble factoring polynomials: please show the steps and explain (simply) what changes were made and why they were made between steps: a few examples of what i am having trouble with: (2x^3) + (3x^2) - 8x - 12 = 0 (x/(x+2) + (1/4) = 5 x^(5/4) - 32 = 0 (6x^4) - (23x^2) + 20 = 0Answer by lwsshak3(6460)   (Show Source): You can put this solution on YOUR website!(2x^3) + (3x^2) - 8x - 12 = 0 factor by grouping (2x^3 - 8x)+ (3x^2 - 12) = 0 2x(x^2-4)+3(x^2-4)=0 factor out common term (x^2-4)(2x+3)=0 solve for x x^2-4=0 x^2=4 x=±√4=±2 or 2x+3=0 2x=-3 x=-3/2 x=-2, -3/2, 2 .. (x/(x+2) + (1/4) = 5 add(-1/4) to each side (x/(x+2) = 20/4-1/4=19/4 (x/(x+2)=19/4 cross-multiply 4x=19(x+2)=19x+38 15x=-38 x=-38/15 x≈2.5333.. .. x^(5/4) - 32 = 0 x^(5/4) = 32 raise both sides to the 4/5 power x=32^(4/5)=16 (take fifth root of 32=2, then raise it to the 4th power=2^4=16 .. (6x^4) - (23x^2) + 20 = 0 6x^4-23x^2+20 = 0 let u=x^2 u^2=x^4 equation: 6u^2-23u+20=0 solve for u, using following quadratic formula: a=6, b=-23, c=20 ans: u=2.5 or u=4/3 solve for x: x^2=u=2.5 x=±√2.5 or x^2=4/3 x=±√(4/3) .. x=-√(4/3), -√(2.5), √2.5, and √(4/3)