Question

4(x+5)(x+6)(x+10)(x+12)-3x²=0

Answer 1

`4(x+5)(x+6)(x+10)(x+12)-3x^(2)=0 \\ \\ 4(x^(2)+11x+30)(x^(2)+22x+120)-3x^(2)=0 \\ \\ 4(x^(4)+22x^(3)+120x^(2)+11x^(3)+242x^(2)+1320x+30x^(2)+660x+3600)- \\ \\ -3x^(2)=0 \\ \\ 4(x^(4)+33x^(3)+392x^(2)+1980x+3600)-3x^(2)=0 \\ \\ 4x^(4)+132x^(3)+1568x^(2)+7920x+14400-3x^(2)=0 \\ \\`


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`\boxed{x\in \lbrace (-35- √(265) )/(4),-8,- (15)/(2), (-35+ √(265) )/(4) \rbrace }`

Answer 2

4 (x+5) (x+6) (x+10) (x+12) − 3x^2 = 0

4x^4 + 132x^3 + 1565x^2 + 7920x + 14400 = 0

Factor left side of equation.

(x+8) (2x+15) (2x^2 + 35x + 120) =0

Set factors equal to 0.

x + 8 = 0 or 2x + 15 = 0 or 2x^2 + 35x + 120 = 0

x = −8

or x = − 15/2

or x = −4.6802948509750735

or x = −12.819705149