1.Solve the following equation
a. 2x^3+3X^2-3X-2=0
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Answer:
2x^3+3x^2-3x-2=0
=> 2x^3-2+3x^2-3x=0
=> 2(x^3-1)+3x(x-1)=0
=> 2(x-1)(x^2+x+1)+3x(x-1)=0
=> (x-1){2(x^2+x+1)+3x}=0
=> (x-1)(2x^2+2x+2+3x)=0
=> (x-1)(2x^2+5x+2)=0
=> (x-1)(2x^2+4x+x+2)=0
=> (x-1){2x(x+2)+1(x+2)}=0
=> (x-1)(x+2)(2x+1)=0
=> x-1=0 => x=1
x+2=0 => x= -2
2x+1=0
=> 2x= -1
=> x= -1/2
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