How do you solve #x^3+3x^2-x-3=0#?

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Answer 1 · 2016-03-18T12:00:59

#x = -1,+1,-3#

You can factor it but here is a shorter trick

#x^3+3x^2-x-3=0#

#x^3+3x^2=x + 3#

#x^2(x + 3)=x + 3#

#x^2cancel((x + 3))=cancel((x + 3))#

#x ^2 = 1#

#x = +-1#

Thats 2 values lets go for the third one

{XD,we actually need to factor it to get the third value}
#x^3+3x^2-x-3=0#

#x^2(x + 3)-1(x+3)=0#

#(x^2 - 1)(x + 3) = 0#

#=> x +3 = 0 => x = -3#

#color(blue)"Now we have got our 3 values and we can be confident we have solved our cubic"#