How do you solve x^3+3x^2-x-3=0x3+3x2x3=0?

1 Answer
Mar 18, 2016

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

Explanation:

You can factor it but here is a shorter trick

x^3+3x^2-x-3=0x3+3x2x3=0

x^3+3x^2=x + 3x3+3x2=x+3

x^2(x + 3)=x + 3x2(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"