Home > Algebra > Comparing Methods for Solving Quadratics

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

Answer 1

The sum of all coefficients of given cubic equation: #x^3-3x^2-x+3=0# is zero hence #x=1# is a root of given cubic equation i.e. #(x-1)# is a factor of #x^3-3x^3-x+3# Now, cubic algebraic polynomial: #x^3-3x^2-x+3# can be factorized as follows #x^3-3x^3-x+3# #=x^2(x-1)-2x(x-1)-3(x-1)# #=(x-1)(x^2-2x-3)# #=(x-1)(x^2-3x+x-3)# #=(x-1)(x(x-3)+(x-3))# #=(x-1)(x-3)(x+1)# hence, the solution of given cubic equation will be given as #(x-1)(x-3)(x+1)=0# #x=-1, 1, 3#

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Eli Assouline

To solve the equation (x^3 - 3x^2 - x + 3 = 0), you can use techniques such as factoring, synthetic division, or the rational root theorem. One method is to try potential rational roots using the rational root theorem and synthetic division, then use factoring or the quadratic formula on the remaining quadratic equation.

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.