Im stuck. Help please: Solve this equation: 2x^3 -10x^2 +13x -2 = 0

Solve this equation:

2x^3 -10x^2 +13x -2 = 0
Hi, I am sorry but we do not solve problems here (read the forum rules) but rather we assist YOU in solving the problems.

Please tell us what you have tried or if you have no idea how to proceed.

I'll assume that you do not know how to start.

My 1st step would be to see if you can factor this by grouping.

If that does not work, then I'd try using the rational root test theorem.

Please let us know how you make out.
 
I have tried factorisation to create a quadratic equation and solve it but could not be solved. I tried the rational root test theroem but cannot understand how to solve it.
 
I have tried factorisation to create a quadratic equation and solve it but could not be solved. I tried the rational root test theroem but cannot understand how to solve it.

By applying rational root theorem:

What were the possible rational roots for this polynomial?
 
I have tried factorisation to create a quadratic equation and solve it but could not be solved. I tried the rational root test theroem but cannot understand how to solve it.
Since an x was not common in all four terms you would not get a quadratic from this cubic equation.

What is troubling you with the rational root test?
 
x = 2 satisfies the given equation. Therefore x = 2 is one of the solution.

Therefore given equation is (x-2)(2x square - 6x + 1) = 0
 
x = 2 satisfies the given equation. Therefore x = 2 is one of the solution.

Therefore given equation is (x-2)(2x square - 6x + 1) = 0

@impel tutors
From your user name I assume you tutor. As such you really should know how to code in LaTeX.

[ tex](x-2)(x^2-6x+1)[ /tex] gives \(\displaystyle (x-2)(x^2-6x+1)\) removing the spaces in [ tex][ /tex]
 
x = 2 satisfies the given equation. Therefore x = 2 is one of the solution.

Therefore given equation is (x-2)(2x square - 6x + 1) = 0
Welcome aboard! As a tutor you should also know to help a student arrive at the result (or an intermediate result) rather than just giving a student an answer. This is also a policy of the forum. Thanks.
 
Thank you to all of you for helping me. I managed to understand it today.
 
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