Can We Solve A Cubic Without Solving It?
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-1 and 2 are factors by inspection. With a bit of thought, you see that -1 is a double factor.
Yeah, this one is kind of obvious.
No you cant solve an equation without solving it...
Aww, man! 🤪
I got √2 + 2i as a solution. I don't see how anything else could work.
Well we can use transformation of roots
form a cubic equation on z = √x and divide that equation by coefficient of z^3.
Negate the coefficient of z^2 to get the desired answer
Now x^3 - 3 x - 2
= ( x + 1) ( x ^2 - x - 2)
= ( x + 1) ( x + 1) ( x - 2)
= ( x + 1) ^2 ( x - 2)
= ( z^2 + 1) ^2 * ( z^2 - 2)
= ( z ^2 + 1) ( z - √2)
* ( z^2 + 1 ) * ( z + √2)
= ( z^3 - √ 2 * z^2 + z - √2)
* ( z^3 + √ 2 * z^2 + z + √2)
Hereby desired answer is
Either √2
Or - √
Hereby
x^3 - 3 x - 2
x^3 + x ^2 - x^2 - x - 2 x - 2
= ( x + 1) ( x^2 - x - 2)
= ( x - 1) ( x - 2) ( x + 1)
Hereby √m + √ n + √k
= √( -1) + √ 2 + √( -1)
= √2 + i * 2
Multivalued roots again?
So is √2 + 2i also a solution.
And √2 - 2i
And don't forget √2 (√2 + i - i = √2 or √2 - i + i = √2)
but all of these answers are not accepted since there's no square root of negative number
Synthetic Division fails for x + 1 = 0 into 1 0 -3 -2 giving a false (x + 1) (x^2 - x - 2) result.
However, choosing the non repeating x - 2 = 0 by going into 1 0 -3 -2 we get x^2 + 2x + 1 = 0 part by Synthetic Division. Synthetic Division works if we choose the correct root that divides into the Polynomial. We then have, easier by Synthetic Division of (x - 2) into 1 0 -3 -2 forms
(x - 2)(x^2 + 2x + 1) = 0 correct answer! 😂🤣
Actually x^2 - x - 2 gives the
(x + 1)(x - 2) Quadratic Equation factoring. I got myself confused erroneously thinking I would get the repeated x + 1 or x^2 + 2x + 1 result. This is when using the easier method Synthetic Division causes confusion that isn't contradictory. I am just guessing why Synthetic Division is not a lectured method 3 we get in high school. 🤯
but without any relationships of m, n, k to the equation?
m,n,k are the roots of the equation
x = 2
Silly to use Latin form of his name, since he was French...