How can I solve $y^{3}-3y^{2}+2=0$?
I am stuck at this equation $y^{3}-3y^{2}+2=0$. How do I solve it without calculator? It might be very trivial so I think I just need a hint. It is actually a substitution $y=\log x$,but I think it shouldn't matter much here.
$\endgroup$ 14 Answers
$\begingroup$The general way to solve these kinds of cubics is to use the rational roots test.
So you have $$y^{3}-3y^{2}+2=0$$
Basically, you take a look at the constant term, in this case $2$ and your coefficient in your highest degree is $1$. So you find the factors of constant term and divide it by coefficient of the highest terms. So you have:
$$\pm 1,\pm 2$$
Since, $1$ is the coefficient of the highest degree term, we don't have to worry about dividing it since it won't matter. Once we have found those values, we plug them into the equation and see which output a zero. In this case, it's $1$. So we have one root as
$$(y-1)$$
Then we take this root and long divide it with the original to get a quadratic. When you long divide it, you will get the following quadratic:
$$y^2 - 2y - 2$$
We can solve this quadratic $y^2 - 2y - 2 = 0$, using the quadratic formula to get the following roots:
$$ y= 1 \pm \sqrt{3} $$.
And that's that! We now have solved the equation to get the following roots
$$y = 1, 1 \pm \sqrt{3}$$
If you have any questions, be sure to comment and I'll try to clarify as much as possible.
A more detailed explanation of how the rational roots test works.
$\endgroup$ 2 $\begingroup$Another way: $y^3-3y^2+3-1 =y^3-1-3(y^2-1)=(y-1)(y^2+y+1)-3(y-1)(y+1)=(y-1)(y^2-2y-2)$
$\endgroup$ $\begingroup$Hint $\ $ It is clear that the sum of the polynomial's coefficients $= 0.$ But this sum is the same as evaluating the polynomial at $\, x = 1,\,$ hence $\,x=1\,$ is a root. By the Factor Theorem, the polynomial is divisible by $\,x-1,\,$ say $\,f(x) = (x-1) g(x).\,$ Since $\,g(x)\,$ is quadratic, you can use the quadratic formula to find its roots.
Alternatively, by the Rational Root Test we know that any rational root is an integer $\,n\,$ (since the lowest terms denominator must divide the lead coefficient $ = 1)$ and, furthermore, the test implies that $\,n\,$ divides $\,f(0) = 2.\,$ Thus there are only a few possible roots to test.
$\endgroup$ $\begingroup$$$y^3-3y^2+2=y^3-y^2-2y^2+2=y^2(y-1)-2(y^2-1)=(y-1)(y^2-2y-2)$$as $y^2-1=(y+1)(y-1)$.
Note the quadratic factor is irreducible over $\mathbb{Q}$ though it is reducible over $\mathbb{Q}(\sqrt3)$.
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