How to derive the polynomial for the squares of the roots of $x^3+4x+1=0$

Key takeaway: This is a neat symmetric-polynomial move. Instead of solving the cubic directly, we track what happens after squaring the roots and rebuild the polynomial from the transformed symmetric sums.

Let the roots of

$x^3+4x+1=0$

be $\alpha,\beta,\gamma$. By Vieta's formulas,

• $\alpha+\beta+\gamma=0$
• $\alpha\beta+\beta\gamma+\gamma\alpha=4$
• $\alpha\beta\gamma=-1$

We want a polynomial whose roots are $\alpha^2,\beta^2,\gamma^2$.

Step 1: Find the sum of the squared roots

$\alpha^2+\beta^2+\gamma^2=(\alpha+\beta+\gamma)^2-2(\alpha\beta+\beta\gamma+\gamma\alpha)$

So

$\alpha^2+\beta^2+\gamma^2=0^2-2(4)=-8.$

Step 2: Find the pairwise product sum

Use

$\alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2=(\alpha\beta+\beta\gamma+\gamma\alpha)^2-2\alpha\beta\gamma(\alpha+\beta+\gamma).$

Since $\alpha+\beta+\gamma=0$,

$\alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2=4^2-2( -1)(0)=16.$

Step 3: Find the product of the squared roots

$\alpha^2\beta^2\gamma^2=(\alpha\beta\gamma)^2=(-1)^2=1.$

Step 4: Build the monic cubic

A monic cubic with roots $\alpha^2,\beta^2,\gamma^2$ is

$y^3-(\alpha^2+\beta^2+\gamma^2)y^2+(\alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2)y-\alpha^2\beta^2\gamma^2=0.$

Substitute the values:

$y^3-(-8)y^2+16y-1=0,$

which gives

$\boxed{y^3+8y^2+16y-1=0}.$

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Pitfalls the pros know 👇 A common slip is to square the original equation term-by-term. That does not preserve the root structure. The safe route is to use Vieta's formulas on the roots themselves, then rebuild the polynomial from the symmetric sums.

What if the problem changes? If the original cubic were $x^3+px+q=0$, the same method still works: first compute $\alpha+\beta+\gamma$, $\alpha\beta+\beta\gamma+\gamma\alpha$, and $\alpha\beta\gamma$, then use them to form the monic polynomial for $\alpha^2,\beta^2,\gamma^2$. The coefficients will change, but the pattern stays the same.

Tags: Vieta's formulas, symmetric polynomials, root transformation