Vieta's Formula

For a quadratic polynomial with roots $r_1$ and $r_2$,

Expanding the right-hand side,

Comparing corresponding coefficients gives

and

Therefore,

For a cubic polynomial with roots $r_1,r_2,r_3$,

Expanding,

Comparing coefficients gives

and

More generally, suppose

We expand the right-hand side and compare corresponding coefficients term by term.

Coefficient of $x^n$: choosing $x$ from every factor gives

Coefficient of $x^{n-1}$: pick the constant term $-r_i$ from exactly one factor and $x$ from all others. Summing over the choice of factor gives

so

Coefficient of $x^{n-2}$: pick a constant term $-r_i$ from exactly two factors and $x$ from all remaining factors. Summing over the pairs gives

so

Coefficient of $x^{n-3}$: pick a constant term $-r_i$ from exactly three factors and $x$ from all remaining factors. Summing over the triples gives

so

Continuing in the same way, the coefficient of $x$ comes from picking the constant term from all but one factor:

and the constant term comes from picking the constant term from every factor:

Let $r_1,r_2$ be the roots of

Find

Find the quadratic polynomials that have roots $3$ and $-5$

The roots of

are $r_1,r_2,r_3$. Find a cubic whose roots are

Let $r_1,r_2,r_3$ be the roots of

Find

which is

Let the roots be $r_1,r_2,r_3,r_4,r_5,r_6$