Vieta's Formulas: Practice

Expanding $(x - r_1)(x - r_2)(x - r_3)$ puts $-(r_1 + r_2 + r_3)$ in front of $x^2$, so for a monic cubic the sum of the roots is the $x^2$ coefficient with its sign flipped.

Here $a = 1$ and $b = -6$:

$r_1 + r_2 + r_3 = -\frac{b}{a} = -\frac{-6}{1} = 6$

The roots are in fact $1$, $2$, and $3$, and they do add up to $6$.

The constant term of $(x - r_1)(x - r_2)(x - r_3)$ is $(-r_1)(-r_2)(-r_3) = -r_1r_2r_3$, so the product of the roots of a cubic is $-\frac{d}{a}$.

Here $a = 1$ and $d = 6$:

$r_1r_2r_3 = -\frac{6}{1} = -6$

Three roots means three minus signs, and $(-1)^3 = -1$, so the sign flips.

Divide by $a$ and expand the factorization. The constant term of $(x - r_1)(x - r_2)(x - r_3)$ is the product of the three constants $-r_1$, $-r_2$, and $-r_3$, which carries three minus signs.

$r_1r_2r_3 = (-1)^3\,\frac{d}{a} = -\frac{d}{a}$

The expression $-\frac{b}{a}$ is the sum of the roots and $\frac{c}{a}$ is the sum of their pairwise products.

Roots are counted with multiplicity, so the complete list is $1$, $2$, $2$: the double root is two entries, not one.

$1 + 2 + 2 = 5$

Vieta agrees, since $-\frac{b}{a} = -\frac{-5}{1} = 5$. Adding only the distinct values $1$ and $2$ would give $3$, and the formula would look broken.

The sum of the roots is $e_1 = -\frac{a_{n-1}}{a_n}$ at every degree; only the name of the coefficient changes.

With $a_4 = 1$ and $a_3 = 2$:

$r_1 + r_2 + r_3 + r_4 = -\frac{2}{1} = -2$

For a quadratic the product of the roots is $\frac{c}{a}$, with no sign change, because two roots contribute two minus signs and $(-1)^2 = 1$.

$r_1r_2 = \frac{12}{1} = 12$

The roots are $-3$ and $-4$: both are negative, and their product is positive.

The sum of the roots is $-\frac{b}{a}$, so the sign of the $x$ coefficient flips.

$r_1 + r_2 = -\frac{7}{1} = -7$

The roots $-3$ and $-4$ add to $-7$.

A monic cubic is $x^3 - e_1x^2 + e_2x - e_3$, so its constant term is $-e_3$, the negative of the product of the roots.

$e_3 = (-1)(2)(4) = -8 \quad\Longrightarrow\quad \text{constant} = -e_3 = 8$

Expanding confirms it: $(x + 1)(x - 2)(x - 4) = x^3 - 5x^2 + 2x + 8$.

Multiply the factorization out and collect the like terms.

$(x - r_1)(x - r_2) = x^2 - r_2x - r_1x + r_1r_2 = x^2 - (r_1 + r_2)x + r_1r_2$

One root gives one minus sign, so the $x$ coefficient is negative; two roots give two minus signs, so the constant is positive.

The root $1$ has multiplicity $3$, so the complete list of roots is $1$, $1$, $1$. All three entries count.

$1 + 1 + 1 = 3$

Vieta agrees: $-\frac{b}{a} = -\frac{-3}{1} = 3$. A polynomial of degree $3$ always has exactly three roots on the list, even when they are all the same number.

The cubic has no $x^2$ term, which does not mean the coefficient is missing. It means the coefficient is zero: $b = 0$.

$r_1 + r_2 + r_3 = -\frac{b}{a} = -\frac{0}{1} = 0$

The three roots add to exactly $0$, which is a real constraint on them, not an absence of information.

For a cubic the product of the roots is $-\frac{d}{a}$, so flip the sign of the constant term and divide by the leading coefficient.

With $a = 2$ and $d = -3$:

$r_1r_2r_3 = -\frac{-3}{2} = \frac{3}{2}$

Two sign slips are on offer here (forgetting the minus, or forgetting to divide by $2$), and each one produces a different wrong answer.