Vieta's Formulas: Practice

The sum of the roots of a cubic is $-\frac{b}{a}$, and here $a = 1$ with $b = k$, so the sum is $-k$.

Set that equal to the given sum:

$-k = 5 \quad\Longrightarrow\quad k = -5$

The other coefficients say nothing about the sum, so $-4$ and $6$ are decoys.

Vieta's first formula is a statement about the whole list of roots, so the third root is whatever makes the total come out right.

$2 + (-3) + r_3 = 4 \quad\Longrightarrow\quad r_3 = 4 + 1 = 5$

Once two roots are known, the sum formula pins the last one exactly.

Compute the three symmetric sums from the roots, then drop them into the monic template $x^3 - e_1x^2 + e_2x - e_3$.

The sums are $e_1 = 1 - 2 + 4 = 3$, then $e_2 = (1)(-2) + (1)(4) + (-2)(4) = -2 + 4 - 8 = -6$, and $e_3 = (1)(-2)(4) = -8$.

$x^3 - 3x^2 + (-6)x - (-8) = x^3 - 3x^2 - 6x + 8$

The signs alternate, so the constant term is $-e_3 = 8$, not $-8$.

For a monic cubic the constant term is $-e_3$, so first find the product of all three roots.

$(1 + 2i)(1 - 2i) = 1 - 4i^2 = 5, \qquad e_3 = 3 \cdot 5 = 15$

Now flip the sign:

$\text{constant} = -e_3 = -15$

The cubic is $x^3 - 5x^2 + 11x - 15$.

Read the symmetric sums off the coefficients: $e_1 = -\frac{-2}{1} = 2$ and $e_2 = \frac{-5}{1} = -5$.

The identity $e_1^2 = (r_1 + r_2 + r_3)^2$ expands into the squares plus twice every pairwise product, so

$r_1^2 + r_2^2 + r_3^2 = e_1^2 - 2e_2 = 4 - 2(-5) = 14$

Check it against the actual roots $1$, $3$, $-2$: $1 + 9 + 4 = 14$.

That sum is $e_2 = (-1)^2\frac{a_2}{a_4} = \frac{a_2}{a_4}$, so read off the coefficient of $x^2$. The quartic has no $x^2$ term, which means $a_2 = 0$, not that the coefficient is unavailable.

$e_2 = \frac{0}{1} = 0$

The six pairwise products of the four roots cancel exactly.

The product of all the roots is $e_4 = (-1)^4\frac{a_0}{a_4}$, and $(-1)^4 = +1$, so no sign flip.

$r_1r_2r_3r_4 = \frac{-2}{1} = -2$

For a cubic the same slot would carry a minus sign; the parity of the degree decides it.

Three roots per term means $k = 3$, so the sign is $(-1)^3 = -1$ and the coefficient is $a_1$, three steps down from the leading one.

$e_3 = (-1)^3\,\frac{a_1}{a_4} = -\frac{7}{1} = -7$

The sum of the roots is $-\frac{a_3}{a_4}$, with the leading coefficient in the denominator.

$e_1 = -\frac{6}{2} = -3$

Skipping the division by $a_4 = 2$ gives $-6$, which is the trap on offer.

Two conditions have to hold at once. The sum of the roots is $-b$, so $-b = 2$ forces $b = -2$, which rules out both options with $+2x^2$.

The product of the roots is $-d$, so $-d = -8$ forces $d = 8$:

$-b = 2 \;\Rightarrow\; b = -2, \qquad -d = -8 \;\Rightarrow\; d = 8$

Only $x^3 - 2x^2 + 5x + 8$ has both. The option $x^3 - 2x^2 + 5x - 8$ gets the sum right but its roots multiply to $+8$.

The roots, counted with multiplicity, are $-1$, $2$, $2$. Both copies of the double root take part in the pairing.

$(-1)(2) + (-1)(2) + (2)(2) = -2 - 2 + 4 = 0$

Vieta agrees, and shows why: the cubic has no $x$ term, so $c = 0$ and $e_2 = \frac{c}{a} = 0$. Using only the distinct roots $-1$ and $2$ would have given $-2$, which is wrong.

For a quartic the constant term is $+e_4$, the product of all four roots, because $(-1)^4 = +1$.

$e_4 = (1)(-1)(2)(-2) = 4$

Expanding confirms it: $(x^2 - 1)(x^2 - 4) = x^4 - 5x^2 + 4$. Note the roots add to $0$, which is why the quartic has no $x^3$ term.