Vieta's Formulas: Additional Practice (Set 1)

Divide by the leading coefficient $a = 3$ in both symmetric sums: $e_1 = -\frac{-6}{3} = 2$ and $e_2 = \frac{9}{3} = 3$.

Now use the identity that comes from squaring the sum of the roots:

$r_1^2 + r_2^2 + r_3^2 = e_1^2 - 2e_2 = 4 - 6 = -2$

A negative sum of squares tells you at once that not all three roots are real.

The cubic is monic, so its complete factorization is $p(x) = (x - r_1)(x - r_2)(x - r_3)$ for every $x$. The product asked for is that expression evaluated at $x = 2$.

$p(2) = 2^3 - 3(2)^2 + 4(2) - 5 = 8 - 12 + 8 - 5 = -1$

Expanding instead gives $8 - 4e_1 + 2e_2 - e_3 = 8 - 12 + 8 - 5 = -1$, the same answer by the longer road.

Expanding the product collects every symmetric sum at once:

$(1 + r_1)(1 + r_2)(1 + r_3)(1 + r_4) = 1 + e_1 + e_2 + e_3 + e_4$

Vieta gives $e_1 = 2$, $e_2 = 1$, $e_3 = -3$, and $e_4 = -5$, so the value is $1 + 2 + 1 - 3 - 5 = -4$.

A faster route: the quartic is monic, so $p(-1) = (-1 - r_1)\cdots(-1 - r_4) = (-1)^4(1 + r_1)\cdots(1 + r_4)$, and $p(-1) = 1 + 2 + 1 - 3 - 5 = -4$.

Over the common denominator $r_1r_2r_3$, the numerator is the sum of the pairwise products, so the answer is $\frac{e_2}{e_3}$.

Vieta gives $e_2 = \frac{c}{a} = -5$ and $e_3 = -\frac{d}{a} = -\frac{-6}{1} = 6$.

$\frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} = \frac{e_2}{e_3} = \frac{-5}{6} = -\frac{5}{6}$

The roots are $-1$, $2$, $-3$, and $-1 + \frac{1}{2} - \frac{1}{3} = -\frac{5}{6}$.

This is $e_2 = (-1)^2\frac{a_3}{a_5}$, so read the coefficient of $x^3$. The quintic has no $x^3$ term, meaning $a_3 = 0$.

$e_2 = \frac{0}{2} = 0$

The absent term is a genuine constraint on the roots: all ten of their pairwise products cancel to zero. The $\frac{1}{2}$ on offer comes from misreading $a_2 = 1$ as the coefficient to use.

Real coefficients force the conjugate $1 - 3i$ to be the third root, so the complete list is $-2$, $1 + 3i$, $1 - 3i$.

$-2 + (1 + 3i) + (1 - 3i) = -2 + 2 = 0$

A sum of zero means the cubic has no $x^2$ term at all.

The third root is the conjugate $1 - 3i$. Take the conjugate pair first, since its product is real.

$(1 + 3i)(1 - 3i) = 1 - 9i^2 = 1 + 9 = 10$

Now bring in the real root:

$(-2)(10) = -20$

The roots are $-2$, $1 + 3i$, and $1 - 3i$. Compute the three symmetric sums.

The sum is $e_1 = 0$, the pairwise sum is $e_2 = (-2)(1 + 3i) + (-2)(1 - 3i) + (1 + 3i)(1 - 3i) = -4 + 10 = 6$, and the product is $e_3 = -20$.

$x^3 - e_1x^2 + e_2x - e_3 = x^3 - 0x^2 + 6x + 20 = x^3 + 6x + 20$

The missing $x^2$ term is not an accident: it is $e_1 = 0$ showing up as a coefficient.

The product of the roots is $e_4 = (-1)^4\frac{a_0}{a_4}$, and the constant term of this quartic is $0$.

$e_4 = \frac{0}{1} = 0$

A product of zero says one of the roots is $0$, which checks out: $p(0) = a_0 = 0$, so $0$ is a root exactly when the constant term vanishes. The quartic factors as $x(x - 1)^2(x - 2)$.

Apply $e_1 = -\frac{a_3}{a_4}$ with $a_3 = -4$ and $a_4 = 1$.

$e_1 = -\frac{-4}{1} = 4$

The roots are $0$, $1$, $1$, and $2$, which do add to $4$: the root $0$ contributes nothing to the sum but still occupies a place on the list.

Cube the sum of the roots and see what has to be removed:

$(r_1 + r_2)^3 = r_1^3 + r_2^3 + 3r_1r_2(r_1 + r_2)$

Rearranged, that says $r_1^3 + r_2^3 = e_1^3 - 3e_2e_1$. Vieta gives $e_1 = 4$ and $e_2 = 1$.

$r_1^3 + r_2^3 = 4^3 - 3(1)(4) = 64 - 12 = 52$

The roots are irrational, and they never had to be written down.

The sum of the pairwise products is $e_2 = \frac{c}{a}$, and here the coefficient of $x$ is $k$ with $a = 2$.

$\frac{k}{2} = 4 \quad\Longrightarrow\quad k = 8$

Forgetting the leading coefficient would leave $k = 4$, which is the distractor on offer.