Vieta's Formulas: Practice

Read the two symmetric sums off the coefficients: $e_1 = -\frac{-4}{1} = 4$ and $e_2 = \frac{5}{1} = 5$.

Squaring $e_1$ produces the squares plus twice every pairwise product, so subtract that surplus:

$r_1^2 + r_2^2 + r_3^2 = e_1^2 - 2e_2 = 16 - 10 = 6$

The cubic is $(x - 1)^2(x - 2)$, so the roots are $1$, $1$, $2$, and $1 + 1 + 4 = 6$ as promised.

Put the four fractions over the common denominator $r_1r_2r_3r_4$. Each numerator term omits one root, so the numerator is the sum of the products of the roots taken three at a time.

$\frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} + \frac{1}{r_4} = \frac{e_3}{e_4}$

Vieta gives $e_3 = (-1)^3\frac{a_1}{a_4} = -\frac{-4}{1} = 4$ and $e_4 = (-1)^4\frac{a_0}{a_4} = 5$, so the sum is $\frac{4}{5}$.

The cubic has no $x^2$ term, so $b = 0$ and the roots sum to zero: $e_1 = 0$. The coefficient of $x$ gives $e_2 = 6$.

$r_1^2 + r_2^2 + r_3^2 = e_1^2 - 2e_2 = 0 - 12 = -12$

A negative answer is not a mistake here. A sum of squares can only be negative if some of the roots are not real, and this cubic does have two complex roots.

The constant term of a monic quartic is $+e_4$, since $(-1)^4 = +1$.

$e_4 = (1)(1)(-2)(3) = -6$

So the constant term is $-6$. Do not flip the sign here: a cubic would flip it, a quartic does not. The quartic is $x^4 - 3x^3 - 3x^2 + 11x - 6$.

Squaring the sum of the roots gives the squares plus twice every pairwise product, so the difference asked for is exactly the doubled cross terms.

$(r_1 + r_2 + r_3)^2 - (r_1^2 + r_2^2 + r_3^2) = 2(r_1r_2 + r_1r_3 + r_2r_3) = 2e_2$

Vieta gives $e_2 = \frac{c}{a} = -10$, so the value is $2(-10) = -20$. Neither $e_1$ nor the roots themselves are needed.

Build the coefficients from the symmetric sums, remembering that the signs alternate.

The sums are $e_1 = 10$, then $e_2 = 6 + 10 + 15 = 31$, and $e_3 = 30$, so $b = -e_1 = -10$, $c = e_2 = 31$, and $d = -e_3 = -30$.

$b + c + d = -10 + 31 - 30 = -9$

A quick check: $p(1) = 1 + b + c + d$, and $p(1) = (1 - 2)(1 - 3)(1 - 5) = -8$, so $b + c + d = -8 - 1 = -9$.

Because the coefficients are real, the conjugate root theorem forces the conjugate $3 + i$ to be a root as well, so the complete list is $2$, $3 - i$, $3 + i$.

$2 + (3 - i) + (3 + i) = 8$

The imaginary parts cancel, which is why a real-coefficient polynomial can have complex roots and still hand you a real sum.

The third root is the conjugate $3 + i$. Multiply the conjugate pair first.

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

Then multiply by the remaining root:

$2 \cdot 10 = 20$

The conjugate $3 + i$ must also be a root, so the roots are $2$, $3 - i$, $3 + i$. Compute the three symmetric sums.

They are $e_1 = 8$, then $e_2 = 2(3 - i) + 2(3 + i) + (3 - i)(3 + i) = 12 + 10 = 22$, and $e_3 = 2 \cdot 10 = 20$.

$x^3 - e_1x^2 + e_2x - e_3 = x^3 - 8x^2 + 22x - 20$

Every coefficient came out real, exactly because the conjugate pair was kept together.

The product of all the roots is $e_n = (-1)^n\frac{a_0}{a_n}$, and here $n = 5$ is odd, so the sign flips.

$e_5 = (-1)^5\,\frac{4}{1} = -4$

The degree, not the shape of the polynomial, decides the sign: an odd degree flips it and an even degree does not.

Two roots per term means $k = 2$, so the sign is $(-1)^2 = +1$ and the coefficient is $a_{n-2} = a_3$, two steps down from the leading one.

$e_2 = (-1)^2\,\frac{a_3}{a_5} = \frac{-3}{1} = -3$

The answer is negative because the coefficient is, not because of any sign attached by the formula.

Compute all four symmetric sums. The roots pair off into $1, -1$ and $i, -i$, and each pair sums to zero, so $e_1 = 0$. Working out the six pairwise products and the four triple products gives $e_2 = 0$ and $e_3 = 0$ as well.

The product of all four is what survives:

$e_4 = (1)(-1)(i)(-i) = (-1)(-i^2) = (-1)(1) = -1$

So the quartic is $x^4 - e_1x^3 + e_2x^2 - e_3x + e_4 = x^4 - 1$, which is indeed $(x - 1)(x + 1)(x - i)(x + i)$.