Vieta's Formulas

Intuition

In the quadratics chapter you learned two facts about $ax^2 + bx + c$ that probably felt like lucky accidents: the two roots add up to $-\frac{b}{a}$ , and they multiply to $\frac{c}{a}$ . Why does one of them carry a minus sign and the other one not? Why should the coefficients know anything about the roots without being asked? Those two formulas were not accidents, and they were not the end of the list. They are the first two members of a family that keeps going for every degree, and the whole family drops out of one honest computation you can already do. Here is that computation: take the complete factorization the Fundamental Theorem of Algebra hands you, multiply it out, and compare it with the coefficients you started with. Everything else in this lesson is bookkeeping on that single idea.

Learning goals

Expand $a_n(x - r_1)\cdots(x - r_n)$ and match coefficients
Write $e_k = (-1)^k \tfrac{a_{n-k}}{a_n}$ for every $k$
Trace the sign to the $k$ minus signs a term drags
Recover the quadratic sum and product as the $n = 2$ case
Include every root, repeats and nonreal ones alike
Compute a symmetric function without finding the roots

Two ways to write the same polynomial

Take any polynomial of degree $n$ with a nonzero leading coefficient,

p(x) = a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0, \qquad a_n \ne 0.

The Fundamental Theorem of Algebra says $p$ has $n$ roots over the complex numbers once they are counted with multiplicity. The factor theorem then turns each of those roots into a factor, so $p$ splits completely:

p(x) = a_n(x - r_1)(x - r_2)\cdots(x - r_n).

Those two lines describe the same object in two different languages. The first one displays the coefficients and hides the roots. The second one displays the roots and hides the coefficients. They are equal, so multiplying out the second and comparing it against the first cannot fail to tell you how the roots and the coefficients are tied to each other. That comparison is the whole content of this lesson.

Start with the degree you have already seen. Expanding two factors is one application of the distributive law:

(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.

Divide $ax^2 + bx + c$ by $a$ so that both sides are monic, and set them side by side:

x^2 + \frac{b}{a}x + \frac{c}{a} = x^2 - (r_1 + r_2)x + r_1r_2.

Reading off the coefficient of $x$ gives $r_1 + r_2 = -\frac{b}{a}$ , and reading off the constant term gives $r_1r_2 = \frac{c}{a}$ . There is the quadratics chapter, recovered in three lines. Now run exactly the same argument one degree higher and watch a formula appear that has no quadratic counterpart at all.

The cubic, expanded slowly

Multiply the three factors in two stages. Stage one is the product you just did, and stage two multiplies it by $(x - r_3)$ , distributing across all three terms:

\begin{aligned} \bigl[x^2 - (r_1 + r_2)x + r_1r_2\bigr](x - r_3) &= x^3 - (r_1 + r_2)x^2 + r_1r_2\,x \\[2pt] &\quad - r_3x^2 + (r_1 + r_2)r_3\,x - r_1r_2r_3. \end{aligned}

Now collect like powers. The two $x^2$ terms merge into $-(r_1 + r_2 + r_3)x^2$ . The two $x$ terms merge into $(r_1r_2 + r_1r_3 + r_2r_3)x$ , because $(r_1 + r_2)r_3 = r_1r_3 + r_2r_3$ :

\begin{aligned} (x - r_1)(x - r_2)(x - r_3) &= x^3 - (r_1 + r_2 + r_3)x^2 \\ &\quad + (r_1r_2 + r_1r_3 + r_2r_3)x - r_1r_2r_3. \end{aligned}

Stop and look at what the coefficients turned out to be, because the whole pattern is already visible. The coefficient of $x^2$ collects the roots one at a time. The coefficient of $x$ collects them two at a time, every possible pair exactly once. The constant term collects all three at once. And the signs alternate: minus, plus, minus.

Matching against a general cubic is now mechanical. Divide $ax^3 + bx^2 + cx + d$ by $a$ and compare:

x^3 + \frac{b}{a}x^2 + \frac{c}{a}x + \frac{d}{a} = x^3 - (r_1 + r_2 + r_3)x^2 + (r_1r_2 + r_1r_3 + r_2r_3)x - r_1r_2r_3.

Equal polynomials have equal coefficients, so

r_1 + r_2 + r_3 = -\frac{b}{a}, \qquad r_1r_2 + r_1r_3 + r_2r_3 = \frac{c}{a}, \qquad r_1r_2r_3 = -\frac{d}{a}.

The first of those is the old sum formula wearing a third root. The last of those is the old product formula, though it has picked up a minus sign the quadratic version did not have. The middle one is where the pattern’s real shape first shows. A quadratic has a “sum of the products of two roots” as well, namely $r_1r_2$ , but a quadratic offers only one pair. So that sum has a single term, and it can pass for nothing more than “the product of the roots”. A cubic offers three pairs, so the same quantity finally has several terms to add up and can no longer be mistaken for anything else. The list of formulas grew by one as well, and it will grow by one more at every degree from here on.

Worked example 1 Read the three symmetric sums off $2x^3 + 5x^2 - 4x + 6$

Match the polynomial to $ax^3 + bx^2 + cx + d$ : here $a = 2$ , $b = 5$ , $c = -4$ , and $d = 6$ . Nothing has to be factored, and no root has to be found.

The sum of the roots is $-\frac{b}{a}$ :

r_1 + r_2 + r_3 = -\frac{5}{2}.

The sum of the products of pairs is $\frac{c}{a}$ :

r_1r_2 + r_1r_3 + r_2r_3 = \frac{-4}{2} = -2.

The product of all three roots is $-\frac{d}{a}$ :

r_1r_2r_3 = -\frac{6}{2} = -3.

Every formula divides by the leading coefficient $a = 2$ , which is the step most often dropped. If the cubic had been monic, the three answers would have been the coefficients themselves with their signs flipped or kept. But this cubic is not monic, so each of the three answers is halved.

Check your understanding

For the cubic $x^3 + 2x^2 - 7x + 4$ , what is the product $r_1r_2r_3$ of its three roots?

4

-4

-7

7

Where the minus signs come from

The alternating sign is the part most students copy down without believing, so watch it being born.

Each factor of the product $(x - r_1)(x - r_2)\cdots(x - r_n)$ offers exactly two things: the $x$ , or the $-r_i$ . Expanding the product means choosing one of those two from every factor, multiplying the $n$ choices together, and adding up the result over all $2^n$ ways to choose. A product of binomials contains nothing else.

Suppose one particular choice takes the root from $k$ of the factors and the $x$ from the other $n - k$ . The factors that gave an $x$ contribute $x^{n-k}$ . The factors that gave a root do not contribute bare roots: each one arrives with the minus sign that sits in front of it inside its own factor. So $k$ roots drag exactly $k$ minus signs along with them, and those multiply to

\underbrace{(-1)(-1)\cdots(-1)}_{k \text{ of them}} = (-1)^k.

A term that uses $k$ roots therefore carries the sign $(-1)^k$ and the power $x^{n-k}$ : negative when $k$ is odd, positive when $k$ is even. The sign has nothing to do with whether you happen to call the result a “sum” or a “product”. It counts how many roots the term used, and that is all it does.

Every term of the expansion takes one entry from each factor. Take the root from two of the three factors and you take two minus signs with it, so the term comes out positive, while the one leftover x fixes the power.

The whole family at once

Give these sums a name so the general result fits on one line. For roots $r_1, r_2, \ldots, r_n$ , define

\begin{aligned} e_1 &= r_1 + r_2 + \cdots + r_n, \\[2pt] e_2 &= r_1r_2 + r_1r_3 + \cdots + r_{n-1}r_n, \\[2pt] e_3 &= r_1r_2r_3 + r_1r_2r_4 + \cdots + r_{n-2}r_{n-1}r_n, \\[2pt] &\;\;\;\vdots \\[2pt] e_n &= r_1r_2\cdots r_n. \end{aligned}

In words: $e_k$ adds up every product of $k$ of the roots, where each product uses $k$ different entries of the list $r_1, \ldots, r_n$ . “Different entries” is not the same as “different values”: if a root appears twice in the list, both copies are available. Those two copies are treated as two separate roots that happen to be equal. These sums are the elementary symmetric functions of the roots. Symmetric, because swapping any two roots leaves every $e_k$ unchanged: you get back the same collection of products, merely written in a different order.

With that name in hand, the expansion argument of the last section says exactly this.

Vieta's formulas for a polynomial of degree $n$ #

Let $p(x) = a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0$ with $a_n \ne 0$ , and let $r_1, \ldots, r_n$ be its roots over the complex numbers, each listed as many times as its multiplicity. The Fundamental Theorem of Algebra, applied repeatedly with the factor theorem, gives the complete factorization $p(x) = a_n(x - r_1)(x - r_2)\cdots(x - r_n)$ . Dividing by $a_n$ makes both sides monic:

x^n + \frac{a_{n-1}}{a_n}x^{n-1} + \cdots + \frac{a_1}{a_n}x + \frac{a_0}{a_n} = (x - r_1)(x - r_2)\cdots(x - r_n).

Expand the right side by the distributive law. Multiplying out $n$ binomials means taking one entry from each factor, either the $x$ or the $-r_i$ . Then multiply those $n$ chosen entries together, and sum the resulting products over every one of the $2^n$ ways to choose. Fix one such choice and suppose it takes the roots from exactly $k$ of the factors. The other $n - k$ factors contribute an $x$ apiece, giving $x^{n-k}$ , and each of the $k$ chosen roots arrives with a minus sign in front of it. So the $k$ minus signs carried by those chosen roots multiply together to $(-1)^k$ . That choice therefore contributes $(-1)^k$ times a product of $k$ of the roots, times $x^{n-k}$ .

Group the choices by the value of $k$ . Every way of taking $k$ of the roots occurs exactly once, and each one produces the same power $x^{n-k}$ and the same sign $(-1)^k$ . So the full coefficient of $x^{n-k}$ is $(-1)^k$ times the sum of all products of $k$ roots, which is $(-1)^ke_k$ by definition. Letting $k$ run from $0$ (take no roots at all, which leaves the leading $x^n$ ) up to $n$ (take every root, which leaves no $x$ at all),

(x - r_1)(x - r_2)\cdots(x - r_n) = x^n - e_1x^{n-1} + e_2x^{n-2} - \cdots + (-1)^ne_n.

One step is left, and it is worth saying why it is legitimate rather than waving it through. If two polynomials are equal, their difference vanishes at every value of $x$ . A nonzero polynomial of degree $d$ has at most $d$ roots, since each root peels off a linear factor and the degree cannot go below zero. So a polynomial that vanishes at infinitely many values of $x$ must be the zero polynomial, and every one of its coefficients is $0$ . Equal polynomials therefore have identical coefficients, one power of $x$ at a time, and comparing them is honest.

So compare the coefficient of $x^{n-k}$ on the two sides of the monic identity. On the left it is $\frac{a_{n-k}}{a_n}$ ; on the right it is $(-1)^ke_k$ . Hence $(-1)^ke_k = \frac{a_{n-k}}{a_n}$ , and multiplying both sides by $(-1)^k$ , which is legal because $(-1)^k(-1)^k = 1$ , gives

e_k = (-1)^k\,\frac{a_{n-k}}{a_n}, \qquad k = 1, 2, \ldots, n.

That one line is the entire family, and every member of it was already sitting inside the factorization waiting to be multiplied out.

The formula is easiest to use as a table. Read the coefficient that sits $k$ steps down from the leading one, divide by $a_n$ , and attach the sign $(-1)^k$ :

$k$	$e_k$ collects	its sign	its value
$1$	every root, added up	$-$	$-\frac{a_{n-1}}{a_n}$
$2$	every product of two roots	$+$	$+\frac{a_{n-2}}{a_n}$
$3$	every product of three roots	$-$	$-\frac{a_{n-3}}{a_n}$
$k$	every product of $k$ roots	$(-1)^k$	$(-1)^k\frac{a_{n-k}}{a_n}$

For monic polynomials the pattern is bare enough to memorize by looking at it:

\begin{aligned} (x - r_1)(x - r_2) &= x^2 - e_1x + e_2, \\[2pt] (x - r_1)(x - r_2)(x - r_3) &= x^3 - e_1x^2 + e_2x - e_3, \\[2pt] (x - r_1)(x - r_2)(x - r_3)(x - r_4) &= x^4 - e_1x^3 + e_2x^2 - e_3x + e_4. \end{aligned}

The signs march minus, plus, minus, plus across every row, and the sign on the constant term depends on the degree. That term is $-e_3$ for the cubic but $+e_4$ for the quartic, because $(-1)^3 = -1$ while $(-1)^4 = +1$ . That single flip is the most common place to go wrong. So read the parity of $k$ every time rather than trusting a memory of “the product of the roots is negative”.

Check your understanding

A monic quartic $x^4 + a_3x^3 + a_2x^2 + a_1x + a_0$ has roots $r_1, r_2, r_3, r_4$ . What is the product $r_1r_2r_3r_4$ ?

a_0

-a_0

-a_1

a_2

The quadratic case, explained

Now set $n = 2$ in the general formula and write the coefficients the way the quadratics chapter did, with $a_2 = a$ , $a_1 = b$ , and $a_0 = c$ :

e_1 = r_1 + r_2 = (-1)^1\frac{a_1}{a_2} = -\frac{b}{a}, \qquad e_2 = r_1r_2 = (-1)^2\frac{a_0}{a_2} = \frac{c}{a}.

Those are precisely the two formulas you were handed earlier, and now they are not two separate facts to memorize but two instances of one rule. The puzzle in the opening callout is answered too. The sum uses one root per term, so it drags along one minus sign and comes out negative. The product uses two roots, so it drags along two, and the two cancel. The asymmetry was never about sums versus products. It was about whether $k$ is odd or even, and a quadratic is simply too short for the pattern to show itself.

Every root belongs on the list

Vieta’s formulas are statements about the complete list of roots that the Fundamental Theorem of Algebra provides: $n$ of them, counted with multiplicity, taken over the complex numbers. Leave any one out and the formulas are not approximately true, they are false. Two cubics that differ only in their constant term make the point.

The first is $x^3 - 5x^2 + 8x - 4$ , which factors as $(x - 1)(x - 2)^2$ . Its roots are $1$ , $2$ , and $2$ : the double root is two entries in the list, not one. Vieta says the roots sum to $-\frac{-5}{1} = 5$ , and indeed $1 + 2 + 2 = 5$ . Had you listed only the distinct values $1$ and $2$ , you would have got $3$ and concluded the formula was broken. Multiplicity is not bookkeeping pedantry; the formula is simply false without it. The other two check out as well: $1 \cdot 2 + 1 \cdot 2 + 2 \cdot 2 = 8 = \frac{c}{a}$ , using both copies of the root $2$ , and $1 \cdot 2 \cdot 2 = 4 = -\frac{d}{a}$ .

The second cubic, $x^3 - 5x^2 + 8x - 6$ , differs only in its constant term, and it has a single real root. That does not mean it has a single root.

Worked example 2 Verify Vieta's formulas for $x^3 - 5x^2 + 8x - 6$ , whose roots are $3$ , $1 + i$ , $1 - i$

Dividing the cubic by $x - 3$ leaves the quadratic factor $x^2 - 2x + 2$ , whose discriminant is $4 - 8 = -4$ , so its two roots are the complex conjugates $1 + i$ and $1 - i$ . The complete list of roots is therefore $3$ , $1 + i$ , $1 - i$ , and Vieta’s formulas describe that list, not just its real member.

The sum of the roots should be $-\frac{b}{a} = 5$ . The two imaginary parts cancel:

3 + (1 + i) + (1 - i) = 3 + 2 = 5.

The sum of the products of pairs should be $\frac{c}{a} = 8$ . Use $i^2 = -1$ on the last product:

\begin{aligned} 3(1 + i) + 3(1 - i) + (1 + i)(1 - i) &= (3 + 3i) + (3 - 3i) + (1 - i^2) \\ &= 6 + 2 = 8. \end{aligned}

The product of all three should be $-\frac{d}{a} = 6$ :

3(1 + i)(1 - i) = 3 \cdot 2 = 6.

All three agree. Now try the same three checks with only the real root $3$ on the list: the sum is $3$ , not $5$ , and the product is $3$ , not $6$ . Every formula fails at once. This is the single most common way Vieta’s formulas are misapplied, and note why the answers came out real despite the complex roots. For a polynomial with real coefficients the non-real roots arrive in conjugate pairs, and a conjugate pair has a real sum ( $2$ here) and a real product ( $2$ here).

Check your understanding

The cubic $x^3 - x^2 - x + 1$ factors as $(x - 1)^2(x + 1)$ . What is the sum of its roots?

0

2

1

-1

Reading the formulas backwards

The formulas run in both directions. Given the coefficients they hand you the symmetric sums, and given the roots they hand you the coefficients. The second direction is how a polynomial gets built from a prescribed set of roots.

Worked example 3 Build the monic cubic whose roots are $2$ , $1 + i$ , and $1 - i$

A monic cubic is $x^3 - e_1x^2 + e_2x - e_3$ , so all three symmetric sums are needed. Compute them directly from the roots.

The sum of the roots:

e_1 = 2 + (1 + i) + (1 - i) = 4.

The sum of the products of pairs, using $(1 + i)(1 - i) = 1 - i^2 = 2$ :

e_2 = 2(1 + i) + 2(1 - i) + (1 + i)(1 - i) = (2 + 2i) + (2 - 2i) + 2 = 6.

The product of all three roots:

e_3 = 2(1 + i)(1 - i) = 2 \cdot 2 = 4.

Substituting into the monic template, with the signs alternating minus, plus, minus:

p(x) = x^3 - 4x^2 + 6x - 4.

Check it the long way if you like: $(x - 2)(x^2 - 2x + 2) = x^3 - 4x^2 + 6x - 4$ , the same polynomial. Every coefficient came out real even though two of the roots were not. The reason is that the conjugate pair was kept together on the list, and its imaginary parts cancelled in each symmetric sum. Drop $1 - i$ and keep $1 + i$ , and no polynomial with real coefficients can have that root set at all.

Symmetric facts without the roots

Here is what the formulas are actually for. A quantity built from the roots that survives any relabelling of them can be rewritten in terms of $e_1, e_2, \ldots, e_n$ , and Vieta hands you those for free, straight off the coefficients. So you can compute such a quantity for a polynomial whose roots you have no hope of finding.

The classic example is the sum of the squares of the roots. Square $e_1$ and see what appears:

(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).

The cross terms are exactly $2e_2$ , so subtracting them off isolates what you want:

r_1^2 + r_2^2 + r_3^2 = e_1^2 - 2e_2.

Nothing in that derivation cared that there were three roots. Squaring a sum always produces the squares plus twice every distinct pair, so the identity holds at any degree, which is why it is worth learning once.

Worked example 4 Find $r_1^2 + r_2^2 + r_3^2$ for $2x^3 - 6x^2 + x - 5$ without finding a single root

First check that the roots really are out of reach. The rational root theorem allows only $\pm 1, \pm 5, \pm\frac{1}{2}, \pm\frac{5}{2}$ as rational candidates, and substituting each one gives a nonzero value, so this cubic has no rational root at all. Finding its roots by hand is not on the table.

The symmetric sums, on the other hand, are two lines of arithmetic. With $a = 2$ , $b = -6$ , $c = 1$ :

e_1 = -\frac{b}{a} = -\frac{-6}{2} = 3, \qquad e_2 = \frac{c}{a} = \frac{1}{2}.

Now apply the identity:

r_1^2 + r_2^2 + r_3^2 = e_1^2 - 2e_2 = 3^2 - 2 \cdot \frac{1}{2} = 9 - 1 = 8.

The three roots are unknown, quite possibly ugly, and two of them may not even be real, and yet the sum of their squares is exactly $8$ . That is the whole point of the method: symmetric information about the roots is available without the roots.

The same trick answers plenty of other questions. The sum of the reciprocals of the roots of a cubic, for instance, needs only a common denominator:

\frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} = \frac{r_2r_3 + r_1r_3 + r_1r_2}{r_1r_2r_3} = \frac{e_2}{e_3}.

That is legal exactly when no root is $0$ , and a root of $0$ happens exactly when the constant term is $0$ , since $p(0) = a_0$ . For the cubic $2x^3 - 6x^2 + x - 5$ above, $e_3 = -\frac{-5}{2} = \frac{5}{2}$ , so the reciprocals of its three unknown roots add up to $\frac{1/2}{5/2} = \frac{1}{5}$ .

Worked example 5 Find the symmetric sums of the quartic $3x^4 - 6x^3 + 2x - 9$

Write out every coefficient first, including the one that is missing. The quartic has no $x^2$ term, which does not mean there is no coefficient there; it means the coefficient is zero:

a_4 = 3, \qquad a_3 = -6, \qquad a_2 = 0, \qquad a_1 = 2, \qquad a_0 = -9.

Now apply $e_k = (-1)^k \frac{a_{4-k}}{a_4}$ four times, watching the sign flip with the parity of $k$ :

\begin{aligned} e_1 &= -\frac{a_3}{a_4} = -\frac{-6}{3} = 2, \\[2pt] e_2 &= +\frac{a_2}{a_4} = \frac{0}{3} = 0, \\[2pt] e_3 &= -\frac{a_1}{a_4} = -\frac{2}{3}, \\[2pt] e_4 &= +\frac{a_0}{a_4} = \frac{-9}{3} = -3. \end{aligned}

Two things are worth pausing on. The missing $x^2$ term is a real constraint, not a gap: it says the four roots pair up so that all six of their pairwise products cancel to exactly $0$ . And the product of the four roots is $-3$ , taken with a plus sign in the formula, because $k = 4$ is even. A cubic in the same position would have carried a minus sign. Read the parity, not your memory.

Common mistakes

Forgetting to divide by the leading coefficient. Every formula has $a_n$ underneath it. For $2x^3 - 6x^2 + x - 5$ the roots sum to $-\frac{-6}{2} = 3$ , not $6$ .
Losing the alternating sign. The sign is $(-1)^k$ , decided by how many roots the term collects. The product of the roots of a cubic is $-\frac{d}{a}$ , but the product of the roots of a quartic is $+\frac{a_0}{a_4}$ . Check the parity of $k$ every single time.
Listing only the distinct roots. A double root is two entries. The roots of $x^3 - 5x^2 + 8x - 4 = (x - 1)(x - 2)^2$ are $1, 2, 2$ and they sum to $5$ , not $3$ .
Leaving the non-real roots out. The formulas describe the complete list over the complex numbers. The roots of $x^3 - 5x^2 + 8x - 6$ are $3, 1 + i, 1 - i$ ; using $3$ alone makes every formula fail.
Treating a missing term as a missing formula. No $x^2$ term means $a_2 = 0$ , so the matching symmetric sum is $0$ . That is information, not an absence of it.
Mistaking $e_2$ for a sum of squares. $e_2$ collects the products of two different entries of the root list ( $r_1r_2 + r_1r_3 + r_2r_3$ ), never $r_1^2 + r_2^2 + r_3^2$ . Two entries may still hold the same value, so a repeated root does contribute its own square: for the roots $1, 2, 2$ the term $2 \cdot 2$ belongs in $e_2$ . The sum of squares is instead the derived quantity $e_1^2 - 2e_2$ .

Key takeaways

The Fundamental Theorem of Algebra gives $p(x) = a_n(x - r_1)\cdots(x - r_n)$ with all $n$ roots listed with multiplicity over the complex numbers. Expanding that product and matching coefficients is Vieta’s formulas. There is no other ingredient.
The coefficient of $x^{n-k}$ collects every product of $k$ of the roots, with the sign $(-1)^k$ , which is the single formula $e_k = (-1)^k \frac{a_{n-k}}{a_n}$ for every $k$ from $1$ to $n$ .
The sign is born in the factor $(x - r_i)$ . A term that uses $k$ roots drags $k$ minus signs along, and they multiply to $(-1)^k$ : nothing to do with sums versus products.
Monic cubic: $x^3 - e_1x^2 + e_2x - e_3$ . Monic quadratic: $x^2 - e_1x + e_2$ , which is the quadratics chapter’s $r_1 + r_2 = -\frac{b}{a}$ and $r_1r_2 = \frac{c}{a}$ , now explained rather than asserted.
Every root must be on the list: repeats repeated, non-real roots included. Drop one and the formulas are false, not approximate.
Vieta gives symmetric information about the roots without finding them, for example $r_1^2 + r_2^2 + \cdots = e_1^2 - 2e_2$ .

Vieta's Formulas

Learning goals

Two ways to write the same polynomial

The cubic, expanded slowly

Where the minus signs come from

The whole family at once

The quadratic case, explained

Every root belongs on the list

Reading the formulas backwards

Symmetric facts without the roots

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Free Response Questions (FRQ)