Welcome! Today we're exploring The Three Cubic Formulas and the Alternating Sign Pattern — the natural extension of what you already know about quadratics.

You derived two formulas for quadratics. Now a cubic has three zeros — $α$ , $β$ , $γ$ — so you need three formulas.

The good news is they follow a beautiful pattern: the signs alternate.

Formula	Sign
First	minus−
Second	plus+
Third	minus−

If you remember that pattern, you never need to memorise three separate formulas.

In this lesson, you will:

Learn all three relationships between roots and coefficients
Practise the tricky one: the sum of pairwise products
Master listing all three pairs without missing any

1. All three cubic formulas with correct signs

You derived two formulas for quadratics. Now a cubic has three zeros — $α$ , $β$ , $γ$ — so you need three formulas.

The good news? They follow a beautiful pattern:

The signs alternate — minus, plus, minus.

📋 Given Info

For a cubic polynomial $a x^{3} + b x^{2} + c x + d$ with zeros $α$ , $β$ , and $γ$ , there are three relationships connecting the zeros to the coefficients.

The key to remembering them is the alternating sign pattern: $-\frac{b}{a}$ , $+\frac{c}{a}$ , $-\frac{d}{a}$

✍️ Question

Your turn ✍️

Write all three zeros-coefficients relationships for a cubic polynomial $a x^{3} + b x^{2} + c x + d$ with zeros $α$ , $β$ , $γ$ .

Remember the alternating sign pattern!

The three cubic formulas are:

Sum of zeros: $\alpha + \beta + \gamma = -\frac{b}{a}$ Don't forget the minus sign in front of b over a

Sum of pairwise products:Take zeros two at a time and add them up $\alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a}$ This is what taking zeros two at a time means
Product of zeros:The product of all three zeros together $\alpha\beta\gamma = -\frac{d}{a}$ Notice we're back to a minus sign here

✍️ MCQ

Choose one

What is the sign pattern in the three cubic formulas:

-\frac{b}{a}

,

+\frac{c}{a}

,

-\frac{d}{a}

?

✓

The sign pattern alternatesThis pattern is your shortcut for remembering: $-\frac{b}{a}$ minusFirst sign is minus, $+\frac{c}{a}$ plusSecond sign is plus, $-\frac{d}{a}$ minusThird sign is minus. Minus, plus, minus.You won't mix up signs even under pressure

This alternating patternDon't memorize blindly — understand where it comes from comes from expanding $(x - α) (x - β) (x - γ)$ : the alternating signs in the factors produce alternating signs in the relationships.

When you multiply out $(x - α) (x - β) (x - γ)$ , each factor has a minuskeyEach factor has a minus sign before its zero before the zero. These minus signs interact during expansion — sometimes they cancel (giving a plus), sometimes they don't (giving a minus). That's exactly why we get: minus, plus, minusThe expansion creates this minus, plus, minus pattern!

✍️ MCQ

Choose one

In the expansion of

(x - \alpha)(x - \beta)(x - \gamma)

, why does the sum of pairwise products

\alpha\beta + \beta\gamma + \gamma\alpha

have a positive sign?

✓

For quadratics we had: $- \frac{b}{a}$ , $+ \frac{c}{a}$ (minus, plusThis alternation is a rule for higher degree polynomials).

Cubics continue the pattern: $-\frac{b}{a}$ Just add one more term with opposite sign, $+\frac{c}{a}$ The whole pattern you need to remember, $-\frac{d}{a}$ Opposite sign again completes the pattern.

✍️ MCQ

Choose one

If we had a quartic polynomial

ax^4 + bx^3 + cx^2 + dx + e

, what sign would appear before

\frac{e}{a}

in the product of all four zeros?

✓

If you ever study quartics: $-\frac{b}{a}$ , $+\frac{c}{a}$ , $-\frac{d}{a}$ , $+\frac{e}{a}$ .

2. Computing the sum of pairwise products (three pairs)

You've seen how quadratics have two zeros and two formulas connecting them to coefficients.

Now we're working with cubics — and a cubic has three zeros: $α$ , $β$ , and $γ$ .

The first and third formulas are pretty straightforward. But the second one — the sum of pairwise productstricky! — is where mistakes tend to happen.

📋 Given Info

Three points labeled alpha, beta, gamma arranged in a triangle, with curved lines connecting each pair, labeled (alpha,beta), (beta,gamma), (gamma,alpha) to show the three pairs

Here's why it's tricky:

For three zeros, there are exactly three pairs:

$(α, β)$
$(β, γ)$
$(γ, α)$

You must compute each product separately and then add them all together.

⚠️ Missing even one pair gives the wrong sum.

✍️ Question

Your turn ✏️

The zeros of a cubic are $3$ , $-1$ , and $-\frac{1}{3}$ .

Compute the sum of pairwise products: $α \cdot β + β \cdot γ + γ \cdot α$

Three zeros produce exactly three pairsThe pattern: three zeros means three pairs. Write them as a checklist:

Pair 1: $α \times β = (3) (- 1) =$ $-3$ result

Pair 2: $β \times γ = (- 1) (- \frac{1}{3}) = + \frac{1}{3}$ (Negative times negative = positive)Sign rule: two negatives make a positive

Pair 3: $γ \times α = (- \frac{1}{3}) (3) =$ $-1$ resultMatching numbers simplify easily

✍️ FIB

Fill in the blank

What is

(-3) + \frac{1}{3} + (-1)

?

✓

Now add all three pairwise products:

-3 + \frac{1}{3} + (-1)</pen> = <pen actions="underline" underline-color="blue" underline-style="single" narrationText="group the whole numbers first">-3 - 1</pen> + \frac{1}{3} = <pen actions="highlight" highlight-color="green" narrationText="negative 4 plus one-third">-4 + \frac{1}{3}

Convert to common denominatorAlways convert to same denominator first:

$- \frac{12}{3} + \frac{1}{3} = - \frac{11}{3}$

So the sum of pairwise productsThis is the second Vieta's formula coefficient $α β + β γ + γ α =$ $-\frac{11}{3}$ answerThe coefficient from Vieta's second formula

✍️ MCQ

Choose one

According to the second cubic formula, the sum of pairwise products equals

\frac{c}{a}

. If our answer is

-\frac{11}{3}

, what does this tell us about the ratio

\frac{c}{a}

in the original cubic?

✓

⚠️ The most common error:Computing only two pairs instead of all three computing only two of the three pairsFormula needs all three, not just two.

If you miss the third pairforgotten!It wraps around so people miss it ( $γ \times α$ ), you get $- 3 + \frac{1}{3} = - \frac{8}{3}$ , which is wrong.

Always use the three-pair checklist.Alpha beta, beta gamma, gamma alpha every time

✍️ MCQ

Choose one

A student computed only

\alpha\beta + \beta\gamma

and got

-\frac{8}{3}

. What did they forget?

✓

3. Complete cubic verification

Complete Cubic Verification 🎯

You've now understood all three formulas linking a cubic's zeros to its coefficients, and you've built the skill of computing pairwise products.

Now it's time to put it all together — we'll verify all three relationships for a single cubic polynomial, comparing what we get from the zeros with what we get from the coefficients.

📋 Given Info

Given:

p(x) = 2x^3 - 4x^2 - 10x + 12

with zeros:

$\alpha = 1$ α
$\beta = -2$ β
$\gamma = 3$ γ

Standard form coefficients:

Coefficient	Value
$a$	$2$
$b$	$-4$
$c$	$-10$
$d$	$12$

✍️ Question

Your Task ✍️

Verify all three zeros-coefficients relationships for the cubic $p (x) = 2 x^{3} - 4 x^{2} - 10 x + 12$ with zeros $1$ , $- 2$ , and $3$ .

For each relationship:

Compute the value from the zeros
Compute the value from the coefficients
Check that they match

Show your work for all three relationships:

Sum of zeros vs $- b / a$
Sum of pairwise products vs $c / a$
Product of zeros vs $- d / a$

Let's go through all three relationshipsSum of zeros, sum of products, and product of all three for the polynomial $2 x^{3} - 4 x^{2} - 10 x + 12$ with zeros $1, - 2,$ and $3$ .

Coefficients: $a = 2$ The leading coefficient comes first, $b = -4$ Following descending powers of x, $c = -10$ (Each coefficient matches its power), $d = 12$ Down to the constant term.

✍️ MCQ

Choose one

For the first relationship (sum of zeros), what is

-\frac{b}{a}

when

a = 2

and

b = -4

?

✓

Relationship 1 (Sum):

From zeros: $1 + (-2) + 3 = 2$ .

From formula: $b = - 4$ , so $-b = +4$ flip signFlipping the sign of b is where most mistakes happen. Thus $\frac{- b}{a} = \frac{4}{2} = 2$ .

$2 = 2$ . ✓ Verified.

✍️ MCQ

Choose one

When

b = -4

, what is

-b

?

✓

Relationship 2 (Pairwise products):

Remember — three zeros means exactly three pairsThree zeros means exactly three pairs to multiply. Let's find each one:

Pair 1: $(1) (- 2) = - 2$

Pair 2: $(- 2) (3) = - 6$

Pair 3: $(3) (1) = 3$

Sum: $- 2 + (- 6) + 3 = - 5$ .

From formula: $\frac{c}{a}$ Get the sum directly from coefficients $= \frac{- 10}{2} = - 5$ .

$- 5 = - 5$ . ✓ Verified.

✍️ MCQ

Choose one

For the third relationship (product of all zeros), the formula gives

-\frac{d}{a}

. With

d = 12

and

a = 2

, what is

-\frac{d}{a}

?

✓

Relationship 3 (Product)Multiplying all three zeros together:

From zeros: $(1)(-2)(3) = -6$ Gives the third piece of information.

From formula: $\frac{-d}{a}$ The key formula to remember = $\frac{- (12)}{2}$ = $- 6$ .

$- 6 = - 6$ . ✓ Verified.

✍️ MCQ

Choose one

We verified all three formulas for

2x^3 - 4x^2 - 10x + 12

. Which relationship uses the alternating sign

-\frac{d}{a}

on the right side?

✓

Watch the Signs! Notice the double negativesDouble negatives turn into positives: $\frac{- b}{a}$ with $b = -4$ Check if coefficient is already negative, and $\frac{- d}{a}$ with $d = 12$ . The first gives a double negative (resolved to positiveBecomes positive when both are negative), the second is straightforward.