Welcome! Today we're looking at Instant Answers from Cubic Coefficients — a powerful shortcut that lets you skip the hard work of finding roots.

Cubic symmetric expressions follow the same principle as quadratic ones:

Go straight from coefficients to the answer.

The three relationships give you:

What you get	Instantly from coefficients
Sum of roots	$α + β + γ$
Sum of pairwise products	$α β + β γ + γ α$
Product of roots	$α β γ$

Combined with one identity, you can also compute:

\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma}

bonus

...in a single line.

1. Reading cubic relationships directly from coefficients

Let's see how well you can read answers directly from cubic coefficients!

The three cubic formulas give instant answers for sum, pairwise products, and product of zeros — no computation beyond reading coefficients and dividing.

For a cubic polynomial $a x^{3} + b x^{2} + c x + d$ with zeros $α, β, γ$ :

Expression	Formula
Sum of zeros	$\alpha + \beta + \gamma = -\frac{b}{a}$
Sum of pairwise products	$\alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a}$
Product of zeros	$\alpha\beta\gamma = -\frac{d}{a}$

✍️ Question

Your turn! 🎯

Consider the cubic polynomial: $2 x^{3} + x^{2} - 13 x + 6$

Find the product of all three zeros.

Use the coefficient relationship directly — no need to factor or find individual zeros!

Product of All Three ZerosYour shortcut for product questions (Cubic Polynomials)

For a cubic polynomial $a x^{3} + b x^{2} + c x + d$ :

\text{Product of all three zeros} = -\frac{d}{a}

(Use this when you only need the product)

Notice the negative sign in the formula!crucial!Students often forget and write d over a instead

Applying to our polynomial: $2 x^{3} + x^{2} - 13 x + 6$

Here, $a = 2$ Identify a from the cubic term and $d = 6$ The constant at the end of the polynomial

$Product of zeros = - \frac{6}{2} = - 3$

Quick method — no factoring needed!Use when question only asks for product, not individual zeros

✍️ MCQ

Choose one

If the polynomial were

3x^3 + x^2 - 13x - 12

, what would be the product of all three zeros?

✓

For $2 x^{3} + x^{2} - 13 x + 6$ :

$a = 2$ , $d = 6$

Product of zeros $= -\frac{d}{a}$ The pattern — just use first and last coefficients $= - \frac{6}{2}$ $= -3$ Answer!No need to find actual zeros

The Power of Direct Formulas

One lineJust one line to get your answer. No factoring, no finding individual zeros. The formula gives the answer directly from the coefficientsAnswers come straight from coefficients, no zeros needed.

Think about what we just did — we found the product of THREE zeros without ever knowing what those zeros actually are! That's the magic of these coefficient relationships.

For the polynomial $2 x^{3} + x^{2} - 13 x + 6$ :

Product of all three zeros $= - \frac{d}{a} =$ $-\frac{6}{2}$ Constant over leading coefficient with sign change $=$ $-3$ AnswerThat's all you need for product of zeros

That's it! The answer is $- 3$ .

✍️ MCQ

Choose one

If a cubic polynomial has

a = 4

and

d = -12

, what is the product of all three zeros?

✓

2. Computing 1/alpha + 1/beta + 1/gamma from coefficients

📋 Given Info

Let's check your understanding of computing reciprocal sums from cubic coefficients.

Here's a useful identity:

$\frac{1}{α} + \frac{1}{β} + \frac{1}{γ} = \frac{sum of pairwise products}{product of roots}$

In terms of coefficients of $a x^{3} + b x^{2} + c x + d$ , this simplifies to:

\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} = \frac{c/a}{-d/a} = -\frac{c}{d}

one-liner

✍️ Question

Your turn 🧮

For the cubic polynomial $x^3 - 6x^2 - x + 30$ , compute:

\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma}

where $α$ , $β$ , $γ$ are the roots.

The identity for finding $\frac{1}{α} + \frac{1}{β} + \frac{1}{γ}$ is:

\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} = \frac{\alpha\beta + \beta\gamma + \gamma\alpha}{\alpha\beta\gamma}

(Pairwise products divided by triple product)

The numerator is $\frac{c}{a}$ Sum of pairwise products from Vieta's (sum of pairwise products) and the denominator is $\frac{-d}{a}$ Product of all three roots (product of roots). Their ratio simplifies to:

$\frac{c / a}{- d / a} = \frac{c}{- d} = - \frac{c}{d}$

(Final Formula) $-\frac{c}{d}$ Memorize this for quick exam answers

✍️ MCQ

Choose one

For the polynomial

x^3 - 6x^2 - x + 30

, what is the value of

\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma}

?

✓

For $x^{3} - 6 x^{2} - x + 30$ , we have $c = -1$ and $d = 30$ .

Applying the formula:

-\frac{c}{d} = -\frac{(-1)}{30} = \frac{1}{30}

(The formula to use every single time)

Notice the double negativeThis is where most mistakes happen: since $c = - 1$ , we get $-c = -(-1) = +1$ Double negative becomes positive. Then $\frac{1}{30}$ .

✍️ MCQ

Choose one

If a cubic polynomial has

c = -5

and

d = 10

, what is

\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma}

?

✓

3. Zero constant term means x = 0 is a zero

Here's a useful special case to keep in your toolkit:

If the constant term $d = 0$ , then the product of zeros is $0$ .

And we know that a product of real numbers is $0$ only when at least one factor is $0$ .

This gives you an instant check for whether $x = 0$ zero! is a zero — no computation needed!

✍️ Question

Without computing anything, how can you instantly tell whether $x = 0$ is a zero of:

$5 x^{3} - 2 x^{2} + 7 x$
$3 x^{3} + x - 4$

The instant check: look at the constant term $d$ Check d to know if zero is a root.

If $d = 0$ When the constant is zero: product of zeros $= - \frac{0}{a} = 0$ . For a product to be 0, at least one zero must be 0Product equals zero means one factor is zero. So $x = 0$ IS a zeroZero is definitely a root.

If $d \neq 0$ When the constant is non-zero: product of zeros $\neq 0$ . None of the zeros can be 0. So $x = 0$ is NOT a zeroZero is not a root.

✍️ Yes/No

Yes or No?

Look at

5x^3 - 2x^2 + 7x

. Is

x = 0

a zero of this polynomial?

✓

✍️ Yes/No

Yes or No?

Now look at

3x^3 + x - 4

. Is

x = 0

a zero of this polynomial?

✓

Example 1: $5x^3 - 2x^2 + 7x$

Written in standard form, this is $5 x^{3} - 2 x^{2} + 7 x + 0$ .

The constant term $d = 0$ No constant written means d is zero. This is the key observation. d is zero.key!Missing constant signals something special

✍️ MCQ

Choose one

Since

d = 0

, what is the product of the zeros of

5x^3 - 2x^2 + 7x

?

✓

Since product of zeros $= - \frac{d}{a} = - \frac{0}{5} = 0$

And a product is zero only when at least one factor is zeroProduct is zero means one factor must be zero...

So $x = 0$ must be a zero!x equals zero is instantly a root

Verification: Factor out $x$ to get $x (5 x^{2} - 2 x + 7)$

Yes! $x = 0$ is clearly a factor ✓ One glance at the constant term told us everything — no calculations needed!Check constant term first to save work

Example 2: $3x^3 + x - 4$

✍️ MCQ

Choose one

What is the constant term

d

in the polynomial

3x^3 + x - 4

?

✓

The constant is $- 4$ . Not zeroIf constant isn't zero, product of zeros isn't zero. So $x = 0$ is not a zeroNone of the zeros can equal zero.

Confirm: $p(0) = 3(0)^3 + (0) - 4 = -4 \neq 0$ ✓verified

✍️ FIB

Fill in the blank

To instantly check if

x = 0

is a zero of a cubic, you only need to look at the ______ term.

✓