Welcome! Today we're tackling Cubic Construction and Scaling to Integer Coefficients — building polynomials from their zeros and cleaning up the results.

Remember the quadratic formula?

$p (x) = x^{2} - S x + P$

This extends to cubics with three symmetric functions:

Symbol	Meaning
$S_{1}$	Sum of zeros
$S_{2}$	Sum of pairwise products
$S_{3}$	Product of all zeros

The signs follow a pattern:

$p (x) = x^{3} - S_{1} x^{2} + S_{2} x - S_{3}$

Minus, plus, minus — the signs alternate.

When the zeros include fractions, the constructed polynomial ends up with fractional coefficients.

That's where the scaling clean-up step comes in — multiplying through to get integer coefficients.

By the end of this lesson, you will master:

Cubic construction — building the polynomial from three zeros
Scaling clean-up — converting fractional coefficients to integers

1. Computing the three symmetric functions for a cubic

Cubic Symmetric Functions 📐

A cubic has three zeros, so you need three symmetric functions to describe them:

Function	Name	Formula
$S_{1}$	Sum of zeros	$α + β + γ$
$S_{2}$	Sum of pairwise products	$α β + β γ + γ α$
$S_{3}$	Product of all zeros	$α β γ$

Computing $S_{2}$ requires listing all three pairskey — the same skill you use when verifying relationships between zeros.

✍️ Question

Your Turn ✏️

Compute $S_1$ , $S_2$ , and $S_3$ for the zeros $3$ , $\frac{1}{2}$ , and $-1$ .

Remember:

$S_1 = \alpha + \beta + \gamma$
$S_2 = \alpha\beta + \beta\gamma + \gamma\alpha$
$S_3 = \alpha\beta\gamma$

Let's compute each symmetric functionStandard combinations of zeros used to build polynomial coefficients for zeros 3, 1/2, -1.

S₁Gives us the coefficient of the second-highest power term (sum): $3 + \frac{1}{2} + (-1)$ (Always add all zeros, including negative ones) = $3 - 1 + \frac{1}{2}$ = $2 + \frac{1}{2}$ = $\frac{5}{2}$

✍️ MCQ

Choose one

Compute

S_2 = (3)(\frac{1}{2}) + (\frac{1}{2})(-1) + (-1)(3)

✓

S₂Sum of all pairwise products of roots (sum of pairwise products — three pairsAlpha-beta, beta-gamma, and gamma-alpha):

Pair 1: $(3) (\frac{1}{2}) = \frac{3}{2}$
Pair 2: $(\frac{1}{2}) (- 1) = - \frac{1}{2}$
Pair 3: $(- 1) (3) = - 3$

Add: $\frac{3}{2} - \frac{1}{2} - 3 = \frac{2}{2} - 3 = 1 - 3 = - 2$

So S₂ = -2AnswerA clean integer for Vieta's formulas

✍️ MCQ

Choose one

Compute

S_3 = (3)\left(\frac{1}{2}\right)(-1)

✓

S₃The third symmetric sum for our cubic (product of all three zeros):

$S_{3} = α \cdot β \cdot γ = (3) (\frac{1}{2}) (- 1)$

Step-by-step calculation:

First, multiply the first two: $(3)\left(\frac{1}{2}\right) = \frac{3}{2}$

Then multiply by the third: $\left(\frac{3}{2}\right)(-1) = -\frac{3}{2}$

Therefore, $S_3 = -\frac{3}{2}$ S₃

✍️ MCQ

Choose one

We found

S_1 = \frac{5}{2}

,

S_2 = -2

, and

S_3 = -\frac{3}{2}

. Which of these symmetric functions gave us an integer value?

✓

⚠️ The most common error in S₂:The mistake students make most often computing only two of the three pairsYou need all three combinations.

Many students forget that "sum of products taken two at a time" means three pairs, not twoAll three pairs must be included!

Always list all three pairs as a checklistList them before adding to catch mistakes before adding:

$\alpha\beta = ?$ 1First pair in your checklist
$\beta\gamma = ?$ 2(Second pair in your checklist)
$\gamma\alpha = ?$ 3Third pair in your checklist

Then sum: $S_2 = \alpha\beta + \beta\gamma + \gamma\alpha$ Adding after listing catches errors

✍️ MCQ

Choose one

A student computing

S_2

for zeros

3, \frac{1}{2}, -1

writes:

(3)(\frac{1}{2}) + (\frac{1}{2})(-1) = \frac{3}{2} - \frac{1}{2} = 1

. What pair did they forget?

✓

2. Applying the cubic construction formula

With $S_{1}$ , $S_{2}$ , $S_{3}$ computed, the cubic construction formula uses alternating signs:

$p (x) = x^{3} - S_{1} x^{2} + S_{2} x - S_{3}$

The signs follow the pattern: minus, plus, minus (alternating).

⚠️ Watch out for the double negative — when $S_{3}$ is negative, you get $- (- something)$ , which becomes positive.

📋 Given Info

The cubic construction formula:

$p (x) = x^{3} - S_{1} x^{2} + S_{2} x - S_{3}$

Given values:

$S_1 = \frac{5}{2}$
$S_2 = -2$
$S_3 = -\frac{3}{2}$

✍️ Question

Your turn ✏️

Using $S_1 = \frac{5}{2}$ , $S_2 = -2$ , $S_3 = -\frac{3}{2}$ , write the cubic polynomial $p(x)$ .

Let's apply the formula term by term:

p(x) = x^3 - S_1 \cdot x^2 + S_2 \cdot x - S_3

(Alternating signs you must memorize)

With $S_1 = \frac{5}{2}$ , $S_2 = -2$ , $S_3 = -\frac{3}{2}$ .

✍️ MCQ

Choose one

When we substitute

S_3 = -\frac{3}{2}

into the term

-S_3

, what do we get?

✓

Building each term:

$x^3$ term:First term is x cubed with coefficient 1 Just $x^{3}$
$-S_1 \cdot x^2$ :Each S value becomes a coefficient Since $S_1 = \frac{5}{2}$ S₁S values pair with decreasing x powers, we get $- \frac{5}{2} x^{2}$

$+ S_{2} \cdot x$ : Since $S_2 = -2$ S₂Plus times negative gives negative result, we get $+(-2)x = -2x$ The result becomes negative
$-S_3$ :(Watch the signs in the constant term) Since $S_3 = -\frac{3}{2}$ S₃Signs are tricky here, we get $-\left(-\frac{3}{2}\right) = +\frac{3}{2}$ Double negative flips to positive — common exam mistake

✍️ MCQ

Choose one

When we compute

-S_3

with

S_3 = -\frac{3}{2}

, we get:

✓

Result:

p(x) = x^3 - \frac{5}{2}x^2 - 2x + \frac{3}{2}

Note the double negativeThis is where most mistakes happen in the last term: $S_3 = -\frac{3}{2}$ S three is already negative, and the formula has $-S_3$ The formula has minus S three = $-\left(-\frac{3}{2}\right)$ double negMinus times minus gives the flip = $+\frac{3}{2}$ The sign flips to positive.

Writing it in two stepsTwo steps makes the sign flip clear makes the sign change visible.

✍️ MCQ

Choose one

In our polynomial

p(x) = x^3 - \frac{5}{2}x^2 - 2x + \frac{3}{2}

, why is the constant term

+\frac{3}{2}

and not

-\frac{3}{2}

?

✓

3. Scaling to clear fractional coefficients

Scaling to Integer Coefficients 📐

We've constructed a polynomial, but it has fractional coefficients. In most problems, we're expected to give our answer with integer coefficients.

The technique is simple: scale by the LCM of all denominators to clear the fractions.

📋 Given Info

Given Information:

For the polynomial $p (x) = x^{3} - \frac{5}{2} x^{2} - 2 x + \frac{3}{2}$

The denominators are:

Term	Denominator
$x^{3}$	1
$- \frac{5}{2} x^{2}$	2
$- 2 x$	1
$+ \frac{3}{2}$	2

LCM of denominators = 2

✍️ Question

Your turn! ✏️

Scale $x^{3} - \frac{5}{2} x^{2} - 2 x + \frac{3}{2}$ to have integer coefficients.

Also answer: Does scaling change the zeros of the polynomial?

Scaling procedure:

Identify all denominators: $x^{3}$ (denominator 1), $\frac{5}{2} x^{2}$ (denominator 2), $2 x$ (denominator 1), $\frac{3}{2}$ (denominator 2).
LCM of 1, 2, 1, 2 = 2key!(The LCM clears all fractions at once).

Multiply EVERY term by 2Not just the fractions — every single term:
- $2 \times x^{3} = 2 x^{3}$
- $2 \times \left(-\frac{5}{2}x^2\right) = -5x^2$ Fractions become whole numbers
- $2 \times (- 2 x) = - 4 x$
- $2 \times \frac{3}{2} = 3$ Left with whole numbers

✍️ MCQ

Choose one

When we multiplied

\frac{3}{2}

by 2, we got 3. What mathematical property allowed the 2s to cancel?

✓

Result: $2 x^{3} - 5 x^{2} - 4 x + 3$

Notice how all the fractions have disappeared — we now have a polynomial with integer coefficientsIntegers make factoring much simpler that's much easier to work with!

✍️ Yes/No

Yes or No?

The original polynomial

p(x) = x^3 - \frac{5}{2}x^2 - 2x + \frac{3}{2}

has zeros at

3, \frac{1}{2}, -1

. Does the scaled polynomial

2x^3 - 5x^2 - 4x + 3

have the same zeros?

✓

⚠️ Critical Rule: When scaling, you must multiply EVERY term by the LCMNot just the fractions — every single term — including the $x^3$ termIt looks fine but you still must multiply it!

This is a common mistakeSkipping any term ruins the whole polynomial students make. Don't just multiply the fractional terms!

Common Mistake:

If you only multiply the fractional terms ( $- \frac{5}{2} x^{2}$ and $\frac{3}{2}$ ) by 2, you get:

x^3 - 5x^2 - 4x + 3

Wrong!(This is what happens when you skip terms)

This is a different polynomial with different zeros — NOT what we want!

Correct approach: Multiply ALL terms by 2 to get $2x^3 - 5x^2 - 4x + 3$ That 2 keeps all the zeros the same.

✍️ MCQ

Choose one

If you scale

x^3 - \frac{5}{2}x^2 - 2x + \frac{3}{2}

by multiplying only the fractional terms by 2, what coefficient does

x^3

have in the result?

✓

Why scaling preserves zeros: If $α$ makes $p (x)$ equal 0, then $2 \times 0 = 0$ Multiplying zero by any number gives zero, so $α$ also makes $2 \cdot p (x)$ equal 0.

✍️ MCQ

Choose one

If

p(3) = 0

, what is

2 \cdot p(3)

?

✓

Key rule: Non-zero constant multiples of a polynomial always have the same zeros.The rule you need to remember

So $p (x) = x^{3} - \frac{5}{2} x^{2} - 2 x + \frac{3}{2}$ and $2p(x) = 2x^3 - 5x^2 - 4x + 3$ Freely multiply to clear fractions have exactly the same zerossame!Zeros stay exactly the same: $3$ , $\frac{1}{2}$ , and $- 1$ .