Welcome! Today we're exploring Deriving Sum and Product of Zeros Formulas — one of the most powerful shortcuts in the polynomials chapter.

Here is something remarkable: if you know the coefficients of a quadratic, you can compute the sum and product of its zeros without ever finding the zeros themselves.

The formulas:

Quantity	Formula
Sum of zeros	$-\frac{b}{a}$
Product of zeros	$\frac{c}{a}$

These are not magic — they come directly from expanding the factored form and comparing coefficients.

Understanding the derivation is important because:

It tells you where the minus signkey! in $- \frac{b}{a}$ comes from.

That minus sign is the source of the most common error in the entire chapter.

1. Stating both formulas correctly

We need two formulas that connect a quadratic's zeros to its coefficients.

These formulas are used in almost every problem in this chapter, so getting them right — especially the signs — is critical.

📋 Given Info

For a quadratic polynomial $a x^{2} + b x + c$ with zeros $\alpha$ and $\beta$ :

There are two relationships connecting the zeros to the coefficients $a$ , $b$ , and $c$ .

✍️ Question

State the formulas for the sum of zeros and the product of zeros in terms of $a$ , $b$ , and $c$ .

The two formulas are:

Sum of zeros: $\alpha + \beta = \frac{-b}{a}$ Use this shortcut to find sum without solving

Product of zeros: $\alpha \cdot \beta = \frac{c}{a}$ Use this when question asks for product of roots

✍️ Yes/No

Yes or No?

In the sum of zeros formula

\alpha + \beta = \frac{-b}{a}

, there is a negative sign. Is there a negative sign in the product of zeros formula

\alpha \cdot \beta = \frac{c}{a}

?

✓

Notice the signs: the sum has a NEGATIVEThis is where most students lose marks sign ( $-b/a$ negativeAlways remember sum has the negative), while the product has no extra negativeProduct does not have the negative sign (just $c/a$ no negativeNo negative sign here). This asymmetryEven if method is correct, wrong signs give wrong answer comes from the derivation.

✍️ MCQ

Choose one

For a quadratic polynomial

2x^2 + 5x + 3

, what is the sum of zeros?

✓

When you expand $a (x - α) (x - β)$ , you get:

ax^2 - a(\alpha + \beta)x + a(\alpha \cdot \beta)

(Sum shows up in middle, product in constant)

Comparing this with $a x^{2} + b x + c$ :

The $x$ coefficient gives: $- a (α + β) = b$ , so $\alpha + \beta = \frac{-b}{a}$ Sum formula — don't forget the negative sign
The constant term gives: $a (α \cdot β) = c$ , so $\alpha \cdot \beta = \frac{c}{a}$ Product formula — c divided by a

✍️ MCQ

Choose one

In the expansion

a(x - \alpha)(x - \beta) = ax^2 - a(\alpha + \beta)x + a(\alpha \cdot \beta)

, why does the sum formula have a negative sign but the product formula doesn't?

✓

⚠️ The negative sign in the sum formulaIt comes directly from the minus signs in the factored form comes from the minus signs in $(x - α) (x - β)$ .

When you expand $(x - α) (x - β)$ , those minus signsThe minus signs create the negative coefficient create a negative coefficient for the $x$ termThat's exactly why we get negative b over a. That's why the sum formula has $-\frac{b}{a}$ correctThe result of expanding with minus signs, not $\frac{b}{a}$ .

Forgetting this sign is the single most common error in the chapter.Forgetting the negative sign is the biggest mistake

Formula	Correct	Common Mistake
Sum of zeros	$\alpha + \beta = -\frac{b}{a}$ ✓The sum formula MUST have the negative	$\alpha + \beta = \frac{b}{a}$ ✗Writing just b over a is incorrect (missing negative)

Students often write $\frac{b}{a}$ instead of $- \frac{b}{a}$ because they forget where the negative comes from.

✍️ MCQ

Choose one

For the quadratic

2x^2 + 5x + 3

, what is the sum of its zeros

\alpha + \beta

?

✓

2. Understanding the derivation through coefficient comparison

Knowing the formulas is one thing; understanding where they come from is what lets you reconstruct them when needed. 🧠

The derivation is a single idea: expand the factored form and match it term-by-term with $ax^2 + bx + c$ .

📋 Given Info

Here's what we know:

If $α$ and $β$ are zeros of $p (x) = a x^{2} + b x + c$ , then we can write:

p(x) = a(x - \alpha)(x - \beta)

The factor $a$ ensures the leading coefficients match.

✍️ Question

Your turn! ✏️

Expand $a(x - \alpha)(x - \beta)$ fully.

Then compare the $x$ -coefficient with $b$ to show that:

\alpha + \beta = -\frac{b}{a}

goal

Let me walk you through the derivation step by step.

Since $α$ and $β$ are zeros of the quadratic, we can write the polynomial as:

p(x) = a(x - \alpha)(x - \beta)

(When you know the zeros, this is the form you can write)

✍️ MCQ

Choose one

When we expand

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

, what will be the coefficient of

x

in the expanded form?

✓

Step 1: Expand $(x - \alpha)(x - \beta)$

Let's multiply these two brackets term by term:

= x^2 - \beta x - \alpha x + \alpha\beta

Now, let's combine the like terms (the two middle terms with $x$ ):

= x^2 - (\alpha + \beta)x + \alpha\beta

(The expansion reveals where the formula comes from)

$(\alpha + \beta)$ sum of zerosSum of zeros appears as the x coefficient

✍️ MCQ

Choose one

In the expansion

x^2 - (\alpha + \beta)x + \alpha\beta

, what is the coefficient of

x

?

✓

Step 2: Multiply by $a$

Now we multiply the entire expression by $a$ :

$a \cdot [x^{2} - (α + β) x + α β]$

= ax^2 - a(\alpha + \beta)x + a(\alpha\beta)

(Multiplying through by a gives us standard form coefficients)

✍️ MCQ

Choose one

Looking at the coefficient of

x

in both forms, what must

-a(\alpha + \beta)

equal?

✓

Step 3: Compare with $ax^2 + bx + c$ , term by term:

$x^{2}$ coefficient: $a = a$ ✓ Consistent.

$x$ coefficient: $- a (α + β) = b$ , so $\alpha + \beta = -\frac{b}{a}$ Sum equals negative b over a

✍️ MCQ

Choose one

In the derivation, why does the sum formula have a negative sign (

-\frac{b}{a}

) instead of just

\frac{b}{a}

?

✓

Constant term: $a (α β) = c$ , so $\alpha\beta = \frac{c}{a}$ Product equals c over a

The negative signThis negative sign comes directly from the expansion in $- \frac{b}{a}$ comes from the minus signs in $(x - \alpha)(x - \beta)$ The minus signs in the factored form create the negative.

When you expand, the sum of zeros appears with a negative in front:

$(x - α) (x - β) = x^{2} - α x - β x + α β$

= x^2 - (\alpha + \beta)x + \alpha\beta

(The sum of zeros gets a negative sign in front)

That minus signculpritThis is the classic exam trap to watch out for before $(α + β)$ is why we get $\alpha + \beta = -\frac{b}{a}$ The formula is negative b over a, not $+\frac{b}{a}$ Not positive b over a.

✍️ MCQ

Choose one

If a quadratic has

a = 1

,

b = 5

, and

c = 6

, what is the sum of its zeros?

✓

3. Applying the formulas to a specific polynomial

The formulas are derived and understood. Now let's apply them to an actual polynomial to see how the signs work in practice.

Pay special attention to the double negative that appears when $b$ is already negative.

📋 Given Info

Consider the polynomial $6x^2 - 7x - 3$ whose zeros are $\frac{3}{2}$ and $-\frac{1}{3}$ .

Here we have:

$a = 6$
$b = -7$
$c = -3$

✍️ Question

Verify the sum formula for $6 x^{2} - 7 x - 3$ whose zeros are $\frac{3}{2}$ and $-\frac{1}{3}$ .

Compute the sum two ways:

From the zeros: Add $\frac{3}{2} + (- \frac{1}{3})$
From the formula: Use $-\frac{b}{a}$

Then compare your results. Show your work for both methods.

The double negativewhere sign errors happen is the danger zone. Let's handle it in two explicit stepsnot in your head.

⚠️ When $b$ is already negative, the formula $- \frac{b}{a}$ becomes $- \frac{(- 7)}{6}$ . That's a negative of a negativethat's where sign errors happen — and this is where most mistakes happen!

For $6 x^{2} - 7 x - 3$ :

Coefficient	Value
$a$	$6$
$b$	$- 7$
$c$	$- 3$

Notice: $b = -7$ , not $7$ Always include the sign when you read off coefficients. The negative sign is PART of the coefficientAlways include the sign!

✍️ MCQ

Choose one

When we compute

-\frac{b}{a}

with

b = -7

, what happens to the sign?

✓

From the zeros ( $\frac{3}{2}$ and $- \frac{1}{3}$ ):

\text{Sum} = \frac{3}{2} + \left(-\frac{1}{3}\right)

Finding common denominator (LCMFirst step when adding fractions with different denominators of 2 and 3 is 6LCMThis technique applies everywhere fractions appear):

$= \frac{9}{6} - \frac{2}{6}$

✍️ MCQ

Choose one

What is

\frac{9}{6} - \frac{2}{6}

?

✓

= \frac{7}{6}

(Computed straight from the zeros)

So the sum of zeros calculated directly is $\frac{7}{6}$ answerNow we have something to check the formula against

From the formula (using the two-step methodFirst identify b, then handle the negative sign separately):

Step 1: $b = - 7$

Step 2: $-b = -(-7) = +7$ Students see negative 7 and forget the formula has a negative sign

✍️ FIB

Fill in the blank

We found

-b = 7

and we know

a = 6

. What is

\frac{-b}{a}

?

✓

Step 3: $\frac{-b}{a} = \frac{7}{6}$

Compare: $\frac{7}{6} = \frac{7}{6}$ Match!You know you've applied the formula correctly. ✓ Verified.Useful for checking your work in exams

⚠️ The trap:This is the key point to focus on If you forget to negate $b$ You must apply the negative sign from the formula, you get $-\frac{7}{6}$ Skipping the negation flips your answer instead of $\frac{7}{6}$ .

Many students see $b = - 7$ and directly write $\frac{b}{a} = \frac{-7}{6}$ wrong!This is where easy marks are lost, completely skipping the negative signWhen b is negative, you must still apply the formula's negative in the formula.

The two-step methodWrite b first, then write negative b separately (write $b$ , then write $- b$ ) makes the sign explicitForces you to consciously deal with the sign change and prevents this error. Always use two steps when $b$ is negative.

✅ Step 1: Write $b = -7$ First step: identify what b equals ✅ Step 2: Write $-b = -(-7) = +7$ Second step: apply the negative from the formula

This forces you to handle the double negative consciouslykey!So you never accidentally skip the sign change.

✍️ MCQ

Choose one

A student claims the sum of zeros for

6x^2 - 7x - 3

is

-\frac{7}{6}

. What mistake did they make?

✓