Welcome to The Capstone: Finding All Zeros of a Quartic — the grand finale of this chapter where everything comes together.

This is the capstone of the chapter. Every skill you have built comes together here:

Definitions
Zeros-coefficients relationships
Factoring
Construction
Division

You are given a degree-3 or degree-4 polynomial with some known zeros and must find all zeros.

The Strategy:

Step	Action
1	Form a factor from the known zeros
2	Divide the polynomial
3	Factor the quotient

1. Forming a quadratic factor from conjugate zeros

🎯 The Capstone: Finding All Zeros of a Quartic

This is the culmination of everything you've learned in this chapter:

Definitions
Zeros-coefficients relationships
Factoring
Construction
Division

All of these skills come together here.

Let's see how ready you are for this challenge!!

Here's a key insight:

When two given zeros are conjugate surds (like $a + \sqrt{b}$ and $a - \sqrt{b}$ ), they form a quadratic factor with integer coefficients.

This is much better than dividing by two irrational linear factors — it keeps all the arithmetic rational and clean!

✍️ Question

Diagnostic Question 🔍

Two zeros of a quartic are $(2 + \sqrt{3})$ and $(2 - \sqrt{3})$ .

What quadratic factor do they give?

For the conjugate pair $(2 + \sqrt{3})$ and $(2 - \sqrt{3})$ :

Sum = $(2 + \sqrt{3}) + (2 - \sqrt{3}) =$ $4$ result

Notice how the $\sqrt{3}$ terms cancel out: $+\sqrt{3}$ and $-\sqrt{3}$ They add to zero, canceling out add to zero.

This is exactly why conjugate pairsThe irrational parts always cancel out when you add them are so useful. The irrational parts always vanishPlus and minus root 3 give zero when you add them.

✍️ MCQ

Choose one

What is

(2 + \sqrt{3})(2 - \sqrt{3})

?

✓

Product = $(2 + \sqrt{3}) (2 - \sqrt{3})$

This is a difference of squaresYour best friend for conjugate surds: $a^{2} - b^{2}$ where $a = 2$ and $b = \sqrt{3}$ .

$= 2^{2} - (\sqrt{3})^{2} =$ $4 - 3$ $=$ $1$ A clean rational number from conjugate pairs

✍️ MCQ

Choose one

Why does

(\sqrt{3})^2

equal

3

and not

\sqrt{9}

?

✓

Quadratic factorYour go-to for conjugate pairs: $x^{2} - (sum) x + product =$ $x^2 - 4x + 1$ Sum and product pattern

Notice: all integer coefficientsConjugates produce integer coefficients. No surds anywhere in the factor. This is why we form the quadratic from the conjugate pairWork with conjugate pairs together instead of dividing by two irrational linear factors.

✍️ MCQ

Choose one

If two zeros of a quartic are

(5 + \sqrt{2})

and

(5 - \sqrt{2})

, what is the constant term of their quadratic factor?

✓

2. Dividing a quartic by a quadratic factor

The Capstone: Finding All Zeros of a Quartic 🏆

This is the culmination of everything you've learned in this chapter.

Every skill you've built comes together here:

Definitions
Zeros-coefficients relationships
Factoring
Construction
Division

With the quadratic factor formed, we now divide the quartic by it. The quotient is another quadratic, which we then factor to find the remaining two zeros.

📋 Given Info

Given Information:

We have the quartic polynomial:

x^4 - 6x^3 - 26x^2 + 138x - 35

And we know that $(x^2 - 4x + 1)$ is a factor of this quartic.

Your task: Divide the quartic by this quadratic factor, then factor the resulting quotient to find the remaining two zerosgoal.

✍️ Question

Your Challenge ✏️

Given that $(x^{2} - 4 x + 1)$ is a factor of $x^{4} - 6 x^{3} - 26 x^{2} + 138 x - 35$ :

Divide the quartic by $(x^2 - 4x + 1)$
Factor the quotient (which will be a quadratic)
Find the remaining two zeros

What are all four zeros of this quartic polynomial?

Let's trace the division carefully.

Cycle 1: $x^4 \div x^2 = x^2$ This is how you find the next piece of your quotient

Multiply(Take what you found and distribute it across the entire divisor): $x^{2} (x^{2} - 4 x + 1) = x^{4} - 4 x^{3} + x^{2}$

SubtractWatch those signs carefully, especially with negatives from dividend: $(x^{4} - 6 x^{3} - 26 x^{2} + 138 x - 35) - (x^{4} - 4 x^{3} + x^{2}) =$ $- 2 x^{3} - 27 x^{2} + 138 x - 35$

✍️ MCQ

Choose one

In cycle 2, we divide

-2x^3

by

x^2

. What is the next term of the quotient?

✓

Cycle 2: $-2x^3 \div x^2 = -2x$

Multiply: $-2x(x^2 - 4x + 1) = -2x^3 + 8x^2 - 2x$ Each cycle, multiply by all three terms of the divisor

Subtract: $(- 2 x^{3} - 27 x^{2} + 138 x - 35) - (- 2 x^{3} + 8 x^{2} - 2 x) =$ $-35x^2 + 140x - 35$ This becomes your new dividend for the next cycle

✍️ MCQ

Choose one

In cycle 3, we divide

-35x^2

by

x^2

. What do we get?

✓

Cycle 3: $-35x^2 \div x^2 = -35$

Multiply: $-35(x^2 - 4x + 1) = -35x^2 + 140x - 35$

Subtract: $0$ . Remainder $= 0$ .Zero confirms our factor is correct ✓ Confirmed.factor confirmedThe division worked perfectly

✍️ MCQ

Choose one

We need two numbers that multiply to

-35

and add to

-2

. What are those numbers?

✓

Quotient: $x^2 - 2x - 35$

To factor: $a \cdot c = 1 \times (- 35) = - 35$ , and we need a pair that sums to $b = - 2$ .

Think about it — what two numbers multiply to give negative 35 and add to give negative 2?

The pair is $-7$ largerWhen a times c is negative, you need opposite signs and $+5$ smaller(The bigger number gets the sign of b, which is negative) because $(- 7) \times 5 = - 35$ and $(- 7) + 5 = - 2$ .

So: $x^{2} - 2 x - 35 = (x - 7) (x + 5)$

✍️ MCQ

Choose one

We factored

x^2 - 2x - 35

as

(x - 7)(x + 5)

. What are the two zeros from this quadratic?

✓

Zeros from the quotient: $x = 7$ and $x = - 5$

So the quartic $x^{4} - 6 x^{3} - 26 x^{2} + 138 x - 35$ has four zerosA quartic polynomial always has four zeros total in total:

From $(x^{2} - 4 x + 1)$ : $x = 2 + \sqrt{3}$ and $x = 2 - \sqrt{3}$ Found using the quadratic formula on the given factor
From $(x - 7) (x + 5)$ : $x = 7$ and $x = -5$ Found by factoring the quotient we got from division

All four zerosThe complete answer for a degree-4 polynomial of the quartic:

$(2 + \sqrt{3})$ irrationalIrrational zeros containing root 3, $(2 - \sqrt{3})$ irrationalIrrational zeros containing root 3, $7$ integerInteger zeros we found, $-5$ integerInteger zeros we found

Summary: We started with a degree-4 polynomialBreaking down the complex polynomial and found all four zeros:

Two irrational zeros from the given factor: $2 + \sqrt{3}$ and $2 - \sqrt{3}$
Two integer zeros from factoring the quotient: $7$ and $- 5$

This is the power of polynomial divisionThe key strategy for solving complex polynomials — breaking a complex problem into manageable piecesThe whole strategy in three words!

✍️ MCQ

Choose one

In the original quartic

x^4 - 6x^3 - 26x^2 + 138x - 35

, how many of the four zeros are irrational?

✓

3. Why use quadratic factor instead of two linear divisions

🎯 The Capstone Challenge

We're at the final insight of this chapter. Every skill you've built comes together here:

Definitions of zeros and factors
Zeros-coefficients relationships
Factoring techniques
Construction of factors from zeros
Polynomial division

Let me set up the scenario for you...

The Situation:

Suppose you've found that a quartic polynomial has two zeros: $2 + \sqrt{3}$ and $2 - \sqrt{3}$ .

Now you need to divide the quartic to find the remaining zeros.

You have two choices:

Option	Approach
Option A	Divide by $(x - (2 + \sqrt{3}))$ first, then divide the result by $(x - (2 - \sqrt{3}))$ — two separate linear divisions
Option B	Form the quadratic factor $x^2 - 4x + 1$ (which has both zeros) and divide once

Something to consider:

If you chose Option A and divided by $(x - (2 + \sqrt{3}))$ first:

$\sqrt{3}$ would appear in every step of the division
Every quotient term and every remainder would contain surds

Then you'd have to divide that messy result by $(x - (2 - \sqrt{3}))$ , again dealing with $\sqrt{3}$ throughout.

This is why forming the quadratic factor with integer coefficients is the smarter approach.

✍️ Question

Think about this strategically 🤔

Why is it better to form the quadratic factor $x^2 - 4x + 1$ and divide once, rather than dividing by $(x - (2 + \sqrt{3}))$ and then by $(x - (2 - \sqrt{3}))$ separately?

Let's compare the two approaches:

Option 1: Divide by $(x - (2 + \sqrt{3}))$ , then divide the result by $(x - (2 - \sqrt{3}))$ .

Problem: The first division has divisor $x - 2 - \sqrt{3}$ The divisor contains an irrational term. Every quotient term involves $\sqrt{3}$ messy!Root 3 infects every term you compute. The second division compounds this with more surd terms. The arithmetic becomes extremely messy.Dividing by surds one at a time creates chaos

✍️ MCQ

Choose one

When we formed the quadratic factor

x^2 - 4x + 1

from the conjugate pair, what happened to all the

\sqrt{3}

terms?

✓

Option 2:Build one quadratic from both roots Form $x^2 - 4x + 1$ Use the conjugate pair to form this from the conjugate pair. Divide once.That's the efficiency gain

Advantage: The factor has integer coefficientsSurds cancel out completely $(1, -4, 1)$ integers!No messy arithmetic. The entire division involves only integersNo messy arithmetic needed. The quotient $x^{2} - 2 x - 35$ has integer coefficients. Clean and fast.When you have a conjugate pair

✍️ MCQ

Choose one

When we multiplied

(2 + \sqrt{3})(2 - \sqrt{3})

, the result was an integer because of which algebraic identity?

✓

This is always the strategy for conjugate pairs: Form the quadratic firstThis eliminates the surds and gives clean integer coefficients, divide onceThat's a trap that creates a mess of surds. Never divide by two irrational linear factors sequentiallyYou'll create surds nearly impossible to simplify.

✍️ MCQ

Choose one

If you're told that

(3 + \sqrt{5})

and

(3 - \sqrt{5})

are zeros of a polynomial, what should your FIRST step be?

✓