Welcome! Today we're exploring The Conjugate Pair Property and Finding All Cubic Zeros — a powerful shortcut that will save you a lot of work.

Before tackling the 'find all zeros' chain, there is a property that saves enormous work:

Irrational zeros of rational-coefficient polynomials come in conjugate pairs.

Here's what this means for you:

What you know	What you get for free
One surd zero	Its conjugate partner
Two zeros	A quadratic factor with rational coefficients

This means the subsequent division involves only rational arithmetic — no messy surds in the middle of long division.

1. The conjugate pair property and forming quadratic factors

Let's explore an important property that can save you a lot of work when finding zeros of polynomials.

When a polynomial has rational coefficients and one of its zeros is a surd like $(a + \sqrt{b})$ :

The conjugate $(a - \sqrt{b})$ must also be a zero.

Together, these conjugate pairs form a quadratic factor with rational coefficients.

📋 Given Info

Here's a useful fact to keep in mind:

If a polynomial with rational coefficients has zero $(3 + \sqrt{5})$ , then $(3 - \sqrt{5})$ must also be a zero.

Their sum = $(3 + \sqrt{5}) + (3 - \sqrt{5}) = 6$ (rational!)(the surds cancel out)
Their product = $(3 + \sqrt{5}) (3 - \sqrt{5}) = 9 - 5 = 4$ (using difference of squares)

✍️ Question

Your turn 🤔

If $(3 + \sqrt{5})$ is a zero of a polynomial with rational coefficients:

What is the other conjugate zero?
Form the quadratic factor from this pair.

Hint: Use the sum and product of the zeros to write the quadratic

The Conjugate Pair PropertyYour automatic trigger when you spot a surd zero:

If $(a + \sqrt{b})$ When you see an irrational zero like this is a zero of a polynomial with rational coefficientsMust have rational coefficients for this to work, then $(a - \sqrt{b})$ Immediately write down the conjugate as another zero is also a zero.

So for $(3 + \sqrt{5})$ , the conjugate is $(3 - \sqrt{5})$ (3 plus root 5 becomes 3 minus root 5).

You change the sign of the surd part ONLYOnly flip the sign of the surd part, never the rational, not the rational partDon't change the rational part — this is where students lose marks.

Zero	Conjugate Zero
$3 + \sqrt{5}$	$3 - \sqrt{5}$

✍️ MCQ

Choose one

What is the conjugate of

(7 + \sqrt{3})

?

✓

⚠️ Common mistake: Writing $(- 3 + \sqrt{5})$ or $(- 3 - \sqrt{5})$ as the conjugate.

These are NOT conjugates — they negate the rational partThe rational part cannot change, which is wrong.

Only the sign in front of the surd changes.(Only the sign before the surd flips, nothing else)

Original	✅ Correct Conjugate	❌ Wrong
$(3 + \sqrt{5})$	$(3 - \sqrt{5})$ ✓The 3 stays, only the plus becomes minus	$(-3 + \sqrt{5})$ ✗Changing the rational part is incorrect

✍️ MCQ

Choose one

What is the conjugate of

(7 - \sqrt{2})

?

✓

From the conjugate pair, we find:

Sum = $(3 + \sqrt{5}) + (3 - \sqrt{5}) = 6$ . The surds cancelcancel!The surds cancel out completely.

✍️ MCQ

Choose one

Why do the surds cancel when we add

(3 + \sqrt{5})

and

(3 - \sqrt{5})

?

✓

Product = $(3 + \sqrt{5}) (3 - \sqrt{5}) = 3^{2} - (\sqrt{5})^{2} = 9 - 5 = 4$ . This uses the difference of squaresa plus b times a minus b equals a squared minus b squared.

✍️ MCQ

Choose one

If

(2 + \sqrt{3})

and

(2 - \sqrt{3})

are conjugate zeros, what is their product?

✓

Quadratic factor:Use sum and product to construct it $x^{2} - (sum) x + product =$ $x^2 - 6x + 4$ The formula for building the quadratic. All integer coefficients!Conjugate pairs always give rational results

2. The complete chain: known zero to all zeros of a cubic

Finding All Zeros of a Cubic 🔍

When you know one zero of a cubic polynomial, you can find all the zeros by following a three-step chain:

Known zero → Factor: If $α$ is a zero, then $(x - α)$ is a factor
Divide: Divide the cubic by $(x - α)$ to get a quadratic quotient
Factor the quadratic: Solve the quadratic to find the remaining two zeros

Let's see if you can execute this complete chain!

✍️ Question

Problem 📝

Given that $1$ is a zero of $f (x) = - x^{3} + 7 x - 6$ , find all zeros.

Show the complete chain:

Step 1: Write the factor from the known zero
Step 2: Perform polynomial division to get the quadratic quotient
Step 3: Factor the quadratic and list all three zeros

Let's trace through the complete chain.

Step 1: Since $1$ When you know a zero, you know a factor is a zero, $(x - 1)$ (That's the Factor Theorem — it works both ways) is a factor.

✍️ MCQ

Choose one

Why does knowing that

1

is a zero tell us that

(x - 1)

is a factor?

✓

Polynomial long division setup: -x^3 + 0x^2 + 7x - 6 divided by (x-1), showing Cycle 1 with -x^2 in quotient, subtraction yielding -x^2 + 7x - 6

Step 2: Divide $- x^{3} + 0 x^{2} + 7 x - 6$ by $(x - 1)$ . Note the $0x^2$ placeholderWrite zero times that power to keep columns aligned.

Cycle 1: $\frac{- x^{3}}{x} = - x^{2}$ . Multiply: $-x^2(x-1) = -x^3 + x^2$ Divide leading term of what remains by leading term of divisor. Subtract: $(0 x^{2} - x^{2}) + 7 x - 6 =$ $- x^{2} + 7 x - 6$ .

Continuation of long division showing Cycles 2-3: quotient building to -x^2 - x + 6, final subtraction yielding remainder 0

Cycle 2: $\frac{- x^{2}}{x} = - x$ . Multiply: $-x(x-1) = -x^2 + x$ . Subtract: $(7 x - x) - 6 =$ $6x - 6$ .

Cycle 3: $\frac{6 x}{x} = 6$ . Multiply: $6 (x - 1) = 6 x - 6$ . Subtract: $0$ zero!.

✍️ MCQ

Choose one

We got a remainder of

0

after dividing by

(x - 1)

. What does this confirm?

✓

Quotient: $- x^{2} - x + 6$ . Remainder: $0$ confirms!Confirms the divisor is indeed a factor (confirms 1 is a zeroThis is how you verify your work in exams).

Step 3: Factor $- x^{2} - x + 6$ .

First, factor out $-1$ Makes factoring easier with positive numbers: $-(x^2 + x - 6)$ Easier to factor when leading coefficient is positive

Now factor $x^{2} + x - 6$ :

ProductFind two numbers with this product $= - 6$
Sum(The same two numbers must have this sum) $= 1$

We need two numbers that multiply to $-6$ and add to $1$ The core pattern for factoring quadratics.

The pair: $3$ and $-2$ ✓These numbers satisfy both conditions ✓

✍️ MCQ

Choose one

We found that

3

and

-2

work. How do we write the factored form of

x^2 + x - 6

?

✓

So $x^2 + x - 6 = (x + 3)(x - 2)$

Therefore: $-x^2 - x + 6 = -(x + 3)(x - 2)$

Zeros: $x = -3$ First zero from the factors and $x = 2$ Second zero from the factors (the $-1$ no zero!Just a constant, no x, can never equal zero factor does not give a zero).

✍️ MCQ

Choose one

Why doesn't the factor

-1

give us a zero of the polynomial?

✓

All three zeros: $1, -3, 2$ .