Welcome! Today we're tackling Quadratic Division and Polynomial Remainders — taking your division skills to the next level.

Dividing by a quadratic is the same cycle as before:

Divide → Multiply → Subtract → Check

But each multiplication step is more involved because you are multiplying by a three-term expression.

The remainder can now be a linear expression like $2x + 3$ example, not just a single number.

Division Type	Remainder Form
By linear	Constant (e.g., 5)
By quadratic	Linear (e.g., $2 x + 3$ )

This skill is essential for the final cluster, where you divide quartics by quadratic factors to find all zeros.

1. Multi-term multiplication in quadratic division

Let's look at dividing by a quadratic.

When dividing by a quadratic, each multiplication step involves multiplying a single term by a trinomial, producing three terms.

This is more arithmetic per step than linear division.

📋 Given Info

Polynomial long division setup with dividend x^4 + 2x^3 + 8x^2 + 12x + 18 under the division bracket and divisor x^2 + 5 outside, with x^2 written above as first quotient term

Here's our problem:

We're dividing $x^4 + 2x^3 + 8x^2 + 12x + 18$ by $(x^2 + 5)$ .

Finding the first quotient term:

Divide the leading terms: $\frac{x^4}{x^2} = x^2$

So the first quotient term is $x^2$ 1st term.

✍️ Question

Your turn ✏️

In dividing $x^{4} + 2 x^{3} + 8 x^{2} + 12 x + 18$ by $(x^{2} + 5)$ , the first quotient term is $x^2$ .

Same division setup with x^2 times (x^2 + 5) = x^4 + 5x^2 written below dividend, subtraction line drawn, showing the multiplication step to perform

Multiply $x^2$ by the divisor and subtract from the dividend. What remains?

When multiplying $x^{2}$ by $(x^{2} + 0 x + 5)$ , include ALL termsInclude all terms when multiplying — even the ones with zero coefficientsDon't skip zero coefficients or columns won't align!

$x^2 \times x^2 = x^4$
$x^2 \times 0x = 0x^3$
$x^2 \times 5 = 5x^2$

✍️ MCQ

Choose one

Why do we include the

0x^3

term even though it equals zero?

✓

Result: $x^{4} + 0 x^{3} + 5 x^{2}$

This is what we subtractThis is where mistakes happen from the dividend $x^{4} + 2 x^{3} + 8 x^{2} + 12 x + 18$ .

Now subtractWhere the actual division happens from the dividend:

$(x^{4} + 2 x^{3} + 8 x^{2} + 12 x + 18)$ - $(x^{4} + 0 x^{3} + 5 x^{2})$

Column by column:

$x^{4} - x^{4} = 0$
$2 x^{3} - 0 x^{3} = 2 x^{3}$
$8 x^{2} - 5 x^{2} = 3 x^{2}$
$12x$ staysThey just come down unchanged (nothing to subtract)
$18$ staysNothing to subtract from these terms (nothing to subtract)

✍️ MCQ

Choose one

When we subtracted

8x^2 - 5x^2

, we got

3x^2

. What coefficient did we get for the

x^3

term after subtraction?

✓

Result: $2x^3 + 3x^2 + 12x + 18$ remainderThis is what we divide in the next round

This is our new dividendThat reduction shows you're making progress for the next step of the division.

2. Completing quadratic division and reading the polynomial remainder

The division continues until the remainder's degree drops below 2. The remainder is then a linear expression (or constant), which some problems ask you to express as $px + q$ .

We've already completed cycle 1 of dividing $x^{4} + 2 x^{3} + 8 x^{2} + 12 x + 18$ by $(x^{2} + 5)$ .

After cycle 1, the remainder is: $2x^3 + 3x^2 + 12x + 18$

✍️ Question

Complete the division 📝

Divide $x^4 + 2x^3 + 8x^2 + 12x + 18$ by $(x^2 + 5)$ .

State the quotient and remainder.

(Remember: After cycle 1, you have $2x^3 + 3x^2 + 12x + 18$ from cycle 1 remaining. Continue from there.)

Cycle 2:

Continuing from $2x^3 + 3x^2 + 12x + 18$ :

DivideLook at leading terms only: $\frac{2 x^{3}}{x^{2}} =$ $2x$ quotient(What you get from dividing leading terms)

✍️ MCQ

Choose one

When we multiply

2x

by

(x^2 + 0x + 5)

, what do we get for the last term?

2x \times 5 = \ ?

✓

Multiply: $2x(x^2 + 0x + 5) = 2x^3 + 0x^2 + 10x$ Must include every term, not just the first

SubtractThe highest power term disappears completely: $(2 x^{3} + 3 x^{2} + 12 x + 18) - (2 x^{3} + 0 x^{2} + 10 x) =$ $3x^2 + 2x + 18$ resultDegree decreases with each cycle

✍️ Yes/No

Yes or No?

After Cycle 2, our remainder is

3x^2 + 2x + 18

. Can we continue dividing this by

x^2 + 5

?

✓

Cycle 3:

Divide: $\frac{3 x^{2}}{x^{2}} =$ $3$ quotient

Multiply: $3 (x^{2} + 0 x + 5) =$ $3x^2 + 0x + 15$

✍️ MCQ

Choose one

When we subtract

(3x^2 + 0x + 15)

from

(3x^2 + 2x + 18)

, what do we get for the

x

term?

✓

Subtract: $(3 x^{2} + 2 x + 18) - (3 x^{2} + 0 x + 15) = 2 x + 3$

Since the degree of $2x + 3$ (which is 1)When remainder degree is lower, we stop is less than the degree of the divisor $x^2 + 5$ (which is 2)The divisor's degree determines when to stop, we stop hereThis is the stopping condition.

This means $2 x + 3$ is our remainderWhat remains after you can't divide anymore.

✍️ T/F

True or False?

We stop the division when the degree of the remainder is less than the degree of the divisor. True or False?

✓

Stopping ConditionTells you when to put down your pen:

$degree (2 x + 3) = 1 < degree (x^{2} + 5) = 2$ . STOPCannot divide when degree drops below divisor.

We cannot continue dividing because the remaining polynomial has a lower degreeWhat's left becomes your remainder than the divisor.

Final Answer:

Quotient: $x^{2} + 2 x + 3$
Remainder(It's a linear expression, not just a number): $2 x + 3$ (a linear expression, not just a numberWhen dividing by quadratic, remainder can be degree 0 or 1)

Sanity check: The quotient has degree $4 - 2 = 2$ . ✓ Correct.

✍️ MCQ

Choose one

If we divided a degree 5 polynomial by a degree 2 polynomial, what would be the degree of the quotient?

✓