EXAM ONE

Polynomial Long Division

You already worked a full problem correctly this week. This is reps, not reteaching — the one spot that bites is the subtract step, where the negative sign has to flip through every term.

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The mnemonic

Dave Matthews Sings Badly

Divide — the leading terms.

Multiply — that result by the whole divisor.

Subtract — flip the sign of every term in the divisor's product, then add. This is where sign errors live.

Bring down — the next term, and repeat.

Worked Example

$$(3x^4 + 5x^3 - 2x^2 + 7x - 4) \div (x^2 + 2x - 1)$$

1 · Divide leading terms: $3x^4 \div x^2 = 3x^2$. First term of the quotient.

2 · Multiply $3x^2$ by the whole divisor: $3x^2(x^2+2x-1) = 3x^4+6x^3-3x^2$.

3 · Subtract — flip every sign, then add. $(3x^4+5x^3-2x^2) - (3x^4+6x^3-3x^2)$ becomes $(3x^4+5x^3-2x^2) + (-3x^4-6x^3+3x^2) = -x^3+x^2$. Bring down the $+7x$: $-x^3+x^2+7x$.

4 · Divide again: $-x^3 \div x^2 = -x$. Second term of the quotient.

5 · Multiply: $-x(x^2+2x-1) = -x^3-2x^2+x$.

6 · Subtract — flip every sign, then add. $(-x^3+x^2+7x) + (x^3+2x^2-x) = 3x^2+6x$. Bring down the $-4$: $3x^2+6x-4$.

7 · Divide: $3x^2 \div x^2 = 3$. Third term of the quotient.

8 · Multiply: $3(x^2+2x-1) = 3x^2+6x-3$.

9 · Subtract — flip every sign, then add. $(3x^2+6x-4)+(-3x^2-6x+3) = -1$. Nothing left to bring down — that's the remainder.

$$3x^2 - x + 3, \quad R = -1$$

The trap: in step 3 and step 6, every single term inside the parentheses flips sign, not just the first one. A double-negative — flipping a sign that was already negative back to positive — is the exact spot a sign error hides. Write out the flip explicitly instead of doing it in your head: "$-(3x^4+6x^3-3x^2)$ becomes $-3x^4-6x^3+3x^2$," term by term.

Now you

Faded — Just the Subtract Step

This step already appears mid-problem. Fill in the flip, then check.

$$(4x^2 - 3x) - (4x^2 - 9x)$$

1 · Flip both signs of the second group: $-(4x^2-9x)$ becomes $\underline{\qquad}$

2 · Combine: $(4x^2-3x) + (\text{your flipped group}) = \underline{\qquad}$

Check it ›

$-(4x^2-9x) = -4x^2+9x$. Combined: $4x^2-3x-4x^2+9x = 6x$.

Practice Set (6) — Work It Out on Paper

Kept free-response here on purpose — a multiple-choice final answer can't show you which step broke, and this is a process skill. Do the full division, then reveal to check every step, not just the final answer.

1 · No sign flips — pure mechanics rep.

$(x^3+5x^2+8x+4) \div (x+1)$

Check it ›

Quotient $x^2+4x+4$, remainder $0$. Every subtract step here has no double-negative — confirm the Divide/Multiply/Bring-down sequence is automatic before adding sign complexity.

2 · One sign flip, isolated.

$(x^3-2x^2-5x+6) \div (x-3)$

Check it ›

Quotient $x^2+x-2$, remainder $0$. Exactly one step needs the flip-and-add move — locate it before you start writing.

3 · Alternating signs — closest to the trap.

$(2x^3-3x^2+4x-5) \div (x-1)$

Check it ›

Quotient $2x^2-x+3$, remainder $-2$. Sign flips required at nearly every subtract step — this is the one to slow all the way down on.

4 · Negative divisor.

$(x^3+2x^2-x-2) \div (x+2)$ — this time watch the divisor: the negative can come from the divisor's own terms too, not just the dividend.

Check it ›

Quotient $x^2-1$, remainder $0$. Same skill, different source of the negative — confirms it's not just "watch the dividend."

5 · Missing-degree term.

$(x^3-8) \div (x-2)$

Check it ›

Quotient $x^2+2x+4$, remainder $0$. There's no $x^2$ or $x$ term in the dividend — write it as $x^3+0x^2+0x-8$ first, so nothing shifts columns by accident.

6 · Fusion — full length, signs and a placeholder.

$(5x^4-3x^2+2x-9) \div (x^2+x+2)$

Check it ›

Write the dividend as $5x^4+0x^3-3x^2+2x-9$ first. Quotient $5x^2-5x-8$, remainder $20x+7$. This one's exam-length — everything you've practiced, in one problem.

Multiple-choice gotcha

Recognizing the Answer Format

The exam is multiple choice — and a division remainder can be shown two different ways. Both mean the same thing; only one might match what's on your screen.

Which of these correctly represents $10x - 6$ with remainder $-2$, divided by $(7x^3+2x^2+10x+7)$?

A) $10x-6, \; R=-2$ B) $10x-6+\dfrac{-2}{7x^3+2x^2+10x+7}$ C) $10x-6-2$

Check it ›

Both A and B are correct — the exam might show either format. A is the "quotient, remainder" style; B folds the remainder into the expression as a fraction over the divisor. They're the same answer written two ways. C just glued the remainder onto the quotient with no fraction — that's not equivalent to either.

Before You Leave — Add This to Your Sheet

"Dave Matthews sings badly — Divide, Multiply, Subtract (flip every sign, then add), Bring down." And: a remainder can show up on the exam as "$R = {-2}$" or as "$+\dfrac{-2}{\text{divisor}}$" — same thing, two formats.

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