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Polynomial Long Division: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 Dividing a fourth-degree dividend

Divide $x^4+x^2+2$ by $x^3+x+1$ . Give the quotient and remainder.
1. Hint 1
  Cancel the dividend's highest power with a multiple of the divisor.
2. Hint 2
  Divide the leading terms to find the first quotient term, then compare the remaining degree with the divisor's degree.
Answer

$Q(x)=x$ ; $R(x)=-x+2$ .

Full solution

Include the missing cubic and linear columns.
Multiplying the divisor by $x$ gives $x^4+x^2+x$ .
Subtracting from the dividend leaves
$R(x)=-x+2$
Its degree is $1<3$ , so division stops with $Q=x$ .
Expanding $(x^3+x+1)x+(-x+2)$ restores $x^4+x^2+2$ .

Answer

$Q(x)=x$ ; $R(x)=-x+2$ .

Key idea

Division stops when the remainder degree falls below the divisor degree, even if powers are missing.
Problem 1: Mark complete
Problem 2 Filling in missing columns

Write $3x^5-4x^2+7$ with every missing power's column shown, ready for long division.
1. Hint 1
  List every power from the polynomial's degree down to the constant term, whether or not it appears in the given expression.
2. Hint 2
  A power that is absent from the polynomial still needs a column; give it coefficient $0$ so the column stays open.
Answer

$3x^5+0x^4+0x^3-4x^2+0x+7$ .

Full solution

The polynomial has degree $5$ , so its columns run from $x^5$ down to $x^0$ , six powers in all.
Only three of them, $x^5$ , $x^2$ , and the constant term, appear in the given expression, so the powers $x^4$ , $x^3$ , and $x^1$ are missing.
$\begin{aligned} 3x^5-4x^2+7 &= 3x^5+0x^4+0x^3 \\ &\quad -4x^2+0x+7 \end{aligned}$
Each inserted $0x^k$ adds nothing to the value, but it opens a column so the tableau's subtractions line up matching powers.

Answer

$3x^5+0x^4+0x^3-4x^2+0x+7$ .

Key idea

Preparing a division starts with detecting which powers are absent, then giving each a coefficient of $0$ so every column stays aligned.
Problem 2: Mark complete
Problem 3 Finding a three-term quotient

Divide $x^3+x^2-10x+13$ by $x+4$ . Give the quotient and remainder.
1. Hint 1
  Divide the leading terms to find the first quotient term, then multiply back and subtract the whole product before repeating.
2. Hint 2
  After each subtraction, compare the new remaining degree with the divisor's degree $1$ to decide whether another pass is needed.
Answer

$Q(x)=x^2-3x+2$ ; $R(x)=5$ .

Full solution

First pass: $x^3\div x=x^2$ , and
$x^2(x+4)=x^3+4x^2$
Subtracting from the dividend leaves $-3x^2-10x+13$ .
Second pass: $-3x^2\div x=-3x$ , and $-3x(x+4)=-3x^2-12x$ ; subtracting leaves $2x+13$ .
Third pass: $2x\div x=2$ , and $2(x+4)=2x+8$ ; subtracting leaves $5$ , of degree $0$ , below the divisor's degree $1$ , so the loop stops.

The quotient is $x^2-3x+2$ and the remainder is $5$ .
Multiplying back, $(x+4)(x^2-3x+2)$ expands to $x^3+x^2-10x+8$ .
Adding the remainder $5$ gives $x^3+x^2-10x+13$ , which restores the dividend.

Answer

$Q(x)=x^2-3x+2$ ; $R(x)=5$ .

Key idea

A quotient of several terms still comes from the same forced move, repeated one leading-term cancellation at a time.
Problem 3: Mark complete
Problem 4 Recovering a rectangular panel's length

A rectangular panel has area $2x^3+7x^2+7x+2$ square meters and width $2x+1$ meters, where $x>0$ . Find its length in polynomial standard form and verify the area by multiplication.
1. Hint 1
  Area equals width times length, so divide the area polynomial by the width.
2. Hint 2
  The successive quotient terms come from canceling each leading term.
Answer

Length $x^2+3x+2$ meters; verification $(2x+1)(x^2+3x+2)=2x^3+7x^2+7x+2$ .

Full solution

Dividing leading terms first gives $x^2$ .
Subtracting $(2x+1)x^2$ leaves $6x^2+7x+2$ .
The next term is $3x$ , leaving $4x+2$ .
The final term is $2$ , leaving zero.

The length is $x^2+3x+2$ .
One contribution to the product is
$2x(x^2+3x+2)=2x^3+6x^2+4x.$
The other contribution is $x^2+3x+2$ .
Adding restores the stated area.
The length and width are both positive for $x>0$ , matching the physical panel.

Answer

Length $x^2+3x+2$ meters; verification $(2x+1)(x^2+3x+2)=2x^3+7x^2+7x+2$ .

Key idea

An exact polynomial division can recover a missing dimension from an area model.
Problem 4: Mark complete
Problem 5 A division after a change

Dividing $P(x)$ by $x^2+2$ gives quotient $x+1$ and remainder $3x-1$ . Find the quotient and remainder when $P(x)+2x^2+5$ is divided by the same divisor.
1. Hint 1
  Use the given identity and separate the added expression into divisor multiples and a smaller remainder.
2. Hint 2
  Two copies of $x^2+2$ account for $2x^2+4$ .
Answer

New quotient $x+3$ ; new remainder $3x$ .

Full solution

The added term satisfies
$2x^2+5=2(x^2+2)+1$
Thus two more copies of the divisor join the quotient, and $1$ joins the remainder.
The quotient becomes $x+3$ and the remainder becomes $3x$ .
Its degree $1$ is below $2$ , so this is finished.
Subtracting the original identity from the new one returns $2x^2+5$ , checking the change.

Answer

New quotient $x+3$ ; new remainder $3x$ .

Key idea

Adding a divisor multiple changes the quotient, while an additional smaller term changes the remainder.
Problem 5: Mark complete
Problem 6 A divided expression

For $x\ne0$ , write $\frac{x^4+2x^2+3}{x^2}$ as a polynomial plus a fraction whose numerator degree is smaller than the denominator degree. Identify its quotient and remainder.
1. Hint 1
  Divide the terms containing at least two powers of $x$ .
2. Hint 2
  The constant term cannot enter the polynomial quotient.
Answer

$x^2+2+\frac3{x^2}$ ; quotient $x^2+2$ , remainder $3$ .

Full solution

The first two terms divide exactly by $x^2$ , while the constant remains over the divisor.
$\frac{x^4+2x^2+3}{x^2}=x^2+2+\frac3{x^2}$
The remainder $3$ has degree $0<2$ .
Multiplying through by the nonzero $x^2$ restores the dividend.

Answer

$x^2+2+\frac3{x^2}$ ; quotient $x^2+2$ , remainder $3$ .

Key idea

Fraction form keeps the remainder over the original nonzero divisor.
Problem 6: Mark complete
Problem 7 A missing division result

A division of $P(x)=2x^4-x^3+3x^2-x+5$ by $D(x)=x^2+1$ has a quotient beginning $2x^2-x$ . Finish the division, and state why the final remainder is allowed.
1. Hint 1
  Subtract the product of the given partial quotient and the divisor.
2. Hint 2
  Continue if the working remainder still has degree at least $2$ .
Answer

$Q(x)=2x^2-x+1$ ; $R(x)=4$ , of degree $0<2$ .

Full solution

The partial product is
$\begin{aligned} D(x)(2x^2-x) &= 2x^4-x^3 \\ &\quad +2x^2-x \end{aligned}$
Subtracting it from $P$ leaves $x^2+5$ .
One more copy of $D$ leaves $4$ , so $Q=2x^2-x+1$ and $R=4$ .
The remainder degree is below the divisor degree.
Adding that product and $4$ reconstructs all five coefficients of $P$ .

Answer

$Q(x)=2x^2-x+1$ ; $R(x)=4$ , of degree $0<2$ .

Key idea

A partial quotient is finished by continuing to cancel the highest remaining power.
Problem 7: Mark complete
Problem 8 A large remainder

A division by $x^2+1$ finishes with remainder $1000x+1000$ . A student rejects it because the remainder is numerically larger than the divisor at $x=1$ . Is that a valid objection? Explain.
1. Hint 1
  Polynomial division measures remainder size by degree.
2. Hint 2
  Compare degrees instead of values at one input.
Answer

No; a degree- $1$ remainder is permitted for a degree- $2$ divisor.

Full solution

The degree condition is $\deg R<\deg D$ , and here
$1<2$
It does not require $R(1)<D(1)$ .
For example, choosing $P=(x^2+1)x+1000x+1000$ produces exactly this remainder, regardless of its large value at one.

Answer

No; a degree- $1$ remainder is permitted for a degree- $2$ divisor.

Key idea

A remainder is small enough in degree, not necessarily in numerical value.
Problem 8: Mark complete
Problem 9 Multiplying a dividend by x

Let $D(x)=x^2+3$ . Polynomials $P$ and $Q$ satisfy $P=DQ+3x+4$ . A student says division of $xP$ by $D$ has remainder $4x-9$ , whatever $Q$ is. Is the claim correct? Find the new quotient in terms of $Q$ .
1. Hint 1
  Multiplying the dividend also multiplies the old remainder, which may need more division.
2. Hint 2
  Multiply the given identity by $x$ , then remove one more multiple of $D$ from the expression outside the divisor product.
Answer

Yes; the new quotient is $xQ+3$ and the remainder is $4x-9$ .

Full solution

Multiplying the given identity by $x$ gives
$xP=D(xQ)+3x^2+4x.$
The expression outside the divisor product now has degree $2$ , so it is not a finished remainder.
Since $D=x^2+3$ , write
$3x^2+4x=3D+(4x-9).$

Combining the divisor multiples gives
$xP=D(xQ+3)+(4x-9).$
The remainder has degree $1$ , below the degree $2$ of $D$ , so this is the completed division.
Expanding the right side restores the multiplied original identity, independently of the choice of $Q$ .

Answer

Yes; the new quotient is $xQ+3$ and the remainder is $4x-9$ .

Key idea

Multiplying a dividend can make its old remainder require another division step.
Problem 9: Mark complete
Problem 10 Scaling both sides of a division

Suppose division of a nonzero polynomial $P$ by a nonzero polynomial $D$ gives quotient $Q$ and remainder $R$ . A learner says division of $3P$ by $3D$ has quotient $Q$ and remainder $R$ . Decide when that statement is correct, and give the general correct remainder.
1. Hint 1
  Scale the entire division identity by $3$ .
2. Hint 2
  Check which part is absorbed into the divisor and which part remains outside the product.
Answer

Correct exactly when $R=0$ ; in general the quotient is $Q$ and the remainder is $3R$ .

Full solution

Multiply $P=DQ+R$ by $3$ .
$3P=(3D)Q+3R$
Scaling by a nonzero constant preserves degree, so $3R$ is still zero or has degree below $3D$ .
Uniqueness makes this the correct result.
It equals the claimed remainder $R$ exactly when $3R=R$ , which forces $R=0$ .

Answer

Correct exactly when $R=0$ ; in general the quotient is $Q$ and the remainder is $3R$ .

Key idea

Scaling both dividend and divisor by the same nonzero constant preserves the quotient and scales the remainder by that constant.
Problem 10: Mark complete

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