Core practice ← Back to lesson

Synthetic Division and the Remainder Theorem: 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)

0 of 10 completed

Progress saved in this browser.

Problem 1 of 10

Problem 1 A row of coefficients

A cubic has coefficient row $3,-2,0,4$ in descending powers. Use synthetic division to divide it by $x+1$ , and give its quotient and remainder.
1. Hint 1
  The corner number is the value that makes $x+1$ zero.
2. Hint 2
  Keep the zero column; bring down, multiply by the corner, and add.
Answer

$Q(x)=3x^2-5x+5$ ; $R=-1$ .

Full solution

The corner is $-1$ .
Bringing down $3$ , the next bottom entries are
$-2+(-1)(3)=-5$
then
$0+(-1)(-5)=5$
and finally
$4+(-1)(5)=-1$
Thus $Q=3x^2-5x+5$ and $R=-1$ .
Expanding $(x+1)Q-1$ gives $3x^3-2x^2+4$ , matching the coefficient row.

Answer

$Q(x)=3x^2-5x+5$ ; $R=-1$ .

Key idea

The final synthetic entry is the remainder; the earlier entries start one degree below the dividend.
Problem 1: Mark complete
Problem 2 An evaluation by tableau

Use synthetic division to evaluate $P(4)$ for $P(x)=2x^3-3x^2-x+5$ .
1. Hint 1
  List the coefficients of $P$ in descending order, then run the tableau with corner $4$ .
2. Hint 2
  The final entry of the bottom row equals $P(4)$ , by the Remainder Theorem.
Answer

$P(4)=81$ .

Full solution

The top row is $2,-3,-1,5$ with corner $4$ .
Bring down $2$ ; then
$-3+4(2)=5$
then
$-1+4(5)=19$
and finally
$5+4(19)=81$
The final entry is the remainder, which by the Remainder Theorem is $P(4)$ .

Direct evaluation confirms it:
$P(4)=2(64)-3(16)-4+5=81$

Answer

$P(4)=81$ .

Key idea

The synthetic tableau's final entry evaluates $P$ at the corner, by the Remainder Theorem.
Problem 2: Mark complete
Problem 3 A divisor label

A synthetic division uses corner $-\frac32$ . Write the monic linear divisor $x-c$ for this division.
1. Hint 1
  The divisor is $x-c$ when the corner is $c$ .
2. Hint 2
  Substitute the corner's value for $c$ , and simplify the resulting sign.
Answer

Divisor $x+\frac32$ .

Full solution

With corner $c=-\frac32$ , the divisor is
$x-c=x-\left(-\frac32\right)=x+\frac32.$

Answer

Divisor $x+\frac32$ .

Key idea

The synthetic corner $c$ names the divisor $x-c$ directly, sign included.
Problem 3: Mark complete
Problem 4 A quartic dividend

Divide $P(x)=x^4+2x^3-5x-6$ by $x-3$ using synthetic division. Give the quotient and remainder, then verify the remainder by evaluating $P(3)$ directly.
1. Hint 1
  Include a placeholder zero for the missing $x^2$ term before starting the tableau.
2. Hint 2
  The Remainder Theorem says the final synthetic entry equals $P$ evaluated at the corner; use that to check your work.
Answer

$Q(x)=x^3+5x^2+15x+40$ ; $R=114$ ; $P(3)=114$ .

Full solution

The top row needs a placeholder zero for the missing $x^2$ term, giving $1,2,0,-5,-6$ with corner $3$ .
Bring down $1$ ; then
$2+3(1)=5$
then
$0+3(5)=15$
then
$-5+3(15)=40$
and finally
$-6+3(40)=114$
Thus $Q(x)=x^3+5x^2+15x+40$ and $R=114$ .

Direct evaluation gives
$P(3)=81+54-15-6=114$
matching the synthetic remainder, as the Remainder Theorem predicts.

Answer

$Q(x)=x^3+5x^2+15x+40$ ; $R=114$ ; $P(3)=114$ .

Key idea

A quartic divided by a linear divisor follows the same bring-down, multiply, add rhythm, with the remainder equal to $P(c)$ .
Problem 4: Mark complete
Problem 5 A nonunit coefficient

Use synthetic division to divide $4x^3+4x^2-x+1$ by $2x+3$ . Give the quotient and remainder and check the division identity.
1. Hint 1
  First divide by the monic factor $x+\frac32$ .
2. Hint 2
  The original divisor has an extra factor of $2$ , so adjust the quotient but not the remainder.
Answer

$Q(x)=2x^2-x+1$ ; $R=-2$ ; $(2x+3)(2x^2-x+1)-2=4x^3+4x^2-x+1$ .

Full solution

Use corner $-\frac32$ on $4,4,-1,1$ .
The successive bottom entries are $4,-2,2,-2$ , since $4-6=-2$ , $-1+3=2$ , and $1-3=-2$ .
This gives quotient $4x^2-2x+2$ for the monic divisor.

Divide that quotient by $2$ to account for $2x+3$ .
The result is $Q(x)=2x^2-x+1$ , with remainder $-2$ .
With $D(x)=2x+3$ , multiplication gives
$D(x)Q(x)=4x^3+4x^2-x+3$
Adding $-2$ restores the dividend.

Answer

$Q(x)=2x^2-x+1$ ; $R=-2$ ; $(2x+3)(2x^2-x+1)-2=4x^3+4x^2-x+1$ .

Key idea

A scaled linear divisor rescales the quotient and leaves the remainder unchanged.
Problem 5: Mark complete
Problem 6 A shifted input rule

A polynomial $P$ leaves remainder $-4$ on division by $x-2$ and remainder $7$ on division by $x+1$ . Define $H(x)=P(x-1)+3$ . Find the remainders when $H$ is divided by $x-3$ and by $x$ .
1. Hint 1
  Each requested remainder is an evaluation of $H$ at a divisor zero.
2. Hint 2
  Translate those inputs into inputs of $P$ using the definition of $H$ .
Answer

Remainder $-1$ for $x-3$ ; remainder $10$ for $x$ .

Full solution

The supplied remainders mean $P(2)=-4$ and $P(-1)=7$ .
For $x-3$ ,
$H(3)=P(2)+3$
This is $-1$ .
For $x$ ,
$H(0)=P(-1)+3$
This is $10$ .
No coefficients of $P$ are needed because both transformed inputs are known.

Answer

Remainder $-1$ for $x-3$ ; remainder $10$ for $x$ .

Key idea

A remainder condition supplies function values that can be carried through a change of input.
Problem 6: Mark complete
Problem 7 An incomplete record

A synthetic division of $2x^2+x-3$ has bottom row $2,5,7$ , where $7$ is the remainder. Recover the corner number and divisor, then show how the middle entry matches the first long-division subtraction.
1. Hint 1
  The middle entry equals the next original coefficient plus the corner times the first entry.
2. Hint 2
  Once the corner is found, multiply the divisor by the first quotient term.
Answer

Corner $2$ ; divisor $x-2$ ; first subtraction leaves $5x-3$ .

Full solution

If the corner is $c$ , the row requires
$1+2c=5$
Hence $c=2$ and the divisor is $x-2$ .
Its first quotient term is $2x$ , giving product $2x^2-4x$ .
Subtracting from $2x^2+x-3$ leaves $5x-3$ , matching the middle bottom entry $5$ .
The final row entry also checks: $-3+2(5)=7$ .

Answer

Corner $2$ ; divisor $x-2$ ; first subtraction leaves $5x-3$ .

Key idea

A synthetic middle entry is the coefficient left by the corresponding long-division subtraction.
Problem 7: Mark complete
Problem 8 A learner's shortcut

In division of $ax^2+bx+d$ by $x-c$ , a learner says the quotient constant is $b+ac$ . Is this correct for real $a,b,c,d$ with $a\ne0$ ? Explain by comparing with one long-division step.
1. Hint 1
  The first quotient term is forced by the leading terms.
2. Hint 2
  Expand $ax(x-c)$ and subtract its entire product.
Answer

Correct; quotient constant $b+ac$ .

Full solution

The first quotient term is $ax$ .
Its product with the divisor is $ax^2-acx$ .
Subtracting this from the dividend leaves
$(b+ac)x+d$
The next quotient term is therefore $b+ac$ .
This is exactly the synthetic rule of adding $c$ times the previous bottom entry $a$ to the next coefficient $b$ .

Answer

Correct; quotient constant $b+ac$ .

Key idea

Synthetic addition incorporates the sign change already present in the divisor $x-c$ .
Problem 8: Mark complete
Problem 9 A sign comparison

A student says the remainders from dividing a polynomial by $x-1$ and by $x+1$ must be opposites. Test the claim using $P(x)=x^2+2x+4$ , and explain your conclusion.
1. Hint 1
  The remainders are $P(1)$ and $P(-1)$ .
2. Hint 2
  Even-power and constant terms do not change sign when the input changes sign.
Answer

False; the two remainders are $7$ and $3$ .

Full solution

Evaluating at the two corners gives
$P(1)=1+2+4$
and
$P(-1)=1-2+4$
The remainders are $7$ and $3$ , which are not opposites.
Changing the corner sign changes the evaluation input, not a fixed sign in the answer.

Answer

False; the two remainders are $7$ and $3$ .

Key idea

Remainders at opposite inputs need not be opposites.
Problem 9: Mark complete
Problem 10 A zero final entry

Knowing only that the final synthetic-division entry is $0$ when some cubic $P$ is divided by $x-c$ , a learner says that proves $P(c)=0$ , but does not determine $Q(c)$ , where $Q$ is the quotient. Is this correct? Support your answer with two possible cubics using $c=0$ and different values of $Q(0)$ .
1. Hint 1
  The identity contains a factor $x-c$ multiplying the quotient.
2. Hint 2
  At $c=0$ , try cubics of the form $x(x^2+k)$ with different constants $k$ .
Answer

Correct; $P=x^3+x$ gives $Q(0)=1$ , while $P=x^3+2x$ gives $Q(0)=2$ . Both have $P(0)=0$ .

Full solution

The zero final entry means $P(x)=(x-c)Q(x)$ , so substitution at $c$ gives $P(c)=0$ regardless of $Q(c)$ .
For $c=0$ , take
$P_1(x)=x(x^2+1)$
and
$P_2(x)=x(x^2+2)$
Their quotients have values $1$ and $2$ at zero, while both dividends vanish there.

Answer

Correct; $P=x^3+x$ gives $Q(0)=1$ , while $P=x^3+2x$ gives $Q(0)=2$ . Both have $P(0)=0$ .

Key idea

A zero remainder fixes the dividend value at the corner without fixing the quotient value there.
Problem 10: Mark complete

← Back to lesson