Polynomial long division and synthetic division: Exercises

This page is a self-test. Read each problem, cover the solution, work it yourself, then open the collapsible [!recall]- block to check every line. The problems climb five levels of difficulty:

Everything here builds directly on Polynomial long division and synthetic division. If a word feels new, follow the wikilinks to the prerequisite notes.

Level 1 — Recognition

Here you only prove you can set up the machinery: order terms, fill missing powers with zeros, and read off the correct value of $c$ .

For $P (x) = 5 + x^{3} - 2 x$ divided by $x - 4$ , write $P (x)$ in descending powers with all zero coefficients shown, list the coefficient row, and state the value of $c$ used in synthetic division.

Recall Solution

WHAT we do first: put every term in order of decreasing power. The powers present are $x^{3}$ and $x^{1}$ and the constant. The power $x^{2}$ is missing, so we insert $0 x^{2}$ . $P (x) = x^{3} + 0 x^{2} - 2 x + 5$ WHY the zero matters: the algorithm lines terms up by place value, exactly like writing $1005$ and not $15$ . Skip a slot and every later column shifts into the wrong power.

Coefficient row: $[1, 0, - 2, 5]$ .

The value of $c$ : synthetic division uses the number that makes the divisor zero. Solve $x - 4 = 0 \Rightarrow x = 4$ . So $c = 4$ .

Exercise 1.2

For division by $x + 7$ , state $c$ . For division by $3 x - 6$ , explain in one sentence why plain synthetic division does not directly apply, and give the equivalent divisor of the form $x - c$ .

Recall Solution

Divisor $x + 7$ : rewrite as $x - (- 7)$ . It equals zero at $x = - 7$ , so $== c = - 7 ==$ .

Divisor $3 x - 6$ : the synthetic-division shortcut only works when the divisor is first degree with leading coefficient 1. Here the leading coefficient is $3$ , not $1$ . The fix: factor out the $3$ : $3 x - 6 = 3 (x - 2)$ . Dividing by $x - 2$ uses $c = 2$ ; you then divide the whole quotient by the extra factor of $3$ at the end.

Level 2 — Application

Now run the full algorithm to a clean finish.

Exercise 2.1

Compute $(x^{3} + 0 x^{2} - 2 x + 5) \div (x - 4)$ by synthetic division. State the quotient and remainder.

Recall Solution

Coefficients $[1, 0, - 2, 5]$ , and $c = 4$ .

  4 |  1    0   -2    5
    |       4   16   56
    |___________________
      1    4   14   61

Walk the steps (WHAT / WHY each column does):

Bring down $1$ . (First quotient coefficient is always the first dividend coefficient.)
$1 \times 4 = 4$ , place under $0$ ; add: $0 + 4 = 4$ .
$4 \times 4 = 16$ , place under $- 2$ ; add: $- 2 + 16 = 14$ .
$14 \times 4 = 56$ , place under $5$ ; add: $5 + 56 = 61$ .

The bottom row is $[1, 4, 14 ∣ 61]$ . The last number is the remainder; the rest are quotient coefficients, one degree lower than the dividend. $Quotient Q (x) = x^{2} + 4 x + 14, R = 61.$ $x^{3} - 2 x + 5 = (x - 4) (x^{2} + 4 x + 14) + 61.$

Exercise 2.2

Divide $2 x^{3} - 3 x^{2} + 0 x - 4$ by $2 x - 1$ using long division (leading coefficient is not 1, so use long division).

Recall Solution

Setup with the divisor $2 x - 1$ :

           x²  -  x  -  1/2
           _____________________
 2x - 1 ) 2x³ - 3x² + 0x - 4
          2x³ -  x²
          ────────
              -2x² + 0x
              -2x² +  x
              ─────────
                    -x - 4
                    -x + 1/2
                    ────────
                       -9/2

Step 1 — divide leading terms: $\frac{2 x ^{3}}{2 x} = x^{2}$ . Multiply $x^{2} (2 x - 1) = 2 x^{3} - x^{2}$ , subtract: $- 3 x^{2} - (- x^{2}) = - 2 x^{2}$ . Step 2: $\frac{- 2 x ^{2}}{2 x} = - x$ . Multiply $- x (2 x - 1) = - 2 x^{2} + x$ , subtract: $0 x - x = - x$ . Step 3: $\frac{- x}{2 x} = - \frac{1}{2}$ . Multiply $- \frac{1}{2} (2 x - 1) = - x + \frac{1}{2}$ , subtract: $- 4 - \frac{1}{2} = - \frac{9}{2}$ . Degree of remainder $- \frac{9}{2}$ (a constant, degree $0$ ) is less than degree of divisor ( $1$ ), so stop. $Q (x) = x^{2} - x - \frac{1}{2}, R = - \frac{9}{2} .$

Level 3 — Analysis

Here the number you get out carries meaning — remainders reveal roots.

Exercise 3.1

Without fully factoring, use synthetic division to decide whether $x = 3$ is a root of $P (x) = x^{3} - 4 x^{2} + x + 6$ . If it is, give the depressed quotient.

Recall Solution

By the Remainder Theorem, dividing by $x - 3$ leaves remainder $P (3)$ . If that remainder is $0$ , then $x = 3$ is a root. Coefficients $[1, - 4, 1, 6]$ , $c = 3$ :

  3 |  1   -4    1    6
    |       3   -3   -6
    |___________________
      1   -1   -2    0

Remainder $= 0$ , so yes, $x = 3$ is a root. The quotient (one degree lower) is $x^{2} - x - 2$ , giving $P (x) = (x - 3) (x^{2} - x - 2) = (x - 3) (x - 2) (x + 1) .$ The last factoring uses ordinary factoring of the depressed quadratic.

Exercise 3.2

$P (x) = 2 x^{3} + 3 x^{2} - 11 x - 6$ . The Rational Root Theorem says any rational root $p / q$ has $p ∣ 6$ and $q ∣ 2$ . Test $x = 2$ by synthetic division; if it works, factor $P$ completely.

Recall Solution

Coefficients $[2, 3, - 11, - 6]$ , $c = 2$ :

  2 |  2    3  -11   -6
    |       4   14    6
    |___________________
      2    7    3    0

Remainder $0$ ⇒ $x = 2$ is a root. Depressed quotient: $2 x^{2} + 7 x + 3$ . Factor it: $2 x^{2} + 7 x + 3 = (2 x + 1) (x + 3)$ . $P (x) = (x - 2) (2 x + 1) (x + 3) .$ Roots: $x = 2, - \frac{1}{2}, - 3$ . All are rational, consistent with the candidates ${\pm 1, \pm 2, \pm 3, \pm 6, \pm \frac{1}{2}, \pm \frac{3}{2}}$ from the Rational Root Theorem.

Level 4 — Synthesis

Combine division with a second idea in one problem.

Exercise 4.1

Simplify $\frac{x ^{3} + 2 x ^{2} - 5 x - 6}{x - 2}$ to a polynomial-plus-proper-remainder form, then confirm the numerator factors fully.

Recall Solution

Coefficients $[1, 2, - 5, - 6]$ , $c = 2$ :

  2 |  1    2   -5   -6
    |       2    8    6
    |___________________
      1    4    3    0

Remainder $0$ , quotient $x^{2} + 4 x + 3 = (x + 1) (x + 3)$ . $\frac{x ^{3} + 2 x ^{2} - 5 x - 6}{x - 2} = x^{2} + 4 x + 3 = (x + 1) (x + 3),$ and the numerator factors as $(x - 2) (x + 1) (x + 3)$ .

Exercise 4.2

Set up (do not fully solve) the partial fraction decomposition of $\frac{2 x ^{3} + 3 x ^{2} - 5 x + 1}{x + 2}$ . Use the fact from the parent note that $2 x^{3} + 3 x^{2} - 5 x + 1 = (x + 2) (2 x^{2} - x - 3) + 7$ .

Recall Solution

A partial fraction decomposition requires a proper fraction (numerator degree $<$ denominator degree). Here the numerator has degree $3$ , denominator degree $1$ — improper. Division fixes this first. Using the parent's result: $\frac{2 x ^{3} + 3 x ^{2} - 5 x + 1}{x + 2} = polynomial part 2 x^{2} - x - 3 + \frac{7}{x + 2} .$ The leftover $\frac{7}{x + 2}$ is already a single proper fraction over a linear factor, so the decomposition is complete — no unknown constants to solve. WHY this matters: every improper rational expression must be divided down before partial fractions; the quotient becomes a polynomial term and only the remainder gets decomposed.

Level 5 — Mastery

Reason about division itself, with symbols standing in for numbers.

Exercise 5.1

Find the value of $k$ so that $x - 2$ is a factor of $P (x) = x^{3} + k x^{2} - 4 x + 4$ .

Recall Solution

" $x - 2$ is a factor" means the remainder on dividing by $x - 2$ is $0$ , i.e. $P (2) = 0$ by the Factor Theorem. $P (2) = 2^{3} + k \cdot 2^{2} - 4 \cdot 2 + 4 = 8 + 4 k - 8 + 4 = 4 k + 4.$ Set to zero: $4 k + 4 = 0 \Rightarrow== k = - 1 ==$ . Check by synthetic division with $k = - 1$ , coefficients $[1, - 1, - 4, 4]$ , $c = 2$ :

  2 |  1   -1   -4    4
    |       2    2   -4
    |___________________
      1    1   -2    0

Remainder $0$ . ✓ And $P (x) = (x - 2) (x^{2} + x - 2) = (x - 2) (x + 2) (x - 1)$ .

Exercise 5.2

A cubic $P (x)$ leaves remainder $3$ when divided by $x - 1$ and remainder $- 1$ when divided by $x + 1$ . Find the remainder when $P (x)$ is divided by $(x - 1) (x + 1) = x^{2} - 1$ .

Recall Solution

Key idea: dividing by a degree- $2$ divisor leaves a remainder of degree at most $1$ , so write it as $R (x) = a x + b$ (unknown constants). Then $P (x) = (x^{2} - 1) Q (x) + a x + b .$ WHY this form: the Remainder Theorem gives us the values of $P$ at $x = 1$ and $x = - 1$ , and at those points the $(x^{2} - 1)$ term vanishes, exposing $a x + b$ .

At $x = 1$ : $P (1) = a (1) + b = a + b = 3$ .
At $x = - 1$ : $P (- 1) = a (- 1) + b = - a + b = - 1$ .

Add the two equations: $2 b = 2 \Rightarrow b = 1$ . Subtract: $2 a = 4 \Rightarrow a = 2$ . $R (x) = 2 x + 1.$ Sanity check: $R (1) = 3$ ✓ and $R (- 1) = - 1$ ✓, matching both given remainders.

Exercise 5.3

Show that for any polynomial $P (x)$ , the remainder on dividing by $x - c$ is always a constant (never degree $\geq 1$ ), and identify that constant.

Recall Solution

The division identity from the parent note is $P (x) = (x - c) Q (x) + R (x), de g R < de g (x - c) .$ The divisor $x - c$ has degree $1$ , so $de g R < 1$ , meaning $de g R = 0$ (a constant) or $R = 0$ . Either way $R$ is a constant; call it $r$ . Substitute $x = c$ : the term $(x - c) Q (x)$ becomes $0 \cdot Q (c) = 0$ , leaving $P (c) = r .$ So the constant remainder is exactly $P (c)$ — this is the Remainder Theorem, and it is why every earlier "test a root" step in this page worked.

[!recall]- One-line summary of every level

Set up right (L1)

order terms, insert zeros, take

c

as the root of the divisor.

Run it clean (L2)

subtract the whole product line every step; stop when

de g R < de g D

Read the result (L3)

remainder

0

means

c

is a root — that's the Factor Theorem.

Combine tools (L4)

always divide down to a proper fraction before partial fractions.

Reason abstractly (L5)

P (c)

is the remainder mod

x - c

; degree of

R

is capped by degree of

D

Exercises — Polynomial long division and synthetic division

Level 1 — Recognition

Exercise 1.1

Exercise 1.2

Level 2 — Application

Exercise 2.1

Exercise 2.2

Level 3 — Analysis

Exercise 3.1

Exercise 3.2

Level 4 — Synthesis

Exercise 4.1

Exercise 4.2

Level 5 — Mastery

Exercise 5.1

Exercise 5.2

Exercise 5.3

[!recall]- One-line summary of every level