13.2 Dividing by Monomials & Binomials

13.2 Dividing by Monomials & Binomials

Quick Answer: Divide a polynomial by a monomial by splitting the fraction term by term. Divide by a binomial with long division (or synthetic division when the divisor is $x - c$). Write any leftover as $\frac{\text{remainder}}{\text{divisor}}$, and always check with $(\text{divisor})(\text{quotient}) + \text{remainder} = \text{dividend}$.

PERT Mathematics explicitly lists dividing by monomials and binomials among polynomial skills. These items reverse multiplication: if you can expand $3x(2x - 5)$, you should also be able to simplify $\frac{6x^2 - 15x}{3x}$. Binomial divisors add structure—long division for general cases, synthetic division as a speed tool for $x - c$.

Part A: Dividing by a Monomial

For a polynomial dividend $P(x)$ and monomial divisor $m(x)$, use the property of fractions:

$\frac{a + b + c}{d} = \frac{a}{d} + \frac{b}{d} + \frac{c}{d}$

Simplify each term using coefficient division and the quotient rule for exponents: $\frac{x^m}{x^n} = x^{m-n}$ (when $x \neq 0$).

Worked Example 1: Clean Monomial Division

Problem: Simplify $\dfrac{12x^4 - 18x^3 + 6x^2}{6x^2}$.

Step 1 — Split term by term:

$\frac{12x^4}{6x^2} - \frac{18x^3}{6x^2} + \frac{6x^2}{6x^2}$

Step 2 — Simplify each:

$2x^{2} - 3x^{1} + 1 = 2x^2 - 3x + 1$

Check by multiplying back: $6x^2(2x^2 - 3x + 1) = 12x^4 - 18x^3 + 6x^2$. Matches the original numerator.

Worked Example 2: Signs and Uneven Powers

Problem: Simplify $\dfrac{10x^3 - 15x^2 - 5x}{5x}$.

Split:

$\frac{10x^3}{5x} - \frac{15x^2}{5x} - \frac{5x}{5x} = 2x^2 - 3x - 1$

Trap: Writing the last term as $-5$ instead of $-1$, or canceling only part of the coefficient. Every term—including the last—must be divided by $5x$.

Worked Example 3: Negative Coefficients in the Divisor

Problem: Simplify $\dfrac{-8x^5 + 12x^3 - 4x}{-4x}$.

Split carefully (negative over negative is positive for coefficients):

$\frac{-8x^5}{-4x} + \frac{12x^3}{-4x} + \frac{-4x}{-4x} = 2x^4 - 3x^2 + 1$

Note that $\frac{12x^3}{-4x} = -3x^2$. Sign tracking on the middle term is where many errors land.

When a Power Does Not Cancel Completely

If the numerator degree of a term is less than the denominator’s power of $x$, leave a fraction for that piece—or rewrite with a negative exponent only if the course allows. On PERT-style multiple choice, prefer simplified polynomial or mixed polynomial-plus-fraction form matching the options.

Example: $\dfrac{6x^2 + 4x + 2}{2x} = 3x + 2 + \dfrac{1}{x}$ (if $x \neq 0$). Some choices write this as $\dfrac{3x^2 + 2x + 1}{x}$; both are equivalent when simplified correctly.

Part B: Long Division by a Binomial

Long division of polynomials mirrors whole-number long division:

1. Arrange dividend and divisor in standard form. Insert $0$ placeholders for missing powers.
2. Divide the leading term of the current dividend by the leading term of the divisor—that is the next quotient term.
3. Multiply that quotient term by the entire divisor.
4. Subtract (add the opposite) from the current dividend.
5. Bring down the next term; repeat until the remainder’s degree is less than the divisor’s degree.
6. Write answer as $\text{quotient} + \dfrac{\text{remainder}}{\text{divisor}}$ if remainder $\neq 0$.

Worked Example 4: Exact Division

Problem: Divide $x^2 + 5x + 6$ by $x + 2$.

Setup:

        x + 3
      _________
 x+2 | x² + 5x + 6

First quotient term: $\frac{x^2}{x} = x$. Multiply $x(x+2) = x^2 + 2x$. Subtract:

$(x^2 + 5x) - (x^2 + 2x) = 3x$

Bring down $+6$: $3x + 6$.

Second quotient term: $\frac{3x}{x} = 3$. Multiply $3(x+2) = 3x + 6$. Subtract: $0$.

Result: $x + 3$ exactly.

Factoring check: $x^2 + 5x + 6 = (x+2)(x+3)$. Division recovered the other factor—exactly what PERT wants you to see connecting division and factoring.

Worked Example 5: Division with Remainder

Problem: Divide $2x^2 - 3x + 5$ by $x - 1$.

First term: $\frac{2x^2}{x} = 2x$. Multiply $2x(x-1) = 2x^2 - 2x$. Subtract from $2x^2 - 3x$:

$(-3x) - (-2x) = -x$

Bring down $+5$: $-x + 5$.

Second term: $\frac{-x}{x} = -1$. Multiply $-1(x-1) = -x + 1$. Subtract from $-x + 5$:

$5 - 1 = 4$

Result: $2x - 1 + \dfrac{4}{x - 1}$.

Check: $(x-1)(2x-1) + 4 = 2x^2 - x - 2x + 1 + 4 = 2x^2 - 3x + 5$. Good.

Worked Example 6: Missing Power Placeholder

Problem: Divide $x^3 - 8$ by $x - 2$.

Rewrite dividend as $x^3 + 0x^2 + 0x - 8$ so columns stay aligned.

Long division outline:

• $\frac{x^3}{x} = x^2$; $x^2(x-2) = x^3 - 2x^2$; subtract → $2x^2$
• Bring down $0x$: $2x^2 + 0x$; $\frac{2x^2}{x} = 2x$; $2x(x-2) = 2x^2 - 4x$; subtract → $4x$
• Bring down $-8$: $4x - 8$; $\frac{4x}{x} = 4$; $4(x-2) = 4x - 8$; subtract → $0$

Result: $x^2 + 2x + 4$.

This is the difference-of-cubes pattern: $x^3 - 8 = (x-2)(x^2 + 2x + 4)$. Placeholders prevented dropping a column and inventing a false remainder.

Part C: Synthetic Division (When the Divisor Is $x - c$)

Synthetic division is a compressed algorithm for divisors of the form $x - c$ (leading coefficient 1, degree 1).

Setup for dividing by $x - c$:

1. Write the value $c$ on the left (note: for $x + 3 = x - (-3)$, use $c = -3$).
2. List coefficients of the dividend in standard form, including zeros for missing powers.
3. Bring down the first coefficient. Multiply by $c$, add to the next coefficient; repeat across the row.
4. The final number is the remainder. The other numbers (left to right) are coefficients of the quotient, one degree lower than the dividend.

Worked Example 7: Synthetic Division

Problem: Divide $x^3 - 4x^2 - 7x + 10$ by $x - 5$ using synthetic division.

Here $c = 5$. Coefficients: $1 ;|; -4 ;|; -7 ;|; 10$.

5 |  1   -4   -7   10
   |      5    5   -10
   -----------------
     1    1   -2    0

• Bring down 1.
• $1 \cdot 5 = 5$; $-4 + 5 = 1$.
• $1 \cdot 5 = 5$; $-7 + 5 = -2$.
• $-2 \cdot 5 = -10$; $10 + (-10) = 0$.

Quotient: $x^2 + x - 2$, remainder $0$. So $x^3 - 4x^2 - 7x + 10 = (x-5)(x^2 + x - 2)$.

(You can factor further: $x^2 + x - 2 = (x+2)(x-1)$, but the division question is complete once the quotient and remainder are correct.)

Worked Example 8: Synthetic with Remainder

Problem: Divide $2x^3 + 3x - 5$ by $x + 1$.

Divisor $x + 1 = x - (-1)$, so $c = -1$. Coefficients: $2,; 0,; 3,; -5$ (zero for missing $x^2$).

-1 |  2    0    3    -5
   |     -2    2    -5
   -------------------
     2   -2    5   -10

Quotient: $2x^2 - 2x + 5$, remainder $-10$.

Answer: $2x^2 - 2x + 5 - \dfrac{10}{x+1}$, or $2x^2 - 2x + 5 + \dfrac{-10}{x+1}$.

Choosing a Method on PERT

Test tip: If answer choices are fully factored or expanded polynomials with no remainder, expect exact division. If choices include a fractional remainder term, plan for a nonzero remainder from the start.

Common Errors to Avoid

1. Canceling only part of a monomial fraction (for example, canceling $x$ from one term of the numerator but not others).
2. Forgetting zero placeholders for missing powers in long/synthetic division.
3. Using the wrong synthetic root for $x + 3$ (use $-3$, not $+3$).
4. Subtracting incorrectly in long division—always change signs of the product row before adding.
5. Leaving the remainder as a bare integer instead of $\dfrac{\text{remainder}}{\text{divisor}}$ when the question asks for the full simplified quotient form.

Division fluency pays off immediately in the next section: factoring is often verified by multiplying, and roots found by factoring can be checked with synthetic division (Remainder Theorem: remainder when dividing by $x - c$ is $P(c)$).