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Amy Ferguson Moncure

Algebra 1 · Unit 7

Divide Polynomials

Divide a polynomial by a monomial, a binomial, or a completely factored divisor.

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Key Ideas Easy Medium Hard Super Challenge
What to Remember

Divide by a monomial

Split the fraction: divide each term by the monomial. Subtract exponents.

Divide by a binomial

If the dividend factors, cancel a matching factor. Otherwise use long division.

A completely factored divisor

Write the dividend in factored form too, then cancel common binomials.

Check

Quotient × divisor should return the dividend (plus a remainder if there is one).

Level 1

Easy

Divide a polynomial by a monomial.

Example
Divide. 6x³ + 9x²3x

6x³3x = 2x²

9x²3x = 3x

2x² + 3x

Answer: 2x² + 3x

1
Divide. 10x⁴ − 5x²5x
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2x³ − x

Answer: 2x³ − x

2
Divide. 8x³ + 4x2x
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4x² + 2

Answer: 4x² + 2

3
Divide. 12x²y + 6xy3x
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4xy + 2y

Answer: 4xy + 2y

4
Divide. x⁵ − x³x²
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x³ − x

Answer: x³ − x

5
Divide. 15x³ − 10x² + 5x5x
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3x² − 2x + 1

Answer: 3x² − 2x + 1

Level 2

Medium

The divisor is a binomial that is a factor.

Example
Divide. x² + 5x + 6x + 2

Numerator (x + 2)(x + 3)

Cancel x + 2.

x + 3

Check: (x + 3)(x + 2) = x² + 5x + 6

Answer: x + 3

1
Divide. x² − 9x − 3
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(x + 3)(x − 3) over (x − 3)

x + 3

Answer: x + 3

2
Divide. x² + 7x + 10x + 5
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(x + 2)(x + 5) over (x + 5)

x + 2

Answer: x + 2

3
Divide. 2x² + 7x + 32x + 1
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(2x + 1)(x + 3) over (2x + 1)

x + 3

Answer: x + 3

4
Divide. x³ − xx
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x² − 1

Answer: x² − 1

5
Divide. 3x² + 12x + 123
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x² + 4x + 4

(x + 2)²

Answer: x² + 4x + 4

Level 3

Hard

A factored divisor, long division, or a remainder.

Example
Divide. x³ + 8x + 2

x³ + 8 = (x + 2)(x² − 2x + 4)

Quotient x² − 2x + 4

Check: (x + 2)(x² − 2x + 4) = x³ + 8

Answer: x² − 2x + 4

1
Divide. x² + 3x + 1x + 1 using long division.
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x + 1 into x² + 3x + 1.

First term x. Multiply: x² + x.

Subtract: 2x + 1.

Next term +2. Multiply: 2x + 2.

Subtract: remainder −1.

x + 2 + −1x + 1

Answer: x + 2 − 1/(x + 1)

2
Divide. 2x³ + 4x² − 6x2x
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x² + 2x − 3

Answer: x² + 2x − 3

3
Divide. (x + 3)(x − 1)(x + 2)(x − 1)(x + 2)
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Cancel the matching factors.

x + 3

Answer: x + 3

4
Divide. x³ − 1x − 1
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x³ − 1 = (x − 1)(x² + x + 1)

x² + x + 1

Answer: x² + x + 1

5
A rectangle’s area is x² + 8x + 15 and one side is x + 3. Find the other side.
Show Answer

(x + 3)(x + 5) = x² + 8x + 15

Other side x + 5

Answer: x + 5

Super Challenge

Super Challenge

1
Show that dividing x² − 5x + 6 by x − 2 and then multiplying the quotient by x − 2 returns the original.
Show Answer

x² − 5x + 6 = (x − 2)(x − 3)

Quotient x − 3

(x − 3)(x − 2) = x² − 5x + 6

Answer: x − 3

2
When does x² − 9x − 3 equal x + 3, and when is it undefined?
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Equal for every x except x = 3, where the original fraction is undefined.

The simplified form x + 3 is defined at 3, but that point is a hole.

Answer: equal except x = 3

3
Divide x³ + 3x² + 3x + 1 by x + 1 two ways: as (x + 1)³ / (x + 1), and by canceling.
Show Answer

(x + 1)³ / (x + 1) = (x + 1)²

x² + 2x + 1

Answer: (x + 1)²

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