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3. Polynomial Expressions, Equations, & Functions
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Algebra 2

3.03 Divide polynomials

Lesson
Practice

Divide polynomials

Recall from our work with rational numbers, that when we divide a sum by a real number, we can use the fact that:\dfrac{a+b}{c} = \dfrac{a}{c} + \dfrac{b}{c}for any real numbers, a, b or c.

We can extend this concept to the division of polynomial expressions. For any polynomial expressions A, B or C:\dfrac{A+B}{C} = \dfrac{A}{C} + \dfrac{B}{C}

The simplest form of division of polynomials is when the divisor is a monomial. The process involves dividing each term of the polynomial by the monomial, then simplifying each individual fraction using the properties of exponents.

When dividing by a monomial that contains a variable, we use the quotient rule of exponents, to simplify each term.

Quotient rule

When dividing two exponential expressions with the same base, subtract the exponents.

Example:

\dfrac{a^{m}}{a^{n}}=a^{m-n}

\dfrac{a^{7}}{a^{4}}=a^{7-4}=a^{3}

Dividing polynomials involves a process known as algebraic manipulation. We can view the dividend as the numerator of a fraction and the divisor as the denominator.

By factoring, we can simplify and divide out any factors that are common between the numerator and denominator.

  1. Completely factor both the numerator and denominator.

  2. Identify all common factors that are present in both the numerator and the denominator. These could be monomial or binomial factors.

  3. Divide out all common factors from the numerator and denominator.

  4. Simplify the resulting expression.

Examples

Example 1

The rectangle has an area of 4 x^{4} - 12 x square units, and its width is 4x units. Find the length of the rectangle.

Worked Solution
Create a strategy

We can begin by identifying the known values and rearranging the formula for the area of the rectangle to highlight the unit we need to solve for.

\displaystyle \text{Area}\displaystyle =\displaystyle \text{Length} \cdot \text{Width}Formula for the Area of a Rectangle
\displaystyle \text{Length}\displaystyle =\displaystyle \dfrac{\text{Area}}{\text{Width}}Solve the formula for length

\text{Area}=4x^{4}-12x and \text{Width}=4x

Apply the idea
\displaystyle \text{Length}\displaystyle =\displaystyle \dfrac{4x^{4} - 12x}{4x}Substitute expressions for area and width
\displaystyle =\displaystyle \dfrac{4x^4}{4x}-\dfrac{12x}{4x}Divide each term by 4x
\displaystyle =\displaystyle x^{3} - 3Simplify the expression

Since there are no negative exponents and the expression is already in standard form, the final answer is x^{3} - 3. The length of the rectangle is x^{3} - 3 units.

Reflect and check

To check the answer we found for the length, we can write the dividend as the product of the quotient and divisor.

\displaystyle \text{Length} \cdot \text{Width}\displaystyle =\displaystyle \text{Area}Formula for the Area of a Rectangle
\displaystyle \left(x^{3} - 3\right)\left(4x\right)\displaystyle =\displaystyle 4x^{4} - 12xDistributive property

The product of the quotient and the divisor does result in the dividend, so our answer is correct.

Example 2

Find the quotient of \left(15s^{4}t^{5}-5s^{3}t^{3}+20s^{2}t\right)\div 40st^{2}.

Worked Solution
Create a strategy

Since the divisor is a monomial, divide each term in the dividend by 40st^{2}.

Apply the idea
\displaystyle \dfrac{15s^{4}t^{5}-5s^{3}t^{3}+20s^{2}t}{40st^{2}}\displaystyle =\displaystyle \dfrac{15s^{4}t^{5}}{40st^{2}}-\dfrac{5s^{3}t^{3}}{40st^{2}}+\dfrac{20s^{2}t}{40st^{2}}Divide each term by 40st^{2}
\displaystyle =\displaystyle \dfrac{3}{8}s^{3}t^{3}-\dfrac{1}{8}s^{2}t+\dfrac{s}{2t}Simplify the expression

The quotient is \dfrac{3}{8}s^{3}t^{3}-\dfrac{1}{8}s^{2}t+\dfrac{s}{2t}.

Reflect and check

Notice the quotient rule was used when dividing each term by 40st^{2}.\dfrac{15s^{4}t^{5}}{40st}=\dfrac{3}{8}s^{4-1}t^{5-2}=\dfrac{3}{8}s^{3}t^{3}

\dfrac{-5s^{3}t^{3}}{40st^{2}}=-\dfrac{1}{8}s^{3-1}t^{3-2}=-\dfrac{1}{8}s^{2}t

\dfrac{20s^{2}t}{40st^{2}}=\dfrac{1}{2}s^{2-1}t^{1-2}=\dfrac{1}{2}st^{-1}

Example 3

Factor the numerators and denominators, then divide the polynomial expressions.

a

\dfrac{2x^{2} + 7x}{6x + 21}

Worked Solution
Create a strategy

Since the denominator is a binomial, we first want to factor the numerator and denominator. Then, we can divide common factors.

Apply the idea

We can begin factoring a GCF. In this case, the numerator has a common factor of x and the denominator has a common factor of 3. Factoring these out gives:\dfrac{2x^{2} + 7x}{6x + 21} = \dfrac{x\left(2x + 7\right)}{3\left(2x + 7\right)}We can now see the common factor of 2x + 7 which can be simplified from the numerator and denominator by the multiplicative inverse property of equality, since \dfrac{\left (2x + 7 \right)}{\left (2x+7 \right)} = 1:\dfrac{x\left(2x + 7\right)}{3\left(2x + 7\right)} = \dfrac{x}{3}

The quotient is \dfrac{x}{3}.

b

\dfrac{y^{2} + 5y - 24}{y^{3} - 27}

Worked Solution
Create a strategy

We first want to factor the numerator and denominator, then divide common factors.

Apply the idea

First, we will factor the numerator.

In this case, the numerator is a quadratic. To factor it by grouping, we need to find two numbers that have a sum of b=5 and a product of ac=-24. The two numbers are 8 and -3, so

\displaystyle \dfrac{y^{2}+5y-24}{y^{3}-27}\displaystyle =\displaystyle \dfrac{y^{2}+8y-3y-24}{y^{3}-27}Rewrite the numerator
\displaystyle =\displaystyle \dfrac{y \left(y + 8 \right) - 3 \left(y + 8 \right)}{y^3-27}Factor out each GCF
\displaystyle =\displaystyle \dfrac{\left(y + 8 \right) \left(y - 3 \right)}{y^3-27}Factor out the common factor

Next, we will factor the denominator. The denominator is a difference of two cubes, since 27 = 3^{3}, so we can factor it using that identity.\dfrac{\left(y + 8 \right) \left(y - 3 \right)}{y^{3}-27}=\dfrac{\left(y + 8 \right) \left(y - 3 \right)}{\left(y-3 \right) \left( y^{2} + 3y + 9 \right)}

Finally, we can divide by the common factor of y - 3.\dfrac{\left(y + 8\right)\cancel{\left(y - 3\right)}}{\cancel{\left(y - 3\right)}\left(y^{2} + 3y + 9\right)} = \dfrac{y + 8}{y^{2} + 3y + 9}

The quotient is \dfrac{y + 8}{y^{2} + 3y + 9}.

Reflect and check

In this case, the simplified form appears to have about the same "complexity" as the original expression, since it has the same number of terms between the numerator and denominator as the original expression did.

The reason that this form is simpler is that the exponents involved are smaller. The numerator and denominator of the original expression had degrees of 2 and 3 respectively, while the simplified expression has degrees of 1 and 2 respectively.

c

\dfrac{3c^{2}d-9cd}{6cd^{2}+3cd}

Worked Solution
Create a strategy

Find and factor out the greatest common factor in the numerator and the denominator. Then, divide by any common factors between the numerator and denominator.

Apply the idea

We can begin factoring a GCF. In this case, the numerator has a common factor of 3cd and the denominator has a common factor of 3cd.

\displaystyle \dfrac{3c^{2}d-9cd}{6cd^{2}+3cd}\displaystyle =\displaystyle \dfrac{3cd\left(c-9\right)}{3cd\left(2d+1\right)}Factor out each GCF
\displaystyle =\displaystyle \dfrac{c-9}{2d+1}Divide by the common factor

Example 4

The rectangle shown has an area of 15 n^{3} + 13 n^{2} + 33 n.

Rectangle with length 'n'
a

Find a polynomial expression for its height.

Worked Solution
Create a strategy

The area of a rectangle is given by the formula \text{Area}= \text{length} \cdot \text{height}. The length is n units, so we can find the height by dividing the area by the length.

\displaystyle \text{Area}\displaystyle =\displaystyle \text{length} \cdot \text{height}
\displaystyle \dfrac{\text{Area}}{\text{length}}\displaystyle =\displaystyle \text{height}
Apply the idea

To find the height, we divide the area by the length: \dfrac{\text{area}}{\text{length}}=\dfrac{15 n^{3} + 13 n^{2} + 33 n}{n}

Since the divisor is a monomial, we can divide each term by n and simplify using the quotient rule. \dfrac{15 n^{3}}{n} + \dfrac{13 n^{2}}{n} + \dfrac{33 n}{n}=15 n^{2} + 13 n + 33 The height of the rectangle is 15 n^{2} + 13 n + 33 units.

b

Write two polynomials that could represent finding the area of a triangle with the same dimensions.

Worked Solution
Create a strategy

To find the area of a triangle with the same base and height as the rectangle, we use the formula \text{Area}=\dfrac{1}{2} \cdot \text{base} \cdot \text{height}. We will create two expressions to represent this area.

Apply the idea

One way to express the area of such a triangle is directly applying the formula with our known base, n, and height, 15 n^{2} + 13 n + 33.

The area of the triangle can be expressed as \dfrac{1}{2} n\left(15 n^{2} + 13 n + 33\right) or \dfrac{n}{2} \left(15 n^{2} + 13 n + 33\right).

Another expression can be found by distributing the coefficient: 7.5 n^{3} + 6.5 n^{2} + 16.5 n

Reflect and check

These polynomial expressions for the triangle's area capture the same geometric relationship in different forms, offering flexibility in how we approach and solve problems involving areas of shapes with similar dimensions.

Idea summary

For dividing polynomials where the divisor is a monomial, we can use the fact that \dfrac{a+b}{c}=\dfrac{a}{c}+\dfrac{b}{c}

To divide polynomials with non-monomial divisors:

  1. Completely factor the numerator and denominator

  2. Divide out all common factors between the numerator and denominator

  3. Simplify the resulting expression (if necessary)

Outcomes

A2.EO.1

The student will perform operations on and simplify rational expressions.

A2.EO.1b

Justify and determine equivalent rational algebraic expressions with monomial and binomial factors. Algebraic expressions should be limited to linear and quadratic expressions.

A2.EO.1d

Represent and demonstrate equivalence of rational expressions written in different forms.

A2.EO.3

The student will perform operations on polynomial expressions in two or more variables and factor polynomial expressions in one and two variables.

A2.EO.3c

Determine the quotient of polynomials in two or more variables, using monomial, binomial, and factorable trinomial divisors

A2.EO.3d

Represent and demonstrate equality of polynomial expressions written in different forms and verify polynomial identities including the difference of squares, sum and difference of cubes, and perfect square trinomials.

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