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4.02 Polynomial division

Lesson

Concept summary

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 rules of exponents.

Remember that when we divide a polynomial by a monomial we can use that: \dfrac{a+b}{c} = \dfrac{a}{c} + \dfrac{b}{c}

In general, when dividing polynomials where the divisor is not a monomial, we can use the process of polynomial long division. Polynomial long division works in a very similar way to long division with whole numbers wherein we divide the first part of the dividend, multiply the answer by the divisor, subtract, bring down the next part of the dividend, and then repeat until done.

Note: Before performing a long division, the terms of the divisor and dividend should first be arranged in descending order of exponents. In cases where there is no term corresponding to an exponent in the dividend, we use a placeholder term with a coefficient of 0.

For special cases when we are dividing a polynomial by a binomial of the form x-a, the following theorems can be used:

Factor theorem

If for a polynomial f\left(x\right) there is a value a such that f\left(a\right)=0, then x-a is a factor of the polynomial.

In addition, there is an efficient method of polynomial division known as synthetic division that can be used only when the divisor is a linear expression of the form x-a. For synthetic division, only the coefficients are needed in the division process.

Worked examples

Example 1

Rewrite \dfrac{x^3 + 7 x^2 + 14 x + 3}{x + 2} as the sum of the quotient and a remainder fraction by using long division.

Solution

We set up and perform the long division as follows.

A figure showing the polynomial long division for x cubed plus 7 x squared plus 14 x plus 3 divided by x plus 2. Speak to your teacher for more information.

Doing so, we get a result of x^2 + 5x + 4 as the quotient and -5 as the remainder. So we can rewrite the expression as \frac{x^3 + 7 x^2 + 14 x + 3}{x + 2} = x^2 + 5x + 4 - \frac{5}{x + 2}

Reflection

We can check the answer by determining if the dividend is equal to the product of the quotient and the divisor plus the remainder.

\displaystyle \text{Dividend}\displaystyle =\displaystyle \left( \text{Quotient} \times \text{Divisor} \right) + \text{Remainder}Formula
\displaystyle x^{3} + 7 x^{2} + 14 x + 3\displaystyle =\displaystyle \left[\left(x^{2} + 5x + 4\right) \left(x + 2\right) \right]-5Check

Example 2

Rewrite \dfrac{2x^3 - 3x^2 + 4x - 1}{x + 1}as the sum of the quotient and a remainder fraction by using synthetic division.

Solution

We set up and perform the synthetic division as follows.

A figure showing a synthetic division with negative 1 in the middle left corner. Top row has digits 2, negative 3, 4, negative 1. Middle row has negative 2 under the negative 3, 5 under the 4, and negative 9 under negative 1. Bottom row has 2, negative 5, 9, and negative 10.

Doing so, we get a result of 2x^2 - 5x + 9 as the quotient and -10 as the remainder. So we can rewrite the expression as \frac{2x^3 - 3x^2 + 4x - 1}{x + 1} = 2x^2 - 5x + 9 - \frac{10}{x + 1}

Reflection

We can check the answer by determining if the dividend is equal to the product of the quotient and the divisor plus the remainder.

\displaystyle \text{Dividend}\displaystyle =\displaystyle \left( \text{Quotient} \times \text{Divisor} \right) + \text{Remainder}Formula
\displaystyle 2x^{3} - 3 x^{2} + 4 x - 1\displaystyle =\displaystyle \left[\left(2x^{2} - 5x + 9\right) \left(x + 1\right) \right]- 10Check

Example 3

Use the Factor theorem to state if the divisor is a factor of the dividend in the expression: \dfrac{2x^3+x^{2}-10x}{x-2}

Approach

By the Factor theorem, if \left(x-a\right) is a factor, then the remainder when f\left(x\right) is divided by a will be 0.

Solution

By the Factor theorem, if \left(x-2\right) is a factor, then the remainder will be equal to zero. Since the divisor is of the form x - a, we can use synthetic division to find the remainder.

A figure showing a synthetic division with 2 in the middle left corner. Top row has digits 2, 1, negative 10, 0. Middle row has 4 under the 1, 10 under the negative 10, and 0 under the 0. Bottom row has 2,  5, 0, and 0.

Since the remainder is 0, \left(x-2\right) is a factor of 2x^3+x^2-10x.

Example 4

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.

Solution

\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}}Rearrange the formula
\displaystyle =\displaystyle \dfrac{4x^4 - 12x}{4x}Substitute known values
\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.

Reflection

Write each division in terms of the quotient and remainder by using long or synthetic division.

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

Outcomes

A2.A.APR.A.1

Know and apply the Factor Theorem: For a polynomial p(x) and a number a, p(a) = 0 if and only if (x – a) is a factor of p(x).

A2.MP1

Make sense of problems and persevere in solving them.

A2.MP3

Construct viable arguments and critique the reasoning of others.

A2.MP5

Use appropriate tools strategically.

A2.MP6

Attend to precision.

A2.MP7

Look for and make use of structure.

A2.MP8

Look for and express regularity in repeated reasoning.

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