Lesson 5.4: Dividing Polynomials

New Concept

Ready to level up your algebra skills? We'll tackle dividing polynomials using methods like long division and the powerful shortcut, synthetic division. This will unlock new ways to work with polynomial functions, including the Remainder and Factor Theorems.

What’s next

Get ready for a series of interactive examples and practice cards that break down each division method, from monomials to synthetic division.

Property

When dividing monomials, rewrite the expression as a fraction. Simplify the numerical coefficients and use the Quotient Property of Exponents for each variable by subtracting the exponents. For variables $a$ and $b$ and integers $m$ and $n$:

Examples

• Find the quotient $36x^6y^3 \div (9x^2y^5)$. Rewrite as a fraction $\frac{36x^6y^3}{9x^2y^5}$. Simplify the coefficients and variables: $\frac{36}{9} \cdot x^{6-2} \cdot y^{3-5} = 4x^4y^{-2}$, which is $\frac{4x^4}{y^2}$.
• Find the quotient $\frac{20a^8b^3}{25a^5b^4}$. Simplify $\frac{20}{25}$ to $\frac{4}{5}$ and subtract the exponents for the variables to get $\frac{4}{5}a^{8-5}b^{3-4} = \frac{4}{5}a^3b^{-1}$, which simplifies to $\frac{4a^3}{5b}$.
• Find the quotient $-42p^{10}q^5 \div (7p^{12}q^2)$. This becomes $\frac{-42}{7} \cdot p^{10-12} \cdot q^{5-2} = -6p^{-2}q^3$, which is written as $\frac{-6q^3}{p^2}$.

Explanation

Think of dividing monomials as simplifying a fraction. You handle the numbers and each variable separately. For the variables, subtracting exponents is a shortcut for cancelling out common factors between the numerator and denominator. It's the inverse of multiplication.

Property

To divide a polynomial by a monomial, divide each term of the polynomial by the monomial. This is based on the property of fraction addition, which states that if $a$, $b$, and $c$ are numbers where $c \neq 0$, then:

Examples

• Find the quotient $(27x^4y - 18xy^3) \div (9xy)$. Divide each term: $\frac{27x^4y}{9xy} - \frac{18xy^3}{9xy}$. This simplifies to $3x^3 - 2y^2$.
• Find the quotient $\frac{40a^5b^3 + 25a^2b^4}{-5a^2b^2}$. Divide each term: $\frac{40a^5b^3}{-5a^2b^2} + \frac{25a^2b^4}{-5a^2b^2}$. This simplifies to $-8a^3b - 5b^2$.
• Find the quotient $(15c^3d^2 - 21c^2d^3 + 6cd^4) \div (3cd^2)$. This becomes $\frac{15c^3d^2}{3cd^2} - \frac{21c^2d^3}{3cd^2} + \frac{6cd^4}{3cd^2}$, which simplifies to $5c^2 - 7cd + 2d^2$.

Explanation

Think of this as distributing the division. The monomial in the denominator must be divided into every single term in the numerator. This breaks one large, complicated fraction into several smaller, simpler fractions that you can solve one by one.

Property

To divide a polynomial by a binomial, use a process similar to long division of numbers.

1. Write the division problem with the dividend in standard form, using placeholders (e.g., $0x^2$) for any missing terms.
2. Divide the leading term of the dividend by the leading term of the divisor to get the first term of the quotient.
3. Multiply the new quotient term by the entire divisor and subtract it from the dividend.
4. Bring down the next term and repeat the process until finished.

Examples

• Find the quotient $(y^2 + 8y + 15) \div (y + 3)$. Long division shows that $y$ goes into $y^2$ $y$ times. Multiply $y(y+3)=y^2+3y$, subtract to get $5y$, bring down 15. $y$ goes into $5y$ 5 times. The result is $y+5$.
• Find the quotient $(x^3 - 2x^2 - 13x + 6) \div (x - 4)$. Using long division, the quotient is $x^2 + 2x - 5$ with a remainder of $-14$. The answer is written as $x^2 + 2x - 5 - \frac{14}{x-4}$.
• Find the quotient $(a^3 - 27) \div (a - 3)$. Use placeholders: $(a^3 + 0a^2 + 0a - 27) \div (a - 3)$. The result is $a^2 + 3a + 9$.

Explanation

This method mirrors the long division you learned for numbers. The key is to focus on dividing the leading terms at each step. Using placeholders for missing powers, like $0x^2$, is crucial to keep all the terms aligned correctly.

Property

Synthetic division is a shorthand method for polynomial division. Synthetic division only works when the divisor is of the form $x - c$.
To perform synthetic division, write the coefficients of the dividend and the value $c$ from the divisor. Bring down the first coefficient, multiply by $c$, add to the next coefficient, and repeat. The last number in the result is the remainder.

Examples

• To divide $x^3 + 4x^2 - 5x - 14$ by $x - 2$, use $c=2$ in synthetic division. The coefficients are $1, 4, -5, -14$. The resulting coefficients are $1, 6, 7$ with a remainder of $0$. The quotient is $x^2 + 6x + 7$.
• To divide $3x^3 + 8x^2 + 5x - 7$ by $x + 2$, use $c=-2$. The coefficients are $3, 8, 5, -7$. The process yields a quotient of $3x^2 + 2x + 1$ and a remainder of $-9$.
• To divide $y^4 - 8y^2 + 16$ by $y + 2$, use $c=-2$ and a placeholder for the $y^3$ and $y$ terms (coefficients $1, 0, -8, 0, 16$). The quotient is $y^3 - 2y^2 - 4y + 8$ with a remainder of $0$.

Explanation

Synthetic division is a fast-track version of long division that gets rid of the variables. It's a cleaner and quicker process, but remember its major limitation: it only works when you are dividing by a simple binomial like $(x-5)$ or $(x+2)$.

Property

For functions $f(x)$ and $g(x)$, where $g(x) \neq 0$, the division of the two functions is defined as:

Examples

• For functions $f(x) = x^2 - 6x - 16$ and $g(x) = x + 2$, find $(\frac{f}{g})(x)$. We calculate $\frac{x^2 - 6x - 16}{x + 2}$. By factoring the numerator to $(x-8)(x+2)$, we simplify to get $(\frac{f}{g})(x) = x - 8$.
• Using the functions from the previous example, find $(\frac{f}{g})(10)$. Substitute $x=10$ into the simplified result: $(\frac{f}{g})(10) = 10 - 8 = 2$.
• For $f(x) = x^3 + 1$ and $g(x) = x + 1$, find $(\frac{f}{g})(x)$. We compute $\frac{x^3+1}{x+1}$. Using polynomial division, the result is $x^2 - x + 1$.

Explanation

The notation $(\frac{f}{g})(x)$ is simply a formal way to express the division of one polynomial function, $f(x)$, by another, $g(x)$. To solve, you set up the division as a fraction and use a method like long division or factoring to find the result.

Property

If the polynomial function $f(x)$ is divided by $x - c$, then the remainder is $f(c)$.

Examples

• To find the remainder when $f(x) = x^3 + 2x^2 - 4x + 5$ is divided by $x - 1$, we calculate $f(1)$. $f(1) = (1)^3 + 2(1)^2 - 4(1) + 5 = 4$. The remainder is 4.
• To find the remainder when $f(x) = 2x^3 - x^2 + 3x - 9$ is divided by $x + 2$, we calculate $f(-2)$. $f(-2) = 2(-2)^3 - (-2)^2 + 3(-2) - 9 = -35$. The remainder is -35.
• When $f(x) = x^4 - 81$ is divided by $x-3$, the remainder is $f(3)$. $f(3) = (3)^4 - 81 = 81 - 81 = 0$. The remainder is 0.

Explanation

This theorem provides a neat shortcut to find a remainder without performing division. To find the remainder when dividing a polynomial by $(x-c)$, just substitute the value $c$ into the polynomial. The result you get is the remainder.

Property

For any polynomial function $f(x)$:

• If $x - c$ is a factor of $f(x)$, then $f(c) = 0$.
• If $f(c) = 0$, then $x - c$ is a factor of $f(x)$.

Examples

• To determine if $x - 2$ is a factor of $f(x) = x^3 - 8$, we check if $f(2) = 0$. We find $f(2) = (2)^3 - 8 = 8 - 8 = 0$. Since the result is 0, $x - 2$ is a factor.
• Is $x+3$ a factor of $f(x) = x^2 + 5x + 7$? We check $f(-3)$. $f(-3) = (-3)^2 + 5(-3) + 7 = 9 - 15 + 7 = 1$. Since $f(-3) \neq 0$, $x+3$ is not a factor.
• Is $x-1$ a factor of $f(x) = 5x^4 - 3x^2 - 2$? We check $f(1)$. $f(1) = 5(1)^4 - 3(1)^2 - 2 = 5 - 3 - 2 = 0$. Since $f(1)=0$, $x-1$ is a factor.

Explanation

The Factor Theorem builds on the Remainder Theorem. It states that if plugging in a value $c$ makes the polynomial equal zero, then $(x-c)$ is a perfect factor. This is because a remainder of zero means the division is exact.