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Dividing Polynomials

Dividing monomials

Dividing Polynomials

Dividing polynomials using long division (IA 5.4.3)
Dividing polynomials using synthetic division (IA 5.4.4)

To divide a polynomial by a binomial, we follow a procedure very similar to long division of numbers. So, let’s look carefully at the steps we take when we divide a 3-digit number, 875, by a 2-digit number, 25.

Example

Try it.

When we divided 875 by 25, we had no remainder. But sometimes division of numbers does leave a remainder.

Vocabulary of the example.

Fill in the blanks.

Try it.

When dividing 69 by 4,

the dividend is_________,

the divisor is________,

the quotient is________,

and the remainder is________.

We check division by multiplying the quotient by the divisor and adding the remainder.

Example

Try it.

Find the quotient: \(({x}^{2}+9x+20)\div (x+5).\)

Solution


Write it as a long division problem. Be sure the dividend is written in descending order of powers, with no missing terms.
Divide \({x}^{2}\) by \(x.\) It may help to ask yourself, “What do I need to multiply \(x\) by to get \({x}^{2}\) ?”
Put the answer, \(x,\) in the quotient over the \(x\) term. Multiply \(x\) times \(x+5.\) Line up the like terms under the dividend.
Subtract \({x}^{2}+5x\) from \({x}^{2}+9x.\) You may find it easier to change the signs and then add. Then bring down the last term, 20.
Divide \(4x\) by \(x.\) It may help to ask yourself, “What do I need to multiply \(x\) by to get \(4x\) ?” Put the answer, \(4\) , in the quotient over the constant term.
Multiply 4 times \(x+5.\)
Subtract \(4x+20\) from \(4x+20.\)
Check: Multiply the quotient by the divisor. \(\ (x+4)(x+5)\) You should get the dividend. \(\ {x}^{2}+9x+20✓\)

Try it.

Divide using long division of polynomials: \(({x}^{2}+10x+21)\div (x+3)\)

Sometimes division of polynomials, just like division of numbers, leaves a remainder. We write the remainder as a fraction with the divisor as the denominator.

Also, if you look back at the dividends in previous examples, you will notice that the terms were written in descending order of degrees, and there were no missing degrees.

Example

Try it.

\((5x+{x}^{4}-{x}^{2}-6)\div (x+2)\)

Solution

Notice, this polynomial is not in descending order and it is missing \({x}^{3}\) term. We need to write it in the correct order and add \(0{x}^{3}\) as a placeholder.

To check, multiply divisor by the quotient and add remainder \((x+2)({x}^{3}-2{x}^{2}+3x-1)-4\)

The result should be \({x}^{4}-{x}^{2}+5x-6\)

Dividing polynomials using long division.

Try it.

\((7x+{x}^{4}-7{x}^{2}+6)\div (x+3)\)

Try it.

\(({x}^{3}-8)\div (x-2)\)

Condensed — the full section is in OpenStax College Algebra 2e.

Using Long Division to Divide Polynomials

We are familiar with the long division algorithm for ordinary arithmetic. We begin by dividing into the digits of the dividend that have the greatest place value. We divide, multiply, subtract, include the digit in the next place value position, and repeat. For example, let’s divide 178 by 3 using long division.

Another way to look at the solution is as a sum of parts. This should look familiar, since it is the same method used to check division in elementary arithmetic.

\[\begin{array}{lll}\text{dividend} & = & (\text{divisor}⋅\text{quotient) + remainder} \\ 178 & = & (3⋅59)+1 \\ & = & 177+1 \\ & = & 178\end{array}\]

We call this the Division Algorithm and will discuss it more formally after looking at an example.

Division of polynomials that contain more than one term has similarities to long division of whole numbers. We can write a polynomial dividend as the product of the divisor and the quotient added to the remainder. The terms of the polynomial division correspond to the digits (and place values) of the whole number division. This method allows us to divide two polynomials. For example, if we were to divide \(2{x}^{3}-3{x}^{2}+4x+5\) by \(x+2\) using the long division algorithm, it would look like this:

We have found

\[\frac{2{x}^{3}-3{x}^{2}+4x+5}{x+2}=2{x}^{2}-7x+18-\frac{31}{x+2}\]

\[2{x}^{3}-3{x}^{2}+4x+5=(x+2)(2{x}^{2}-7x+18)-31\]

We can identify the dividend, the divisor, the quotient, and the remainder.

Condensed — the full section is in OpenStax College Algebra 2e.

Using Synthetic Division to Divide Polynomials

As we’ve seen, long division of polynomials can involve many steps and be quite cumbersome. Synthetic division is a shorthand method of dividing polynomials for the special case of dividing by a linear factor whose leading coefficient is 1.

To illustrate the process, recall the example at the beginning of the section.

Divide \(2{x}^{3}-3{x}^{2}+4x+5\) by \(x+2\) using the long division algorithm.

The final form of the process looked like this:

There is a lot of repetition in the table. If we don’t write the variables but, instead, line up their coefficients in columns under the division sign and also eliminate the partial products, we already have a simpler version of the entire problem.

Synthetic division carries this simplification even a few more steps. Collapse the table by moving each of the rows up to fill any vacant spots. Also, instead of dividing by 2, as we would in division of whole numbers, then multiplying and subtracting the middle product, we change the sign of the “divisor” to –2, multiply and add. The process starts by bringing down the leading coefficient.

We then multiply it by the “divisor” and add, repeating this process column by column, until there are no entries left. The bottom row represents the coefficients of the quotient; the last entry of the bottom row is the remainder. In this case, the quotient is \(2{x}^{2}-7x+18\) and the remainder is \(-31.\) The process will be made more clear in .

Example

Try it.

Use synthetic division to divide \(4{x}^{3}+10{x}^{2}-6x-20\) by \(x+2.\)

Solution

The binomial divisor is \(x+2\) so \(k=-2.\) Add each column, multiply the result by –2, and repeat until the last column is reached.

The result is \(4{x}^{2}+2x-10.\) The remainder is 0. Thus, \(x+2\) is a factor of \(4{x}^{3}+10{x}^{2}-6x-20.\)

Condensed — the full section is in OpenStax College Algebra 2e.

Using Polynomial Division to Solve Application Problems

Polynomial division can be used to solve a variety of application problems involving expressions for area and volume. We looked at an application at the beginning of this section. Now we will solve that problem in the following example.

Example

Try it.

The volume of a rectangular solid is given by the polynomial \(3{x}^{4}-3{x}^{3}-33{x}^{2}+54x.\) The length of the solid is given by \(3x\) and the width is given by \(x-2.\) Find the height, \(h,\) of the solid.

Solution

There are a few ways to approach this problem. We need to divide the expression for the volume of the solid by the expressions for the length and width. Let us create a sketch as in .

We can now write an equation by substituting the known values into the formula for the volume of a rectangular solid.

\[\begin{array}{lll}V & = & l⋅w⋅h \\ 3{x}^{4}-3{x}^{3}-33{x}^{2}+54x & = & 3x⋅(x-2)⋅h\end{array}\]

To solve for \(h,\) first divide both sides by \(3x.\)

\[\begin{array}{lll}\frac{3x⋅(x-2)⋅h}{3x} & = & \frac{3{x}^{4}-3{x}^{3}-33{x}^{2}+54x}{3x} \\ (x-2)h & = & {x}^{3}-{x}^{2}-11x+18\end{array}\]

Now solve for \(h\) using synthetic division.

\[h=\frac{{x}^{3}-{x}^{2}-11x+18}{x-2}\]

The quotient is \({x}^{2}+x-9\) and the remainder is 0. The height of the solid is \({x}^{2}+x-9.\)

Key Concepts

Polynomial long division can be used to divide a polynomial by any polynomial with equal or lower degree. See and .
The Division Algorithm tells us that a polynomial dividend can be written as the product of the divisor and the quotient added to the remainder.
Synthetic division is a shortcut that can be used to divide a polynomial by a binomial in the form \(x-k.\) See , , and .
Polynomial division can be used to solve application problems, including area and volume. See .

Dividing Monomials

We are now familiar with all the properties of exponents and used them to multiply polynomials. Next, we’ll use these properties to divide monomials and polynomials.

Example

Try it.

Find the quotient: \(54{a}^{2}{b}^{3}\div (-6a{b}^{5}).\)

Solution

When we divide monomials with more than one variable, we write one fraction for each variable.

	\(54{a}^{2}{b}^{3}\div (-6a{b}^{5})\)
Rewrite as a fraction.	\(\frac{54{a}^{2}{b}^{3}}{-6a{b}^{5}}\)
Use fraction multiplication.	\(\frac{54}{-6}\cdot \frac{{a}^{2}}{a}\cdot \frac{{b}^{3}}{{b}^{5}}\)
Simplify and use the Quotient Property.	\(-9\cdot a\cdot \frac{1}{{b}^{2}}\)
Multiply.	\(-\frac{9a}{{b}^{2}}\)

Once you become familiar with the process and have practiced it step by step several times, you may be able to simplify a fraction in one step.

Example

Try it.

Find the quotient: \(\frac{14{x}^{7}{y}^{12}}{21{x}^{11}{y}^{6}}.\)

Solution

Be very careful to simplify \(\frac{14}{21}\) by dividing out a common factor, and to simplify the variables by subtracting their exponents.

	\(\frac{14{x}^{7}{y}^{12}}{21{x}^{11}{y}^{6}}\)
Simplify and use the Quotient Property.	\(\frac{2{y}^{6}}{3{x}^{4}}\)

Divide a Polynomial by a Monomial

Now that we know how to divide a monomial by a monomial, the next procedure is to divide a polynomial of two or more terms by a monomial.

The method we’ll use to divide a polynomial by a monomial is based on the properties of fraction addition. So we’ll start with an example to review fraction addition. The sum \(\frac{y}{5}+\frac{2}{5}\) simplifies to \(\frac{y+2}{5}.\)

Now we will do this in reverse to split a single fraction into separate fractions. For example, \(\frac{y+2}{5}\) can be written \(\frac{y}{5}+\frac{2}{5}.\)

This is the “reverse” of fraction addition and it states that if a, b, and c are numbers where \(c\ne 0,\) then \(\frac{a+b}{c}=\frac{a}{c}+\frac{b}{c}.\) We will use this to divide polynomials by monomials.

Example

Try it.

Find the quotient: \((18{x}^{3}y-36x{y}^{2})\div (-3xy).\)

Solution

	\((18{x}^{3}y-36x{y}^{2})\div (-3xy)\)
Rewrite as a fraction.	\(\frac{18{x}^{3}y-36x{y}^{2}}{-3xy}\)
Divide each term by the divisor. Be careful with the signs!	\(\frac{18{x}^{3}y}{-3xy}-\frac{36x{y}^{2}}{-3xy}\)
Simplify.	\(-6{x}^{2}+12y\)

Divide Polynomials Using Long Division

Divide a polynomial by a binomial, we follow a procedure very similar to long division of numbers. So let’s look carefully the steps we take when we divide a 3-digit number, 875, by a 2-digit number, 25.

We check division by multiplying the quotient by the divisor.

If we did the division correctly, the product should equal the dividend.

\[\begin{array}{l}35\cdot 25 \\ 875✓\end{array}\]

Now we will divide a trinomial by a binomial. As you read through the example, notice how similar the steps are to the numerical example above.

Example

Try it.

Find the quotient: \(({x}^{2}+9x+20)\div (x+5).\)

Solution


Write it as a long division problem. Be sure the dividend is in standard form.
Divide \({x}^{2}\) by \(x.\) It may help to ask yourself, “What do I need to multiply \(x\) by to get \({x}^{2}\)?”
Put the answer, \(x,\) in the quotient over the \(x\) term. Multiply \(x\) times \(x+5.\) Line up the like terms under the dividend.
Subtract \({x}^{2}+5x\) from \({x}^{2}+9x.\) You may find it easier to change the signs and then add. Then bring down the last term, 20.
Divide \(4x\) by \(x.\) It may help to ask yourself, “What do I need to multiply \(x\) by to get \(4x\)?” Put the answer, \(4\), in the quotient over the constant term.
Multiply 4 times \(x+5.\)
Subtract \(4x+20\) from \(4x+20.\)
Check: Multiply the quotient by the divisor. \(\ (x+4)(x+5)\) You should get the dividend. \(\ {x}^{2}+9x+20✓\)

When we divided 875 by 25, we had no remainder. But sometimes division of numbers does leave a remainder. The same is true when we divide polynomials. In the next example, we’ll have a division that leaves a remainder. We write the remainder as a fraction with the divisor as the denominator.

Look back at the dividends in previous examples. The terms were written in descending order of degrees, and there were no missing degrees. The dividend in this example will be \({x}^{4}-{x}^{2}+5x-6.\) It is missing an \({x}^{3}\) term. We will add in \(0{x}^{3}\) as a placeholder.

In the next example, we will divide by \(2a+3.\) As we divide, we will have to consider the constants as well as the variables.

Example

Try it.

Find the quotient: \((8{a}^{3}+27)\div (2a+3).\)

Solution

This time we will show the division all in one step. We need to add two placeholders in order to divide.

To check, multiply \((2a+3)(4{a}^{2}-6a+9).\)

The result should be \(8{a}^{3}+27.\)

Condensed — the full section is in OpenStax Intermediate Algebra 2e.

Divide Polynomials using Synthetic Division

As we have mentioned before, mathematicians like to find patterns to make their work easier. Since long division can be tedious, let’s look back at the long division we did in and look for some patterns. We will use this as a basis for what is called synthetic division. The same problem in the synthetic division format is shown next.

Synthetic division basically just removes unnecessary repeated variables and numbers. Here all the \(x\) and \({x}^{2}\) are removed. as well as the \(\text{-}{x}^{2}\) and \(-4x\) as they are opposite the term above.

The first row of the synthetic division is the coefficients of the dividend. The \(-5\) is the opposite of the 5 in the divisor.

The second row of the synthetic division are the numbers shown in red in the division problem.

The third row of the synthetic division are the numbers shown in blue in the division problem.

Notice the quotient and remainder are shown in the third row.

\[\text{Synthetic division only works when the divisor is of the form}\ x-c.\]

The following example will explain the process.

Example

Try it.

Use synthetic division to find the quotient and remainder when \(2{x}^{3}+3{x}^{2}+x+8\) is divided by \(x+2.\)

Solution

Write the dividend with decreasing powers of \(x.\)
Write the coefficients of the terms as the first row of the synthetic division.
Write the divisor as \(x-c\) and place c in the synthetic division in the divisor box.
Bring down the first coefficient to the third row.
Multiply that coefficient by the divisor and place the result in the second row under the second coefficient.
Add the second column, putting the result in the third row.
Multiply that result by the divisor and place the result in the second row under the third coefficient.
Add the third column, putting the result in the third row.
Multiply that result by the divisor and place the result in the third row under the third coefficient.
Add the final column, putting the result in the third row.
The quotient is \(2{x}^{2}-1x+3\) and the remainder is 2.

The division is complete. The numbers in the third row give us the result. The \(2\ -1\ 3\) are the coefficients of the quotient. The quotient is \(2{x}^{2}-1x+3.\) The 2 in the box in the third row is the remainder.

Check:

\(\ \begin{array}{lll} \\ \\ \text{(quotient)(divisor)}+\text{remainder} & = & \text{dividend} \\ (2{x}^{2}-1x+3)(x+2)+2 & \overset{?}{=} & 2{x}^{3}+3{x}^{2}+x+8 \\ 2{x}^{3}-{x}^{2}+3x+4{x}^{2}-2x+6+2 & \overset{?}{=} & 2{x}^{3}+3{x}^{2}+x+8 \\ 2{x}^{3}+3{x}^{2}+x+8 & = & 2{x}^{3}+3{x}^{2}+x+8✓\end{array}\)

Condensed — the full section is in OpenStax Intermediate Algebra 2e.

Divide Polynomial Functions

Just as polynomials can be divided, polynomial functions can also be divided.

Example

Try it.

For functions \(f(x)={x}^{2}-5x-14\) and \(g(x)=x+2,\) find: ⓐ \((\frac{f}{g})(x)\) ⓑ \((\frac{f}{g})(-4).\)

Solution

ⓐ

Substitute for \(f(x)\) and \(g(x).\)	\(\ (\frac{f}{g})(x)=\frac{{x}^{2}-5x-14}{x+2}\)
Divide the polynomials.	\(\ (\frac{f}{g})(x)=x-7\)

ⓑ In part ⓐ we found \((\frac{f}{g})(x)\) and now are asked to find \((\frac{f}{g})(-4).\)

	\(\ (\frac{f}{g})(x)=x-7\)
To find \((\frac{f}{g})(-4),\) substitute \(x=-4.\)	\((\frac{f}{g})(-4)=-4-7\)
	\((\frac{f}{g})(-4)=-11\)

Use the Remainder and Factor Theorem

Let’s look at the division problems we have just worked that ended up with a remainder. They are summarized in the chart below. If we take the dividend from each division problem and use it to define a function, we get the functions shown in the chart. When the divisor is written as \(x-c,\) the value of the function at \(c,f(c),\) is the same as the remainder from the division problem.

Dividend	Divisor \(x-c\)	Remainder	Function	\(f(c)\)
\({x}^{4}-{x}^{2}+5x-6\)	\(x-(-2)\)	\(-4\)	\(f(x)={x}^{4}-{x}^{2}+5x-6\)	\(-4\)
\(3{x}^{3}-2{x}^{2}-10x+8\)	\(x-2\)	4	\(f(x)=3{x}^{3}-2{x}^{2}-10x+8\)	4
\({x}^{4}-16{x}^{2}+3x+15\)	\(x-(-4)\)	3	\(f(x)={x}^{4}-16{x}^{2}+3x+15\)	3

To see this more generally, we realize we can check a division problem by multiplying the quotient times the divisor and add the remainder. In function notation we could say, to get the dividend \(f(x),\) we multiply the quotient, \(q(x)\) times the divisor, \(x-c,\) and add the remainder, r.


If we evaluate this at \(c,\) we get:

This leads us to the Remainder Theorem.

Example

Try it.

Use the Remainder Theorem to find the remainder when \(f(x)={x}^{3}+3x+19\) is divided by \(x+2.\)

Solution

To use the Remainder Theorem, we must use the divisor in the \(x-c\) form. We can write the divisor \(x+2\) as \(x-(-2).\) So, our \(c\) is \(-2.\)

To find the remainder, we evaluate \(f(c)\) which is \(f(-2).\)


To evaluate \(f(-2),\) substitute \(x=-2.\)
Simplify.

	The remainder is 5 when \(f(x)={x}^{3}+3x+19\) is divided by \(x+2.\)
Check: Use synthetic division to check.

The remainder is 5.

When we divided \(8{a}^{3}+27\) by \(2a+3\) in the result was \(4{a}^{2}-6a+9.\) To check our work, we multiply \(4{a}^{2}-6a+9\) by \(2a+3\) to get \(8{a}^{3}+27\).

\[(4{a}^{2}-6a+9)(2a+3)=8{a}^{3}+27\]

Written this way, we can see that \(4{a}^{2}-6a+9\) and \(2a+3\) are factors of \(8{a}^{3}+27.\) When we did the division, the remainder was zero.

Whenever a divisor, \(x-c,\) divides a polynomial function, \(f(x),\) and resulting in a remainder of zero, we say \(x-c\) is a factor of \(f(x).\)

Condensed — the full section is in OpenStax Intermediate Algebra 2e.

Key Concepts

Division of a Polynomial by a Monomial
- To divide a polynomial by a monomial, divide each term of the polynomial by the monomial.
Division of Polynomial Functions
- For functions \(f(x)\) and \(g(x),\) where \(g(x)\ne 0,\)
  \((\frac{f}{g})(x)=\frac{f(x)}{g(x)}\)
Remainder Theorem
- If the polynomial function \(f(x)\) is divided by \(x-c,\) then the remainder is \(f(c).\)
Factor Theorem: 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 \(f(x)\)

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

When we divided 875 by 25, we had no remainder. But sometimes division of numbers does leave a remainder.
When dividing 69 by 4,
the dividend is_________,
the divisor is________,
the quotient is________,
and the remainder is________.
We check division by multiplying the quotient by the divisor and adding the remainder.

Find the quotient: \(({x}^{2}+9x+20)\div (x+5).\)

Atbildēt uz šo jautājumu


Write it as a long division problem. Be sure the dividend is written in descending order of powers, with no missing terms.
Divide \({x}^{2}\) by \(x.\) It may help to ask yourself, “What do I need to multiply \(x\) by to get \({x}^{2}\) ?”
Put the answer, \(x,\) in the quotient over the \(x\) term. Multiply \(x\) times \(x+5.\) Line up the like terms under the dividend.
Subtract \({x}^{2}+5x\) from \({x}^{2}+9x.\) You may find it easier to change the signs and then add. Then bring down the last term, 20.
Divide \(4x\) by \(x.\) It may help to ask yourself, “What do I need to multiply \(x\) by to get \(4x\) ?” Put the answer, \(4\) , in the quotient over the constant term.
Multiply 4 times \(x+5.\)
Subtract \(4x+20\) from \(4x+20.\)
Check: Multiply the quotient by the divisor. \(\ (x+4)(x+5)\) You should get the dividend. \(\ {x}^{2}+9x+20✓\)

Divide using long division of polynomials: \(({x}^{2}+10x+21)\div (x+3)\)
\((5x+{x}^{4}-{x}^{2}-6)\div (x+2)\)

Atbildēt uz šo jautājumu

Notice, this polynomial is not in descending order and it is missing \({x}^{3}\) term. We need to write it in the correct order and add \(0{x}^{3}\) as a placeholder.
To check, multiply divisor by the quotient and add remainder \((x+2)({x}^{3}-2{x}^{2}+3x-1)-4\)
The result should be \({x}^{4}-{x}^{2}+5x-6\)
\((7x+{x}^{4}-7{x}^{2}+6)\div (x+3)\)
\(({x}^{3}-8)\div (x-2)\)
Use synthetic division to find the quotient and remainder when \({x}^{4}-16{x}^{2}+3x+12\) is divided by x+4. Note that the divisor is in the form x-(-4), so use c as the divisor.

Atbildēt uz šo jautājumu

The polynomial \({x}^{4}-16{x}^{2}+3x+12\) has its term in order with descending degree but we notice there is no x3term. We will add a 0 as a placeholder for the \({x}^{3}\) term.
We divided a 4th degree polynomial by a 1st degree polynomial so the quotient will be a 3rd degree polynomial. Reading from the third row, the quotient has the coefficients 1,−4,0, and 3, which is \({x}^{3}-4{x}^{2}+0x+3\) . The remainder is 0.
Let \(f(x)={x}^{4}-5{x}^{2}+4x+12\)
1. ⓐ Find \(f(2)\)
2. ⓑ Divide \(f(x)\) by \((x-2)\) . What is the quotient? What is the remainder?
Let \(f(x)={x}^{3}+2{x}^{2}-5x-6\)
1. ⓐ Find \(f(-3)\)
2. ⓑ Divide \(f(x)\) by \((x+3)\) . What is the quotient? What is the remainder?
What is the connection between \(f(c)\) and the remainder when \(f(x)\) is divided by \((x-c)\) ? Summarize your findings.
Divide \(5{x}^{2}+3x-2\) by \(x+1.\)

Atbildēt uz šo jautājumu

The quotient is \(5x-2.\) The remainder is 0. We write the result as
\[\frac{5{x}^{2}+3x-2}{x+1}=5x-2\]
or
\[5{x}^{2}+3x-2=(x+1)(5x-2)\]
Divide \(6{x}^{3}+11{x}^{2}-31x+15\) by \(3x-2.\)

Atbildēt uz šo jautājumu

There is a remainder of 1. We can express the result as:
\[\frac{6{x}^{3}+11{x}^{2}-31x+15}{3x-2}=2{x}^{2}+5x-7+\frac{1}{3x-2}\]
Divide \(16{x}^{3}-12{x}^{2}+20x-3\) by \(4x+5.\)

Atbildēt uz šo jautājumu

\(4{x}^{2}-8x+15-\frac{78}{4x+5}\)
Use synthetic division to divide \(5{x}^{2}-3x-36\) by \(x-3.\)

Atbildēt uz šo jautājumu

Begin by setting up the synthetic division. Write \(k\) and the coefficients.
Bring down the lead coefficient. Multiply the lead coefficient by \(k.\)
Continue by adding the numbers in the second column. Multiply the resulting number by \(k.\) Write the result in the next column. Then add the numbers in the third column.
The result is \(5x+12.\) The remainder is 0. So \(x-3\) is a factor of the original polynomial.
Use synthetic division to divide \(4{x}^{3}+10{x}^{2}-6x-20\) by \(x+2.\)

Atbildēt uz šo jautājumu

The binomial divisor is \(x+2\) so \(k=-2.\) Add each column, multiply the result by –2, and repeat until the last column is reached.
The result is \(4{x}^{2}+2x-10.\) The remainder is 0. Thus, \(x+2\) is a factor of \(4{x}^{3}+10{x}^{2}-6x-20.\)
Use synthetic division to divide \(-9{x}^{4}+10{x}^{3}+7{x}^{2}-6\) by \(x-1.\)

Atbildēt uz šo jautājumu

Notice there is no x-term. We will use a zero as the coefficient for that term.
The result is \(-9{x}^{3}+{x}^{2}+8x+8+\frac{2}{x-1}.\)
Use synthetic division to divide \(3{x}^{4}+18{x}^{3}-3x+40\) by \(x+7.\)

Atbildēt uz šo jautājumu

\(3{x}^{3}-3{x}^{2}+21x-150+\frac{1,090}{x+7}\)
The volume of a rectangular solid is given by the polynomial \(3{x}^{4}-3{x}^{3}-33{x}^{2}+54x.\) The length of the solid is given by \(3x\) and the width is given by \(x-2.\) Find the height, \(h,\) of the solid.

Atbildēt uz šo jautājumu

There are a few ways to approach this problem. We need to divide the expression for the volume of the solid by the expressions for the length and width. Let us create a sketch as in .
We can now write an equation by substituting the known values into the formula for the volume of a rectangular solid.
\[\begin{array}{lll}V & = & l⋅w⋅h \\ 3{x}^{4}-3{x}^{3}-33{x}^{2}+54x & = & 3x⋅(x-2)⋅h\end{array}\]
To solve for \(h,\) first divide both sides by \(3x.\)
\[\begin{array}{lll}\frac{3x⋅(x-2)⋅h}{3x} & = & \frac{3{x}^{4}-3{x}^{3}-33{x}^{2}+54x}{3x} \\ (x-2)h & = & {x}^{3}-{x}^{2}-11x+18\end{array}\]
Now solve for \(h\) using synthetic division.
\[h=\frac{{x}^{3}-{x}^{2}-11x+18}{x-2}\]
The quotient is \({x}^{2}+x-9\) and the remainder is 0. The height of the solid is \({x}^{2}+x-9.\)
The area of a rectangle is given by \(3{x}^{3}+14{x}^{2}-23x+6.\) The width of the rectangle is given by \(x+6.\) Find an expression for the length of the rectangle.

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\(3{x}^{2}-4x+1\)
If division of a polynomial by a binomial results in a remainder of zero, what can be conclude?

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The binomial is a factor of the polynomial.
If a polynomial of degree \(n\) is divided by a binomial of degree 1, what is the degree of the quotient?
\(({x}^{2}+5x-1)\div (x-1)\)

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\(x+6+\frac{5}{x-1}\text{,}\ \text{quotient:}\ x+6\text{,}\ \text{remainder:}\ \text{5}\)
\((2{x}^{2}-9x-5)\div (x-5)\)
\((3{x}^{2}+23x+14)\div (x+7)\)

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\(3x+2\text{,}\ \text{quotient: }3x+2\text{,}\ \text{remainder: 0}\)
\((4{x}^{2}-10x+6)\div (4x+2)\)
\((6{x}^{2}-25x-25)\div (6x+5)\)

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\(x-5\text{,}\ \text{quotient:}\ x-5\text{,}\ \text{remainder:}\ \text{0}\)
\((-{x}^{2}-1)\div (x+1)\)
\((2{x}^{2}-3x+2)\div (x+2)\)

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\(2x-7+\frac{16}{x+2}\text{,}\ \text{quotient:}\text{}\ 2x-7\text{,}\ \text{remainder:}\ \text{16}\)
\(({x}^{3}-126)\div (x-5)\)
\((3{x}^{2}-5x+4)\div (3x+1)\)

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\(x-2+\frac{6}{3x+1}\text{,}\ \text{quotient:}\ x-2\text{,}\ \text{remainder:}\ \text{6}\)
\(({x}^{3}-3{x}^{2}+5x-6)\div (x-2)\)
\((2{x}^{3}+3{x}^{2}-4x+15)\div (x+3)\)

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\(2{x}^{2}-3x+5\text{,}\ \text{quotient:}\ 2{x}^{2}-3x+5\text{,}\ \text{remainder:}\ \text{0}\)
\((3{x}^{3}+2{x}^{2}-x+4)\div (x-3)\)
\((2{x}^{3}-6{x}^{2}-7x+6)\div (x-4)\)

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\(2{x}^{2}+2x+1+\frac{10}{x-4}\)
\((6{x}^{3}-10{x}^{2}-7x-15)\div (x+1)\)
\((4{x}^{3}-12{x}^{2}-5x-1)\div (2x+1)\)

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\(2{x}^{2}-7x+1-\frac{2}{2x+1}\)
\((9{x}^{3}-9{x}^{2}+18x+5)\div (3x-1)\)
\((3{x}^{3}-2{x}^{2}+x-4)\div (x+3)\)

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\(3{x}^{2}-11x+34-\frac{106}{x+3}\)
\((-6{x}^{3}+{x}^{2}-4)\div (2x-3)\)
\((2{x}^{3}+7{x}^{2}-13x-3)\div (2x-3)\)

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\({x}^{2}+5x+1\)

Symbols used here

\pm

plus or minus
Both signs at once: x = 3 ± 2 means 5 and 1.

\neq

not equal
The two sides are different.

\leq,\ \geq

less/greater than or equal
Inequalities that allow equality; < and > exclude it.

\sqrt{x},\ \sqrt[n]{x}

square root, n-th root
The non-negative number whose square (n-th power) is x.

|x|

absolute value / modulus
Distance from zero: |−3| = 3. For a complex number, distance from the origin.

imaginary unit
i² = −1.

\log_b x,\ \ln x

logarithm, natural log
The exponent b must be raised to for x; ln uses base e.

\mathbb{N},\ \mathbb{Z},\ \mathbb{Q},\ \mathbb{R},\ \mathbb{C}

number sets
Naturals, integers, rationals, reals, complex numbers.

How to: Dividing Polynomials

Dividing monomials
Dividing a polynomial by a monomial
Dividing polynomials using long division
Dividing polynomials using synthetic division
Dividing polynomial functions
Use the remainder and factor theorems
if
if

Questions people ask

What does it mean to solve an equation?

To find every value of the unknown that makes both sides equal. Each step is an operation applied to both sides that keeps the solution set the same, until the unknown stands alone.

Why do I sometimes get two answers?

A quadratic can cross the axis twice, so it can have two solutions. A degree-n polynomial has up to n. The graph shows where each one comes from.

How do I know whether to factor or use the quadratic formula?

Try factoring for a few seconds: look for two numbers that multiply to a·c and add to b. If nothing obvious appears, the discriminant b² − 4ac tells you how many real roots there are, and the formula finds them without guessing.

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Parts of this page are adapted from OpenStax College Algebra 2e (CC BY-NC-SA 4.0), OpenStax Intermediate Algebra 2e (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

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