How To Find The Quotient Of A Polynomial

1.6 Polynomials and Their Operations

Learning Objectives

1. Identify a polynomial and make up one's mind its degree.
2. Add together and subtract polynomials.
3. Multiply and divide polynomials.

Definitions

A polynomialAn algebraic expression consisting of terms with real number coefficients and variables with whole number exponents. is a special algebraic expression with terms that consist of real number coefficients and variable factors with whole number exponents. Some examples of polynomials follow:

$3x^2$

$7\mathrm{ten}y+5$

$\frac{3}{2}{x}^3+3x^2-\frac{\mathrm{i}}{\mathrm{two}}\mathrm{10}+1$

$\mathrm{six}{\mathrm{10}}^{2}y-4\mathrm{10}{y}^{\mathrm{three}}+7$

The degree of a termThe exponent of the variable. If at that place is more than i variable in the term, the degree of the term is the sum their exponents. in a polynomial is divers to be the exponent of the variable, or if there is more than one variable in the term, the degree is the sum of their exponents. Think that $x^0=1$ ; any constant term can be written every bit a product of $x^0$ and itself. Hence the degree of a abiding term is 0.

Term	Degree
$3x^2$	$2$
$6{\mathrm{10}}^{2}y$	$2+1=3$
$7a^{\mathrm{ii}}{b}^{\mathrm{three}}$	$\mathrm{ii}+\mathrm{iii}=5$
$\mathrm{eight}$	$0$ , since $8=8x^0$
$\mathrm{two}\mathrm{10}$	i, since $2x=2{\mathrm{10}}^1$

The degree of a polynomialThe largest degree of all of its terms. is the largest degree of all of its terms.

Polynomial	Degree
$4{\mathrm{ten}}^{\mathrm{v}}-3x^3+\mathrm{ii}x-\mathrm{ane}$	$5$
$6{\mathrm{10}}^{2}y-5x{y}^3+7$	$4$ , because $5x{y}^3$ has degree 4.
$\frac{1}{2}x+\frac{5}{4}$	1, because $\frac{1}{2}\mathrm{10}=\frac{1}{2}{x}^1$

Of item involvement are polynomials with i variableA polynomial where each term has the form $a_n{\mathrm{10}}^n$ , where $a_{\mathrm{due\; north}}$ is any real number and northward is whatsoever whole number. , where each term is of the form $a_n{\mathrm{10}}^n.$ Hither $a_n$ is whatever existent number and n is whatsoever whole number. Such polynomials have the standard form:

$$a_n{\mathrm{ten}}^{\mathrm{due\; north}}+a_{n-1}{x}^{n-1}+\cdots +a_{\mathrm{ane}}\mathrm{10}+a_0$$

Typically, we arrange terms of polynomials in descending guild based on the degree of each term. The leading coefficientThe coefficient of the term with the largest degree. is the coefficient of the variable with the highest power, in this case, $a_n.$

Example 1

Write in standard form: $3x-4x^2+\mathrm{five}{x}^3+7-\mathrm{ii}{x}^4.$

Solution:

Since terms are defined to be separated past addition, nosotros write the post-obit:

$$\begin{array}{l}3\mathrm{10}-4x^2+5{\mathrm{10}}^3+\mathrm{vii}-2{\mathrm{10}}^{\mathrm{four}}\hfill \\ =3\mathrm{10}+\left(-\mathrm{four}\right)x^2+5x^3+7+\left(-2\right)x^4\hfill \end{array}$$

In this course, nosotros can see that the subtraction in the original corresponds to negative coefficients. Considering improver is commutative, nosotros tin can write the terms in descending order based on the degree as follows:

$$\begin{array}{l}=\left(-\mathrm{two}\right)x^{\mathrm{iv}}+5x^3+\left(-4\right)x^2+3x+7\\ =-2{\mathrm{ten}}^{\mathrm{iv}}+\mathrm{v}{x}^3-\mathrm{iv}{\mathrm{10}}^2+3x+\mathrm{vii}\end{array}$$

Answer: $-\mathrm{two}{x}^{\mathrm{iv}}+5x^{\mathrm{iii}}-4{\mathrm{10}}^2+3\mathrm{10}+7$

Nosotros classify polynomials past the number of terms and the degree:

Expression	Nomenclature	Degree
$\mathrm{v}{\mathrm{10}}^7$	Monomial (one term)	seven
$8{\mathrm{10}}^{\mathrm{vi}}-1$	Binomial (two terms)	half dozen
$-\mathrm{iii}{\mathrm{ten}}^2+x-1$	Trinomial (3 terms)	2
$5{\mathrm{ten}}^3-\mathrm{two}{x}^2+3x-\mathrm{vi}$	Polynomial (many terms)	3

Polynomial with i term.

Polynomial with two terms.

Polynomial with three terms.

We tin can further classify polynomials with one variable past their degree:

Polynomial	Proper name
5	Abiding (degree 0)
$2\mathrm{10}+\mathrm{one}$	Linear (degree 1)
$3x^2+\mathrm{five}\mathrm{ten}-3$	Quadratic (degree 2)
$x^3+x^{\mathrm{ii}}+\mathrm{ten}+\mathrm{ane}$	Cubic (degree 3)
$7{\mathrm{10}}^{\mathrm{iv}}+3x^{\mathrm{iii}}-\mathrm{seven}x+8$	Quaternary-caste polynomial

A polynomial with caste 0.

A polynomial with degree one.

A polynomial with degree ii.

A polynomial with caste iii.

In this text, nosotros call whatsoever polynomial of caste $n\ge 4$ an nthursday-caste polynomial. In other words, if the degree is iv, nosotros call the polynomial a fourth-degree polynomial. If the degree is v, we phone call information technology a fifth-caste polynomial, and then on.

Example 2

State whether the following polynomial is linear or quadratic and give the leading coefficient: $25+\mathrm{iv}\mathrm{ten}-x^2.$

Solution:

The highest power is 2; therefore, information technology is a quadratic polynomial. Rewriting in standard form we have

$$-x^2+\mathrm{iv}\mathrm{ten}+25$$

Hither $-x^2=-1x^2$ and thus the leading coefficient is −1.

Answer: Quadratic; leading coefficient: −ane

Adding and Subtracting Polynomials

We begin by simplifying algebraic expressions that look like $+\left(a+b\right)$ or $-\left(a+b\right).$ Here, the coefficients are actually unsaid to be +1 and −i respectively and therefore the distributive property applies. Multiply each term within the parentheses by these factors every bit follows:

$$\begin{array}{lllllll}+\left(a+b\right)\hfill & =\hfill & +1\left(a+b\right)\hfill & =\hfill & \left(+1\right)a+\left(+1\right)b\hfill & =\hfill & a+b\hfill \\ -\left(a+b\right)\hfill & =\hfill & -1\left(a+b\right)\hfill & =\hfill & \left(-1\right)a+\left(-1\right)b\hfill & =\hfill & -a-b\hfill \end{array}$$

Use this thought as a ways to eliminate parentheses when adding and subtracting polynomials.

Instance iii

Add: $9x^{\mathrm{ii}}+\left(x^2-\mathrm{v}\right).$

Solution:

The belongings $+\left(a+b\right)=a+b$ allows us to eliminate the parentheses, after which we can then combine like terms.

$$\begin{array}{ccc}\hfill 9x^2+\left(x^{\mathrm{two}}-5\right)& =& 9{\mathrm{10}}^{\mathrm{ii}}+{\mathrm{ten}}^{\mathrm{ii}}-5\hfill \\ \hfill & =& 10x^2-5\hfill \end{array}$$

Answer: $\mathrm{ten}{x}^2-5$

Case iv

Add: $\left(3x^{\mathrm{two}}{y}^{\mathrm{two}}-4\mathrm{10}y+\mathrm{nine}\right)+\left(2x^2{y}^{\mathrm{two}}-6\mathrm{10}y-7\right).$

Solution:

Remember that the variable parts have to exist exactly the same before we tin can add the coefficients.

$$\begin{array}{c}\left(\mathrm{iii}{\mathrm{ten}}^{\mathrm{ii}}{y}^2-4xy+9\right)+\left(\mathrm{two}{x}^2{y}^2-6xy-7\right)\hfill \\ =\underset{–}{\mathrm{three}{\mathrm{10}}^2{y}^{\mathrm{ii}}}-\underset{–}{\underset{–}{4xy}}+\underset{–}{\underset{–}{\underset{–}{9}}}+\underset{–}{\mathrm{two}{x}^2{y}^{\mathrm{two}}}-\underset{–}{\underset{–}{6xy}}-\underset{–}{\underset{–}{\underset{–}{7}}}\\ =5x^2{y}^2-10xy+2\hfill \end{array}$$

Respond: $\mathrm{v}{x}^2{y}^2-10xy+2$

When subtracting polynomials, the parentheses go very important.

Example 5

Decrease: $4x^2-\left(\mathrm{iii}{x}^{\mathrm{two}}+\mathrm{five}x\right).$

Solution:

The property $-\left(a+b\right)=-a-b$ allows united states to remove the parentheses after subtracting each term.

$$\begin{array}{ccc}\hfill 4x^2-\left(3x^{\mathrm{two}}+5\mathrm{10}\right)& =& 4{\mathrm{ten}}^2-3x^2-\mathrm{v}x\hfill \\ \hfill & =& x^2-5x\hfill \end{array}$$

Answer: $x^2-5x$

Subtracting a quantity is equivalent to multiplying it by −1.

Example 6

Subtract: $\left(3x^2-2xy+y^2\right)-\left(\mathrm{two}{\mathrm{ten}}^2-xy+3y^2\right).$

Solution:

Distribute the −1, remove the parentheses, and and then combine like terms. Multiplying the terms of a polynomial by −1 changes all the signs.

$$\begin{array}{ccc}\hfill & =& \mathrm{three}{x}^2-\mathrm{ii}xy+y^2-\mathrm{ii}{x}^2+\mathrm{ten}y-3y^{\mathrm{two}}\hfill \\ \hfill & =& {\mathrm{ten}}^2-xy-2y^2\hfill \end{array}$$

Respond: $x^{\mathrm{two}}-xy-2y^2$

Try this! Subtract: $\left(\mathrm{vii}{a}^2-2ab+b^2\right)-\left(a^2-2ab+\mathrm{five}{b}^{\mathrm{two}}\right).$

Answer: $6a^2-4b^{\mathrm{two}}$

Multiplying Polynomials

Use the product dominion for exponents, $x^{\mathrm{thou}}\cdot x^n={\mathrm{ten}}^{\mathrm{one\; thousand}+\mathrm{north}}$ , to multiply a monomial times a polynomial. In other words, when multiplying two expressions with the same base, add the exponents. To find the product of monomials, multiply the coefficients and add the exponents of variable factors with the same base. For instance,

$$\begin{array}{cccc}\hfill \mathrm{vii}{x}^4\cdot \mathrm{viii}{x}^3& =& 7\cdot 8\cdot x^4\cdot x^{\mathrm{three}}\hfill & Co\mathrm{thousand}\mathrm{yard}utati\mathrm{five}eproperty\hfill \\ \hfill & =& 56x^{4+3}\hfill & Productrul\mathrm{due\; east}for\mathrm{east}xpon\mathrm{due\; east}\mathrm{north}ts\hfill \\ \hfill & =& 56x^7\hfill \end{array}$$

To multiply a polynomial by a monomial, use the distributive belongings, and then simplify each term.

Example vii

Multiply: $\mathrm{five}x{y}^{\mathrm{two}}\left(2x^2{y}^2-xy+1\right).$

Solution:

Apply the distributive property and then simplify.

$$\begin{array}{ccc}\hfill & =& \mathrm{five}\mathrm{ten}{y}^{\mathrm{two}}\cdot \mathrm{ii}{x}^2{y}^2-5x{y}^{\mathrm{two}}\cdot \mathrm{ten}y+\mathrm{v}x{y}^{\mathrm{ii}}\cdot \mathrm{ane}\hfill \\ \hfill & =& 10x^{\mathrm{three}}{y}^4-5x^{\mathrm{two}}{y}^3+5x{y}^2\hfill \end{array}$$

Answer: $10x^3{y}^{\mathrm{iv}}-5x^2{y}^3+5x{y}^2$

To summarize, multiplying a polynomial past a monomial involves the distributive belongings and the product rule for exponents. Multiply all of the terms of the polynomial by the monomial. For each term, multiply the coefficients and add together exponents of variables where the bases are the same.

In the same manner that we used the distributive belongings to distribute a monomial, nosotros utilise information technology to distribute a binomial. $$\begin{array}{ccc}\hfill \left(a+b\right)\left(c+d\right)& =& \left(a+b\right)\cdot c+\left(a+b\right)\cdot d\hfill \\ \hfill & =& ac+bc+ad+bd\hfill \\ \hfill & =& ac+ad+bc+bd\hfill \end{array}$$ Here we utilise the distributive property multiple times to produce the final event. This same result is obtained in ane stride if we apply the distributive property to a and b separately as follows:

This is often called the FOIL method. Multiply the first, outer, inner, and then terminal terms.

Example 8

Multiply: $\left(6\mathrm{ten}-\mathrm{one}\right)\left(\mathrm{three}x-5\right).$

Solution:

Distribute vix and −1 and then combine similar terms.

$$\begin{array}{ccc}\hfill \left(6x-1\right)\left(3x-5\right)& =& 6\mathrm{10}\cdot \mathrm{iii}\mathrm{10}-6x\cdot \mathrm{v}+\left(-1\right)\cdot 3x-\left(-1\right)\cdot 5\hfill \\ \hfill & =& 18x^2-30\mathrm{10}-3\mathrm{10}+5\hfill \\ \hfill & =& 18x^2-33x+5\hfill \end{array}$$

Answer: $\mathrm{eighteen}{\mathrm{ten}}^2-33x+5$

Consider the following two calculations:

$\begin{array}{ccc}\hfill {\left(a+b\right)}^2& =& \left(a+b\right)\left(a+b\right)\hfill \\ \hfill & =& a^2+ab+ba+b^{\mathrm{ii}}\hfill \\ \hfill & =& a^2+ab+ab+b^2\hfill \\ \hfill & =& a^2+2ab+b^{\mathrm{two}}\hfill \end{array}$

$\begin{array}{ccc}\hfill {\left(a-b\right)}^{\mathrm{two}}& =& \left(a-b\right)\left(a-b\right)\hfill \\ \hfill & =& a^2-ab-ba+b^{\mathrm{two}}\hfill \\ \hfill & =& a^2-ab-ab+b^2\hfill \\ \hfill & =& a^{\mathrm{two}}-\mathrm{ii}ab+b^{\mathrm{two}}\hfill \end{array}$

This leads united states to two formulas that describe perfect square trinomialsThe trinomials obtained by squaring the binomials ${\left(a+b\right)}^2=a^2+2ab+b^2$ and ${\left(a-b\right)}^2=a^{\mathrm{two}}-\mathrm{ii}ab+b^2.$ :

$$\begin{array}{c}{\left(a+b\right)}^2=a^{\mathrm{two}}+\mathrm{ii}ab+b^{\mathrm{two}}\\ {\left(a-b\right)}^2=a^2-2ab+b^2\end{array}$$

We tin can use these formulas to chop-chop square a binomial.

Instance 9

Multiply: ${\left(3x+\mathrm{v}\right)}^{\mathrm{two}}.$

Solution:

Here $a=3x$ and $b=\mathrm{v}.$ Utilise the formula:

Reply: $\mathrm{ix}{\mathrm{ten}}^2+\mathrm{xxx}x+25$

This process should become routine enough to exist performed mentally. Our third special product follows: $$\begin{array}{ccc}\hfill \left(a+b\right)\left(a-b\right)& =& a^2-ab+ba-b^2\hfill \\ \hfill & =& a^2-ab+ab-b^2\hfill \\ \hfill & =& a^2-b^2\hfill \end{array}$$ This production is called difference of squaresThe special product obtained by multiplying conjugate binomials $\left(a+b\right)\left(a-b\right)=a^2-b^{\mathrm{ii}}.$ :

$$\left(a+b\right)\left(a-b\right)=a^2-b^2$$

The binomials $\left(a+b\right)$ and $\left(a-b\right)$ are chosen conjugate binomialsThe binomials $\left(a+b\right)$ and $\left(a-b\right).$ . When multiplying conjugate binomials the heart terms are opposites and their sum is zero; the production is itself a binomial.

Example 10

Multiply: $\left(\mathrm{iii}xy+1\right)\left(3xy-1\right).$

Solution:

$$\begin{array}{ccc}\hfill \left(3\mathrm{10}y+1\right)\left(\mathrm{iii}xy-1\right)& =& {\left(\mathrm{iii}\mathrm{10}y\right)}^{\mathrm{ii}}-3xy+3\mathrm{10}y-{\mathrm{one}}^{\mathrm{two}}\hfill \\ \hfill & =& 9x^{\mathrm{two}}{y}^2-\mathrm{ane}\hfill \end{array}$$

Answer: $9{\mathrm{ten}}^2{y}^2-\mathrm{i}$

Try this! Multiply: $\left({\mathrm{10}}^{\mathrm{ii}}+5y^2\right)\left(x^2-5y^2\right).$

Answer: $\left({\mathrm{ten}}^{\mathrm{iv}}-25y^4\right)$

Case eleven

Multiply: ${\left(5\mathrm{ten}-2\right)}^3.$

Solution:

Here we perform one product at a time.

Respond: $125x^2-150x^{\mathrm{ii}}+\mathrm{sixty}x-8$

Dividing Polynomials

Use the quotient rule for exponents, $\frac{x^m}{{\mathrm{10}}^{\mathrm{northward}}}={\mathrm{ten}}^{m-n}$ , to split a polynomial by a monomial. In other words, when dividing two expressions with the same base, subtract the exponents. In this section, nosotros will presume that all variables in the denominator are nonzero.

Example 12

Divide: $\frac{24x^{\mathrm{seven}}{y}^{\mathrm{v}}}{\mathrm{eight}{x}^{\mathrm{three}}{y}^{\mathrm{ii}}}.$

Solution:

Divide the coefficients and apply the quotient rule by subtracting the exponents of the like bases.

$$\begin{array}{ccc}\hfill \frac{24{\mathrm{ten}}^7{y}^{\mathrm{v}}}{8{\mathrm{10}}^3{y}^2}& =& \frac{24}{8}{x}^{7-\mathrm{three}}{y}^{\mathrm{five}-\mathrm{ii}}\hfill \\ \hfill & =& 3{\mathrm{10}}^4{y}^{\mathrm{three}}\hfill \end{array}$$

Answer: $\mathrm{iii}{x}^{\mathrm{iv}}{y}^3$

When dividing a polynomial by a monomial, we may treat the monomial as a common denominator and pause up the fraction using the following belongings: $$\frac{a+b}{c}=\frac{a}{c}+\frac{b}{c}$$ Applying this property volition consequence in terms that can be treated as quotients of monomials.

Instance thirteen

Divide: $\frac{-5x^4+25x^3-\mathrm{fifteen}{x}^{\mathrm{ii}}}{\mathrm{five}{\mathrm{10}}^2}.$

Solution:

Interruption up the fraction past dividing each term in the numerator by the monomial in the denominator, and then simplify each term.

$$\begin{array}{ccc}\hfill \frac{-\mathrm{five}{\mathrm{10}}^4+25{\mathrm{10}}^3-\mathrm{fifteen}{\mathrm{10}}^2}{\mathrm{five}{x}^2}& =& -\frac{5x^{\mathrm{iv}}}{5x^{\mathrm{two}}}+\frac{25x^3}{5{\mathrm{ten}}^2}-\frac{15x^2}{5x^2}\hfill \\ \hfill & =& -\frac{\mathrm{v}}{5}{\mathrm{10}}^{\mathrm{iv}-\mathrm{ii}}+\frac{25}{\mathrm{five}}{\mathrm{ten}}^{3-2}-\frac{\mathrm{fifteen}}{\mathrm{v}}{x}^{2-\mathrm{ii}}\hfill \\ \hfill & =& -1x^2+\mathrm{five}{x}^1-3x^0\hfill \\ \hfill & =& -{\mathrm{10}}^2+5x-3\cdot 1\hfill \end{array}$$

Respond: $-{\mathrm{10}}^2+5x-3$

We can check our partition by multiplying our respond, the caliber, past the monomial in the denominator, the divisor, to see if we obtain the original numerator, the dividend.

$\frac{Divid\mathrm{eastward}nd}{Divi\mathrm{southward}or}=Quotie\mathrm{north}t$	$\frac{-\mathrm{five}{\mathrm{ten}}^4+25x^{\mathrm{iii}}-15x^2}{5{\mathrm{ten}}^{\mathrm{ii}}}=-{\mathrm{10}}^{\mathrm{two}}+5x-3$
or	or
$Divide\mathrm{northward}d=Divi\mathrm{southward}or\cdot Quotient$	$-\mathrm{five}{x}^{\mathrm{four}}+25x^3-\mathrm{fifteen}{\mathrm{10}}^2=5x^2\left(-{\mathrm{ten}}^{\mathrm{ii}}+5x-3\right)\text{✓}$

The same technique outlined for dividing by a monomial does non work for polynomials with two or more terms in the denominator. In this section, we will outline a process chosen polynomial long divisionThe procedure of dividing ii polynomials using the division algorithm. , which is based on the division algorithm for existent numbers. For the sake of clarity, we will assume that all expressions in the denominator are nonzero.

Example 14

Divide: $\frac{{\mathrm{ten}}^{\mathrm{iii}}+3x^{\mathrm{ii}}-\mathrm{eight}\mathrm{ten}-4}{x-\mathrm{ii}}.$

Solution:

Here $x-\mathrm{two}$ is the divisor and ${\mathrm{ten}}^{\mathrm{three}}+3{\mathrm{ten}}^2-\mathrm{viii}x-\mathrm{iv}$ is the dividend. To determine the first term of the quotient, carve up the leading term of the dividend by the leading term of the divisor.

Multiply the first term of the quotient by the divisor, remembering to distribute, and line up similar terms with the dividend.

Subtract the resulting quantity from the dividend. Take care to subtract both terms.

Bring down the remaining terms and repeat the process.

Observe that the leading term is eliminated and that the issue has a degree that is one less. The complete process is illustrated below:

Polynomial long sectionalization ends when the degree of the remainder is less than the degree of the divisor. Hither, the remainder is 0. Therefore, the binomial divides the polynomial evenly and the answer is the quotient shown to a higher place the sectionalization bar.

$$\frac{x^{\mathrm{three}}+3x^2-8\mathrm{10}-4}{\mathrm{ten}-2}=x^2+5x+2$$

To check the answer, multiply the divisor by the quotient to see if yous obtain the dividend as illustrated below:

$$x^{\mathrm{three}}+3{\mathrm{10}}^2-8x-\mathrm{iv}=\left(\mathrm{ten}-2\right)\left(x^2+\mathrm{v}x+2\right)$$

This is left to the reader every bit an exercise.

Answer: $x^{\mathrm{two}}+5x+2$

Next, we demonstrate the case where at that place is a nonzero remainder.

But every bit with real numbers, the final respond adds to the quotient the fraction where the balance is the numerator and the divisor is the denominator. In general, when dividing nosotros have: $$\frac{Divid\mathrm{eastward}nd}{Divisor}=Quotie\mathrm{due\; north}t+\frac{Remai\mathrm{north}d\mathrm{east}r}{Divisor}$$ If we multiply both sides by the divisor we obtain, $$Divid\mathrm{eastward}\mathrm{due\; north}d=Quoti\mathrm{due\; east}nt\times Divisor+R\mathrm{east}\mathrm{thou}ai\mathrm{north}d\mathrm{due\; east}r$$

Case 15

Split: $\frac{\mathrm{half\; dozen}{\mathrm{10}}^{\mathrm{ii}}-5x+\mathrm{three}}{\mathrm{two}x-1}.$

Solution:

Since the denominator is a binomial, begin by setting upwards polynomial long division.

To start, make up one's mind what monomial times $2x-1$ results in a leading term $\mathrm{half\; dozen}{x}^2.$ This is the caliber of the given leading terms: $\left(6{\mathrm{ten}}^2\right)÷\left(2x\right)=3\mathrm{10}.$ Multiply $\mathrm{iii}x$ times the divisor $\mathrm{two}x-\mathrm{i}$ , and line upward the effect with like terms of the dividend.

Subtract the result from the dividend and bring down the constant term +3.

Subtracting eliminates the leading term. Multiply $2x-\mathrm{i}$ by −1 and line upward the outcome.

Decrease again and observe that we are left with a remainder.

The constant term 2 has degree 0 and thus the division ends. Therefore,

$$\frac{6x^2-5x+\mathrm{three}}{2\mathrm{10}-\mathrm{i}}=3x-\mathrm{i}+\frac{2}{\mathrm{ii}x-\mathrm{i}}$$

To check that this effect is right, nosotros multiply as follows:

$$\begin{array}{ccc}\hfill quotie\mathrm{due\; north}t\times di\mathrm{five}i\mathrm{due\; south}or+r\mathrm{east}\mathrm{1000}ainder& =& \left(3\mathrm{10}-\mathrm{one}\right)\left(2\mathrm{10}-\mathrm{one}\right)+2\hfill \\ \hfill & =& \mathrm{six}{x}^2-3\mathrm{10}-2\mathrm{ten}+1+\mathrm{two}\hfill \\ \hfill & =& \mathrm{half\; dozen}{x}^{\mathrm{two}}-5x+2=di5id\mathrm{due\; east}\mathrm{due\; north}d\text{✓}\hfill \end{array}$$

Respond: $\mathrm{iii}x-1+\frac{2}{\mathrm{two}\mathrm{10}-\mathrm{ane}}$

Occasionally, some of the powers of the variables appear to exist missing inside a polynomial. This can lead to errors when lining upward like terms. Therefore, when first learning how to split up polynomials using long partition, fill in the missing terms with zero coefficients, chosen placeholdersTerms with cypher coefficients used to fill in all missing exponents within a polynomial. .

Example 16

Split up: $\frac{27{\mathrm{ten}}^3+64}{3x+4}.$

Solution:

Notice that the binomial in the numerator does non have terms with degree 2 or 1. The division is simplified if nosotros rewrite the expression with placeholders:

$$27x^{\mathrm{three}}+64=27x^3+0x^2+0\mathrm{10}+64$$

Prepare polynomial long division:

We begin with $27x^3÷\mathrm{iii}x=9x^2$ and work the remainder of the division algorithm.

Answer: $9{\mathrm{10}}^2-12\mathrm{ten}+16$

Case 17

Divide: $\frac{3x^{\mathrm{four}}-\mathrm{two}{x}^3+6x^2+23\mathrm{10}-7}{x^{\mathrm{two}}-2x+\mathrm{five}}.$

Solution:

Begin the process by dividing the leading terms to determine the leading term of the caliber $\mathrm{iii}{\mathrm{ten}}^4÷x^{\mathrm{ii}}=3x^2.$ Accept care to distribute and line up the like terms. Continue the process until the remainder has a degree less than 2.

The remainder is $\mathrm{10}-\mathrm{ii}.$ Write the reply with the residue:

$$\frac{\mathrm{iii}{x}^4-2x^{\mathrm{three}}+6x^2+23\mathrm{10}-7}{x^{\mathrm{ii}}-2x+5}=3{\mathrm{10}}^2+\mathrm{four}x-\mathrm{one}+\frac{\mathrm{10}-2}{x^2-\mathrm{two}\mathrm{10}+\mathrm{five}}$$

Answer: $3{\mathrm{10}}^2+4\mathrm{10}-1+\frac{\mathrm{ten}-2}{x^{\mathrm{two}}-\mathrm{ii}x+5}$

Polynomial long division takes time and practise to principal. Work lots of problems and recall that you may bank check your answers past multiplying the quotient by the divisor (and calculation the remainder if present) to obtain the dividend.

Try this! Split up: $\frac{\mathrm{half-dozen}{x}^4-\mathrm{xiii}{\mathrm{ten}}^{\mathrm{iii}}+9x^2-14\mathrm{ten}+\mathrm{half\; dozen}}{3x-2}.$

Answer: $\mathrm{ii}{x}^{\mathrm{iii}}-3x^2+x-4-\frac{2}{3x-2}$

Key Takeaways

• Polynomials are special algebraic expressions where the terms are the products of existent numbers and variables with whole number exponents.
• The degree of a polynomial with one variable is the largest exponent of the variable found in whatever term. In addition, the terms of a polynomial are typically arranged in descending gild based on the caste of each term.
• When calculation polynomials, remove the associated parentheses and and then combine similar terms. When subtracting polynomials, distribute the −1, remove the parentheses, and then combine like terms.
• To multiply polynomials apply the distributive property; multiply each term in the first polynomial with each term in the second polynomial. Then combine like terms.
• When dividing by a monomial, dissever all terms in the numerator past the monomial and and so simplify each term.
• When dividing a polynomial by some other polynomial, apply the division algorithm.

Topic Exercises

Office A: Definitions

Write the given polynomials in standard form.

1. $1-\mathrm{10}-x^2$

2. $y-\mathrm{v}+y^2$

3. $y-3y^{\mathrm{two}}+5-y^3$

4. $8-12a^2+a^3-a$

5. $2-x^{\mathrm{two}}+6x-\mathrm{v}{\mathrm{ten}}^3+{\mathrm{ten}}^4$

6. $a^3-5+a^2+\mathrm{two}{a}^4-a^{\mathrm{five}}+6a$

Classify the given polynomial as a monomial, binomial, or trinomial and country the degree.

7. $x^{\mathrm{two}}-\mathrm{10}+2$

8. $5-10x^3$

9. ${\mathrm{ten}}^{\mathrm{two}}{y}^{\mathrm{ii}}+\mathrm{five}xy-6$

10. $-\mathrm{ii}{x}^{\mathrm{three}}{y}^2$

11. ${\mathrm{10}}^4-1$

12. 5

State whether the polynomial is linear or quadratic and give the leading coefficient.

13. $1-\mathrm{ix}{x}^2$

14. $\mathrm{x}{\mathrm{ten}}^{\mathrm{two}}$

15. $\mathrm{two}x-\mathrm{three}$

16. $100\mathrm{ten}$

17. $5{\mathrm{10}}^2+3\mathrm{ten}-1$

18. $x-\mathrm{ane}$

19. $\mathrm{10}-6-2x^2$

20. $1-5x$

Part B: Adding and Subtracting Polynomials

Simplify.

21. $\left(5x^{\mathrm{ii}}-3\mathrm{10}-\mathrm{two}\right)+\left(\mathrm{two}{\mathrm{ten}}^2-6\mathrm{10}+7\right)$

22. $\left(x^2+\mathrm{seven}x-12\right)+\left(2x^2-\mathrm{ten}+\mathrm{three}\right)$

23. $\left(x^{\mathrm{two}}+5x+\mathrm{ten}\right)+\left({\mathrm{10}}^2-10\right)$

24. $\left(x^2-\mathrm{i}\right)+\left(\mathrm{four}\mathrm{10}+\mathrm{ii}\right)$

25. $\left(\mathrm{ten}{\mathrm{ten}}^2+3x-\mathrm{ii}\right)-\left(x^2-6x+1\right)$

26. $\left(x^2-3x-\mathrm{viii}\right)-\left(2x^2-\mathrm{iii}\mathrm{10}-8\right)$

27. $\left(\frac{2}{\mathrm{iii}}{x}^2+\frac{3}{4}x-1\right)-\left(\frac{\mathrm{ane}}{6}{\mathrm{10}}^2+\frac{5}{2}\mathrm{ten}-\frac{1}{2}\right)$

28. $\left(\frac{\mathrm{iv}}{5}{\mathrm{ten}}^2-\frac{\mathrm{five}}{8}\mathrm{10}+\frac{10}{\mathrm{vi}}\right)-\left(\frac{3}{\mathrm{x}}{x}^2-\frac{\mathrm{two}}{\mathrm{iii}}x+\frac{3}{\mathrm{five}}\right)$

29. $\left({\mathrm{10}}^{\mathrm{ii}}{y}^2+7xy-5\right)-\left(2x^{\mathrm{two}}{y}^2+\mathrm{five}\mathrm{ten}y-\mathrm{iv}\right)$

30. $\left(x^2-y^2\right)-\left({\mathrm{ten}}^{\mathrm{two}}+6xy+y^{\mathrm{two}}\right)$

31. $\left(a^2{b}^{\mathrm{two}}+\mathrm{five}ab-2\right)+\left(7ab-2\right)-\left(4-a^2{b}^2\right)$

32. $\left(a^2+9ab-6b^{\mathrm{two}}\right)-\left(a^2-b^2\right)+7ab$

33. $\left(10x^{\mathrm{ii}}y-8\mathrm{10}y+5x{y}^2\right)-\left(x^{2}y-4xy\right)+\left(\mathrm{ten}{y}^{\mathrm{ii}}+\mathrm{iv}xy\right)$

34. $\left(\mathrm{two}{m}^{2}n-6mn+\mathrm{nine}m{n}^{\mathrm{ii}}\right)-\left({\mathrm{thou}}^{2}n+10\mathrm{chiliad}\mathrm{north}\right)-m^{2}n$

35. $\left(8{\mathrm{10}}^2{y}^{\mathrm{two}}-5\mathrm{10}y+2\right)-\left(x^2{y}^2+5\right)+\left(2xy-3\right)$

36. $\left(x^2-y^2\right)-\left(5x^2-2xy-y^2\right)-\left(x^{\mathrm{two}}-\mathrm{vii}xy\right)$

37. $\left(\frac{\mathrm{i}}{\mathrm{six}}{a}^{\mathrm{ii}}-2ab+\frac{3}{4}{b}^2\right)-\left(\frac{5}{3}{a}^2+\frac{4}{\mathrm{five}}{b}^2\right)+\frac{11}{8}ab$

38. $\left(\frac{\mathrm{v}}{2}{x}^2-2y^2\right)-\left(\frac{7}{5}{x}^2-\frac{\mathrm{ane}}{2}xy+\frac{7}{3}{y}^2\right)-\frac{\mathrm{i}}{2}xy$

39. $\left({\mathrm{10}}^{2\mathrm{north}}+\mathrm{five}{x}^n-2\right)+\left(\mathrm{ii}{x}^{2n}-3{\mathrm{10}}^{\mathrm{due\; north}}-1\right)$

40. $\left(7{\mathrm{10}}^{\mathrm{ii}n}-x^n+5\right)-\left(\mathrm{vi}{x}^{\mathrm{two}\mathrm{northward}}-x^n-\mathrm{eight}\right)$

41. Subtract $\mathrm{iv}y-\mathrm{iii}$ from $y^2+7y-10.$

42. Decrease $x^2+\mathrm{three}x-2$ from $2x^2+4x-1.$

43. A right circular cylinder has a height that is equal to the radius of the base, $h=r.$ Find a formula for the surface area in terms of h.

44. A rectangular solid has a width that is twice the superlative and a length that is 3 times that of the height. Find a formula for the surface expanse in terms of the top.

Office C: Multiplying Polynomials

Multiply.

45. $-8x^2\cdot \mathrm{two}\mathrm{ten}$

46. $-10{\mathrm{10}}^{\mathrm{ii}}y\cdot 5x^3{y}^{\mathrm{ii}}$

47. $2x\left(5\mathrm{ten}-\mathrm{one}\right)$

48. $-4x\left(\mathrm{three}\mathrm{ten}-5\right)$

49. $\mathrm{vii}{\mathrm{10}}^2\left(\mathrm{ii}\mathrm{10}-\mathrm{vi}\right)$

50. $-3x^{\mathrm{ii}}\left({\mathrm{ten}}^2-x+\mathrm{three}\right)$

51. $-\mathrm{v}{y}^{\mathrm{four}}\left(y^{\mathrm{ii}}-2y+3\right)$

52. $\frac{\mathrm{five}}{2}{a}^3\left(24a^2-6a+\mathrm{four}\right)$

53. $2xy\left(x^2-\mathrm{vii}xy+y^2\right)$

54. $$-2a^{2}b\left(a^2-3ab+5b^2\right)$$

55. $x^{\mathrm{north}}\left({\mathrm{ten}}^{\mathrm{two}}+x+\mathrm{ane}\right)$

56. ${\mathrm{10}}^n\left(x^{\mathrm{ii}n}-x^{\mathrm{due\; north}}-\mathrm{i}\right)$

57. $\left(\mathrm{ten}+\mathrm{four}\right)\left(x-\mathrm{five}\right)$

58. $\left(\mathrm{10}-7\right)\left(x-6\right)$

59. $\left(\mathrm{ii}x-\mathrm{three}\right)\left(\mathrm{iii}\mathrm{10}-1\right)$

60. $\left(9x+\mathrm{one}\right)\left(3x+2\right)$

61. $\left(\mathrm{iii}{\mathrm{10}}^2-y^2\right)\left(x^2-5y^2\right)$

62. $\left(\mathrm{five}{y}^2-x^2\right)\left(2y^2-\mathrm{three}{x}^{\mathrm{two}}\right)$

63. $\left(3x+\mathrm{v}\right)\left(3\mathrm{ten}-5\right)$

64. $\left(\mathrm{ten}+\mathrm{half\; dozen}\right)\left(\mathrm{10}-\mathrm{six}\right)$

65. $\left(a^2-b^2\right)\left(a^2+b^{\mathrm{two}}\right)$

66. $\left(ab+7\right)\left(ab-7\right)$

67. $\left(4x-5y^2\right)\left(3x^{\mathrm{two}}-y\right)$

68. $\left(xy+5\right)\left(x-y\right)$

69. $\left(x-\mathrm{five}\right)\left(x^{\mathrm{two}}-\mathrm{three}\mathrm{10}+8\right)$

70. $\left(2\mathrm{10}-7\right)\left(\mathrm{three}{\mathrm{10}}^{\mathrm{ii}}-\mathrm{ten}+\mathrm{one}\right)$

71. $\left(x^{\mathrm{ii}}+7x-1\right)\left(\mathrm{ii}{x}^2-3\mathrm{10}-1\right)$

72. $\left(\mathrm{iv}{x}^{\mathrm{two}}-x+6\right)\left(\mathrm{five}{\mathrm{10}}^2-4\mathrm{10}-\mathrm{iii}\right)$

73. ${\left(x+8\right)}^2$

74. ${\left(x-3\right)}^{\mathrm{two}}$

75. ${\left(2\mathrm{10}-\mathrm{v}\right)}^{\mathrm{ii}}$

76. ${\left(\mathrm{iii}x+1\right)}^{\mathrm{two}}$

77. ${\left(a-3b\right)}^2$

78. ${\left(\mathrm{vii}a-b\right)}^2$

79. $${\left({\mathrm{10}}^2+2y^{\mathrm{two}}\right)}^2$$

80. $${\left(x^{\mathrm{ii}}-6y\right)}^{\mathrm{two}}$$

81. $${\left(a^{\mathrm{ii}}-a+5\right)}^{\mathrm{ii}}$$

82. $${\left(x^2-\mathrm{iii}x-1\right)}^2$$

83. ${\left(x-3\right)}^3$

84. ${\left(x+2\right)}^3$

85. ${\left(3\mathrm{10}+1\right)}^3$

86. ${\left(\mathrm{two}\mathrm{ten}-3\right)}^3$

87. ${\left(x+2\right)}^{\mathrm{four}}$

88. ${\left(\mathrm{ten}-3\right)}^4$

89. ${\left(\mathrm{two}x-1\right)}^4$

90. ${\left(3x-1\right)}^4$

91. $$\left({\mathrm{ten}}^{2n}+5\right)\left(x^{2n}-5\right)$$

92. $$\left(x^{\mathrm{due\; north}}-\mathrm{one}\right)\left(x^{2\mathrm{northward}}+4x^{\mathrm{northward}}-3\right)$$

93. $${\left(x^{\mathrm{two}n}-1\right)}^2$$

94. $${\left(x^{3n}+1\right)}^{\mathrm{ii}}$$

95. Notice the product of $3\mathrm{10}-\mathrm{two}$ and $x^{\mathrm{two}}-5x-\mathrm{ii}.$

96. Find the product of $x^2+4$ and $x^3-1.$

97. Each side of a square measures $\mathrm{three}{\mathrm{10}}^{\mathrm{three}}$ units. Determine the area in terms of x.

98. Each border of a cube measures $\mathrm{ii}{x}^{\mathrm{ii}}$ units. Make up one's mind the volume in terms of x.

Role D: Dividing Polynomials

Split up.

99. $$\frac{125x^5{y}^2}{25x^{\mathrm{four}}{y}^{\mathrm{ii}}}$$

100. $$\frac{256x^{\mathrm{two}}{y}^3{z}^5}{64{\mathrm{ten}}^{2}y{z}^2}$$

101. $$\frac{20{\mathrm{ten}}^3-12x^{\mathrm{ii}}+\mathrm{four}x}{\mathrm{iv}\mathrm{10}}$$

102. $$\frac{15{\mathrm{10}}^4-75{\mathrm{ten}}^{\mathrm{iii}}+18x^2}{\mathrm{three}{\mathrm{ten}}^2}$$

103. $$\frac{12a^{2}b+28a{b}^{\mathrm{two}}-\mathrm{four}ab}{4ab}$$

104. $$\frac{-2a^4{b}^{\mathrm{iii}}+\mathrm{sixteen}{a}^2{b}^{\mathrm{ii}}+8a{b}^{\mathrm{iii}}}{\mathrm{ii}a{b}^2}$$

105. $$\frac{{\mathrm{10}}^3+x^2-3x+\mathrm{nine}}{\mathrm{ten}+\mathrm{iii}}$$

106. $$\frac{x^3-4{\mathrm{ten}}^{\mathrm{ii}}-\mathrm{nine}\mathrm{10}+20}{\mathrm{10}-\mathrm{v}}$$

107. $$\frac{\mathrm{vi}{x}^{\mathrm{three}}-11x^2+7x-\mathrm{half-dozen}}{\mathrm{ii}\mathrm{ten}-3}$$

108. $$\frac{9x^3-\mathrm{ix}{x}^{\mathrm{ii}}-\mathrm{10}+\mathrm{ane}}{3\mathrm{10}-\mathrm{i}}$$

109. $$\frac{16{\mathrm{ten}}^3+\mathrm{eight}{\mathrm{10}}^2-39\mathrm{10}+17}{4x-3}$$

110. $$\frac{12x^{\mathrm{iii}}-56{\mathrm{ten}}^2+55x+30}{\mathrm{ii}\mathrm{10}-5}$$

111. $$\frac{6x^4+13x^{\mathrm{iii}}-9x^2-x+6}{3x+\mathrm{two}}$$

112. $$\frac{25{\mathrm{ten}}^4-\mathrm{x}{\mathrm{10}}^3+11x^{\mathrm{two}}-\mathrm{vii}\mathrm{ten}+\mathrm{one}}{5\mathrm{10}-1}$$

113. $$\frac{20x^{\mathrm{iv}}+12{\mathrm{ten}}^3+9x^{\mathrm{ii}}+10x+5}{\mathrm{two}x+\mathrm{i}}$$

114. $$\frac{25x^{\mathrm{iv}}-45x^3-26{\mathrm{ten}}^2+36x-11}{\mathrm{v}x-\mathrm{two}}$$

115. $$\frac{3x^{\mathrm{iv}}+x^2-1}{x-\mathrm{two}}$$

116. $$\frac{x^{\mathrm{four}}+x-3}{\mathrm{ten}+\mathrm{iii}}$$

117. $$\frac{{\mathrm{ten}}^3-10}{x-2}$$

118. $$\frac{{\mathrm{ten}}^{\mathrm{three}}+\mathrm{xv}}{\mathrm{10}+3}$$

119. $$\frac{y^5+\mathrm{ane}}{y+\mathrm{one}}$$

120. $$\frac{y^6+1}{y+1}$$

121. $$\frac{x^4-\mathrm{iv}{x}^3+6x^{\mathrm{two}}-7\mathrm{10}-1}{x^2-x+2}$$

122. $$\frac{6{\mathrm{10}}^4+x^{\mathrm{three}}-\mathrm{ii}{x}^2+2x+\mathrm{iv}}{3x^2-x+\mathrm{one}}$$

123. $$\frac{\mathrm{two}{\mathrm{ten}}^3-7x^2+\mathrm{eight}\mathrm{ten}-\mathrm{iii}}{x^2-\mathrm{two}\mathrm{ten}+\mathrm{i}}$$

124. $$\frac{2x^4+\mathrm{three}{x}^3-\mathrm{half-dozen}{x}^2-\mathrm{iv}x+\mathrm{iii}}{x^2+x-3}$$

125. $$\frac{x^4+4x^{\mathrm{iii}}-2x^2-\mathrm{iv}x+\mathrm{one}}{{\mathrm{ten}}^2-\mathrm{i}}$$

126. $$\frac{x^{\mathrm{four}}+\mathrm{ten}-1}{x^2+\mathrm{ane}}$$

127. $$\frac{x^{\mathrm{iii}}+6{\mathrm{10}}^{2}y+\mathrm{iv}x{y}^2-y^3}{x+y}$$

128. $$\frac{\mathrm{ii}{x}^3-3x^{\mathrm{ii}}y+4x{y}^2-3y^3}{x-y}$$

129. $$\frac{8a^3-b^{\mathrm{iii}}}{2a-b}$$

130. $$\frac{a^3+27b^3}{a+3b}$$

131. Find the quotient of $10x^2-\mathrm{eleven}\mathrm{10}+\mathrm{three}$ and $2x-\mathrm{i}.$

132. Find the caliber of $12{\mathrm{10}}^2+\mathrm{10}-11$ and $3x-2.$

Answers

1. $-x^2-x+\mathrm{ane}$

3. $-y^3-\mathrm{three}{y}^2+y+\mathrm{five}$

5. $x^4-\mathrm{five}{\mathrm{ten}}^3-x^{\mathrm{two}}+6x+\mathrm{two}$

7. Trinomial; degree 2

9. Trinomial; degree iv

11. Binomial; degree iv

13. Quadratic, −ix

15. Linear, ii

17. Quadratic, 5

19. Quadratic, −2

21. $\mathrm{vii}{\mathrm{10}}^{\mathrm{two}}-\mathrm{nine}x+\mathrm{five}$

23. $\mathrm{ii}{x}^2+5x$

25. $9x^2+9\mathrm{ten}-3$

27. $\frac{1}{2}{x}^2-\frac{7}{\mathrm{iv}}x-\frac{1}{2}$

29. $-x^2{y}^2+2xy-1$

31. $2a^{\mathrm{two}}{b}^2+12ab-\mathrm{eight}$

33. $9{\mathrm{10}}^{2}y+6x{y}^2$

35. $7{\mathrm{10}}^{\mathrm{two}}{y}^2-3xy-6$

37. $-\frac{3}{2}{a}^2-\frac{5}{8}ab-\frac{\mathrm{ane}}{20}{b}^{\mathrm{ii}}$

39. $\mathrm{three}{x}^{2n}+2x^n-3$

41. $y^2+3y-\mathrm{vii}$

43. $SA=\mathrm{four}\mathit{\pi }{h}^{\mathrm{ii}}$

45. $-\mathrm{sixteen}{\mathrm{ten}}^3$

47. $10{\mathrm{ten}}^2-2x$

49. $14x^3-42x^2$

51. $-\mathrm{five}{y}^6+\mathrm{ten}{y}^5-15y^4$

53. $2{\mathrm{ten}}^{\mathrm{three}}y-14x^2{y}^2+2x{y}^3$

55. ${\mathrm{ten}}^{n+2}+{\mathrm{10}}^{n+1}+x^{\mathrm{north}}$

57. $x^2-x-20$

59. $\mathrm{vi}{x}^{\mathrm{two}}-\mathrm{xi}\mathrm{10}+\mathrm{three}$

61. $3x^4-\mathrm{sixteen}{x}^2{y}^2+\mathrm{v}{y}^4$

63. $\mathrm{ix}{x}^2-25$

65. $a^{\mathrm{four}}-b^4$

67. $12x^{\mathrm{iii}}-15x^2{y}^2-\mathrm{four}\mathrm{ten}y+5y^{\mathrm{three}}$

69. ${\mathrm{ten}}^{\mathrm{three}}-\mathrm{viii}{x}^2+23\mathrm{10}-40$

71. $2{\mathrm{10}}^4+\mathrm{eleven}{\mathrm{10}}^3-24x^2-4x+1$

73. $x^2+16x+64$

75. $4x^2-\mathrm{twenty}x+25$

77. $a^2-\mathrm{six}ab+9b^2$

79. $x^4+4x^2{y}^2+4y^4$

81. $a^4-2a^3+\mathrm{eleven}a-10a+25$

83. $x^3-9x^2+27x-27$

85. $27x^3+27{\mathrm{ten}}^2+9x+1$

87. $x^4+8x^{\mathrm{iii}}+24x^2+32\mathrm{ten}+\mathrm{xvi}$

89. $16x^{\mathrm{iv}}-32x^3+24x^2-8\mathrm{ten}+1$

91. ${\mathrm{10}}^{4\mathrm{northward}}-25$

93. $x^{\mathrm{four}\mathrm{north}}-\mathrm{ii}{x}^{\mathrm{ii}\mathrm{northward}}+1$

95. $3x^3-17{\mathrm{ten}}^{\mathrm{ii}}+\mathrm{iv}x+4$

97. $9{\mathrm{ten}}^{\mathrm{half\; dozen}}$ foursquare units

99. $\mathrm{v}x$

101. $\mathrm{five}{x}^{\mathrm{ii}}-\mathrm{iii}x+1$

103. $\mathrm{three}a+7b-1$

105. ${\mathrm{10}}^2-\mathrm{ii}x+3$

107. $3x^{\mathrm{ii}}-x+\mathrm{two}$

109. $\mathrm{four}{x}^{\mathrm{two}}+5x-\mathrm{vi}-\frac{1}{4x-3}$

111. $2x^3+3x^{\mathrm{ii}}-5x+3$

113. $10{\mathrm{10}}^{\mathrm{iii}}+x^2+4\mathrm{10}+3+\frac{2}{\mathrm{two}x+1}$

115. $3x^{\mathrm{three}}+6x^{\mathrm{two}}+\mathrm{thirteen}x+26+\frac{51}{x-2}$

117. ${\mathrm{ten}}^{\mathrm{ii}}+\mathrm{two}x+4-\frac{2}{x-\mathrm{two}}$

119. $y^4-y^{\mathrm{three}}+y^2-y+\mathrm{i}$

121. $$x^{\mathrm{ii}}-\mathrm{three}x+\mathrm{ane}-\frac{3}{x^2-x+2}$$

123. $\mathrm{ii}\mathrm{10}-\mathrm{three}$

125. $x^{\mathrm{two}}+4\mathrm{10}-1$

127. $x^2+\mathrm{v}xy-y^2$

129. $\mathrm{iv}{a}^2+\mathrm{two}ab+b^2$

131. $5x-3$

Source: https://saylordotorg.github.io/text_intermediate-algebra/s04-06-polynomials-and-their-operatio.html

Posted by: alfordtheyetage.blogspot.com

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