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## The polynomial x3 + 8 is equal to

Find an answer to your question โ โThe polynomial x3 + 8 is equal to …โ in ๐ Mathematics if you’re in doubt about the correctness of the answers or there’s no answer, then try to use the smart search and find answers to the similar questions.

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Partial fraction decomposition online calculator – Our online calculator finds partial fraction decomposition of any (proper, improper) rational fraction. If initial fraction is the improper one, (i.e. order of polynomial in the numerator greaters of equals to the order of polynomial in the denominator)…Two monomials are equal if they have the same variable and the same degree. Example: 3×2 and -5×2 ; ยฝx4 and 2โ3×4. The sum of unequal monomials is called a polynomial. The general form of the polynomial with only one variable p(x)=anxn+an-1xn-1+…+a1x1 +a0 whereClassifying Polynomials: Polynomials can be classified two different ways – by the number of terms and by their degree. It is 0 degree because x0=1. So technically, 5 could be written as 5×0. 3x2y5 Since both variables are part of the same term, we must add their exponents together to determine…

Polynomial Long Division – Polynomials will show up in pretty much every section of every chapter in the remainder of this material and so it is important that you understand them. Also, polynomials can consist of a single term as we see in the third and fifth example. We should probably discuss the final example a little more.What is Polynomial equation, Roots of an equation & Solution set of an equation ? State Fundamental theorem of algebra. Draw the Graph of the Polynomial when roots of the polynomial is given . Role of domain and solution with examples :: Refer-Cengage 2.4 :: Refer-Miscellaneous.Find an answer to your question โ "The polynomial x3 + 8 is equal to" in Mathematics if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.

Classifying Polynomials – Online polynomial roots calculator finds the roots of any polynomial and creates a graph of the resulting polynomial. For Polynomials of degree less than 5, the exact value of the roots are returned. Calculator displays the work process and the detailed explanation.Multiple Choice Questions on Class 9 Polynomials are important for the upcoming annual exams.Problem Statement: EE Board April 1996, EE Board March 1998. What is the remainder when the Polynomial x^3+4x^2-3x+8 is divided by x-5? The remainder is equal to 218.

**Parametric Representation of the Solution Set to a Linear Equation** – Welcome to a lesson on the parametric representation of a solution set for a linear equation.

We should already be familiar

with linear equations in the form of, let's say,

Y equals 2/3 X minus two, which is a slope-intercept form of a line, as well as two X plus three Y equals 12, which is a standard form

of a linear equation. But to define a linear

equation more formally, a linear equation with N

unknowns, X sub one, X sub two, all the way through X sub N, is an equation that can

be put in standard form, or this form here. Where X sub one, X sub two,

all the way through X sub N, and B are real numbers, the As

are called the coefficients, and B is called the constant term. Notice the As are being

multiplied by the unknowns, and B is the number without the variable. A solution to a linear equation

is a set of all real numbers C sub one, C sub two, and so on, such that X sub one equals C sub one, X sub two equals C sub two, X

sub three equals C sub three, and so on, satisfy the

given linear equation. The set of all solutions

is called the solution set, and to describe a solution

set for a linear equation, parametric equations are often used. A parametric equation is a

way of defining a relation using parameters or other variables. And then finally, a linear

equation with N variables has N minus one free variables, or variables that can

take on any real value, which will assign the parameters

or their variables, too. Let's take a look at an example. We want to solve the linear equation and use a parametric

representation for the solution. We have two times X sub one minus four times X sub two equals 12. So the first step is to

solve this for X sub one or X sub two. Let's go ahead and solve

this for X sub one. So our first step is to

add four times X sub two to both sides of the equation, that would give us two

times X sub one equals 12 plus four times X sub two. Now let's go and divide everything by two. So now we have X sub

one equals six plus two times X sub two. Because we have two

unknowns in one equation, we have one free variable,

so we should recognize that we are going to have an

infinite number of solutions. And because of this, we'll

introduce a third variable, or a parameter, which will

assign to our free variable. Let's let X sub two be our free variable. Let's also let T be any real number. And since we're letting X

sub two be our free variable, to represent our solution parametrically, we're going to let X sub two equal T which means X sub one would have to be six plus two times T,

replacing X sub two with T. So this is one way to

represent the solution to our linear equation parametrically. And I say one way because

the parametric representation of a solution is not unique. So just to illustrate this,

let's say we let X sub one be our free variable. And again, we let T be any real number. But in this case to form our

parametric representation, let's let X sub one equal T and therefore, using this

equation here the second equation would be T equals six

plus two times X sub two. And now we'd have to solve

this equation for X sub two, let's go ahead and do that. We'd subtract six on both sides, that would give us T minus six

equals two times X sub two, divide everything by two. We'd have 1/2 T minus

three equals X sub two. So we could represent the

same solution parametrically as X sub one equals T, and X sub two equals 1/2 T minus three. These are two different ways to represent the solution parametrically. Often in algebra class, we are asked to graph our linear equations,

and we could do the same here using the parametric equations rather than the original equation. Let's go ahead and do

that using the solution expressed using these equations here. So we'd make a column for T which can take on the

value of any real number, then we have X sub one,

and then we have X sub two. Well the first thing we should recognize is that whatever T value we select, it's also going to be

the value of X sub two, and then since we know

that X sub one equals six plus two times T, we can

substitute the same values for T to get the values for X sub one. Every X sub one and X sub

two forms an ordered pair, which represents one possible solution, which we can graph on

the coordinate plane. Again, thinking of the X

axis as the X sub one axis, and thinking of the Y axis

as the X sub two axis. If we plot these five points,

we can sketch our line, which is another way

to represent a solution to a linear equation with two unknowns. Let's take a look at a second example. Notice here we have one

equation with three unknowns. And three minus one is equal to two, so here we'll have two free variables which means we'll

introduce two parameters, or two new variables to represent the solution parametrically. Looks like it's going to be

easiest to solve this for X, so we'd add two Y to both sides

as well as subtract three Z, so we would have X equals

six and then plus two Y, minus three Z. So in this case, let's let Y

and Z be our free variables, so we'll introduce two

parameters, or two new variables, let's let S be any real number, and T be any real number. So we'll go ahead and let

Y equal S, and Z equal T, which means X is going to be equal to, using this equation here, we would have six plus two times Y, but

Y is S, so plus two S, minus three times Z, but Z is T. So this would be the

parametric representation of the solution to our linear equation. And because we have one

equation with three unknowns, the graph of this would

be three-dimensional, or would be a plane in space. Let's go and show how I would do this using the parametric equations. Again, we'd start with a column

for S and a column for T, which we'd let be any real numbers that would be convenient to

perform the calculations, and since S equals Y, this column and this column are the same, and since Z equals T, this column and this column are the same. And then we're left with finding X by performing substitution for S and T, which has been done here. Notice here each solution

consists of a value for X, Y, and Z, so we

have ordered triples, which would be points in space, which as we see here would

be a plane graphed in space. Okay, I hope you found this helpful. .

**Ex: Find the Taylor Series of x^3** – We want to find the first several terms of the Taylor Series for

F of X equals X cubed centered at X equals negative three and then give the first few coefficients, C sub zero through C sub four.

The Taylor series for

any function is equal to the infinite series, where

we have the summation from N equals zero to infinity

of the Nth derivative of F evaluated at C, where the

series is centered at C, divided by N factorial

times the quantity X minus C raised to the power of N. If we look at the expansion of the power series here on the right, notice how the first two terms don't have a denominator because when N is zero, we have zero factorial,

which is equal to one, and when N is one, we have one factorial, which is equal to one. So the Taylor series

for our function F of X would be the summation from

N equals zero to infinity of the Nth derivative of F,

evaluated at negative three, divided by N factorial times the quantity X minus negative three or X

plus three raised to the Nth. So when N is zero, we just

have F of negative three. When N is one, we have F

prime of negative three times the quantity X plus

three raised to the first. And then when N is two,

we have F double prime of negative three divided by two factorial times the quantity X plus

three raised to the second, plus, when N is three,

we have F triple prime of negative three divided

by three factorial times the quantity X

plus three to the third. We need to find at least one more term to find the first five coefficients. So we'll have plus the

fourth derivative of F, evaluated at negative three

divided by four factorial times the quantity X plus

three to the fourth and so on. Now, for the next step, we'll find all these function values. F of negative three, F

prime of negative three, F double prime of negative three, F triple prime of negative

three, and finally, the fourth derivative of F

evaluated at negative three. So starting with F of X equals X cubed, well F prime of X would be

equal to three X squared. F double prime of X

would be equal to six X. F triple prime of X would be equal to six. And the fourth derivative

of F would be equal to zero. So F of negative three would

be equal to negative 27. F prime of negative three

would be equal to positive 27. F double prime of negative

three would be equal to negative 18. F triple prime of negative three is six. And the fourth derivative,

F negative three is equal to zero. Now I'll perform substitution. So our series is equal to,

we have F of negative three, that's negative 27, and then F prime of negative three is equal to 27, so we'd have plus 27 times

the quantity X plus three, plus here we have F double

prime of negative three divided by two factorial. Two factorial is equal

to two, so we'd have negative 18 divided by two,

which is negative nine. So let's write this as minus nine times the quantity X plus three squared. Next, we have plus F triple

prime of negative three is equal to six. Three factorial is also equal to six, so we'll just have one times

the quantity X plus three to the third. And the fourth derivative of

F evaluated at negative three is actually zero, so here

we just have plus zero times the quantity X plus

three to the fourth and so on. Of course, this product here is zero. Let's leave it in this form so we can find C sub zero through C sub four. C sub zero is negative 27. C sub one is positive 27. C sub two is negative nine. C sub three is positive one. And C sub four is zero. Again, we have negative 27, positive 27, negative nine, one, and zero. So because the original function is just a basic polynomial function, when using a Taylor series to

approximate this function, notice how the fourth

derivative and higher would always be zero, and

therefore in this case, the Taylor series is

not an infinite series. It actually ends after this product here. So we can go ahead and just

erase this part of the series. And because each term

in the remaining series would be zero, these

terms of the Taylor series are exactly equal to the

original function F of X. So if we multiplied this

out and simplified it, it would simplify

perfectly to just X cubed. Of course, this is a special case because the original function F of X

is a basic polynomial function. To verify this, let's go ahead and graph these terms of the Taylor series, which would be called

a 'Taylor polynomial' and compare it to the original function. The original function

is graphed here in blue, and our Taylor polynomial

is graphed here in red. As you can see graphically,

they're exactly the same. I hope you found this helpful. .

**Find the Vertical, Horizontal and Slant Asymptote** – .