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Artyom0805 [142]
3 years ago
12

HELP PLEASE!!!! ASAP!!! Describe, with examples of your own, when you would use long division and synthetic division and how to

check polynomial division.


Mathematics
1 answer:
Elan Coil [88]3 years ago
5 0

Synthetic division is a shorthand, or shortcut, method of polynomial division in the special case of dividing by a linear factor -- and it only works in this case. Synthetic division is generally used, however, not for dividing out factors but for finding zeroes (or roots) of polynomials. More about this later.

If you are given, say, the polynomial equation y = x2 + 5x + 6, you can factor the polynomial as y = (x + 3)(x + 2). Then you can find the zeroes of y by setting each factor equal to zero and solving. You will find that x = –2 and x = –3 are the two zeroes of y.

You can, however, also work backwards from the zeroes to find the originating polynomial. For instance, if you are given that x = –2 and x = –3 are the zeroes of a quadratic, then you know that x + 2 = 0, so x + 2 is a factor, and x + 3 = 0, so x + 3 is a factor. Therefore, you know that the quadratic must be of the form y = a(x + 3)(x + 2).

(The extra number "a" in that last sentence is in there because, when you are working backwards from the zeroes, you don't know toward which quadratic you're working. For any non-zero value of "a", your quadratic will still have the same zeroes. But the issue of the value of "a" is just a technical consideration; as long as you see the relationship between the zeroes and the factors, that's all you really need to know for this lesson.)

Anyway, the above is a long-winded way of saying that, if x – n is a factor, then x = n is a zero, and if x = n is a zero, then x – n is a factor. And this is the fact you use when you do synthetic division.

Let's look again at the quadratic from above: y = x2 + 5x + 6. From the Rational Roots Test, you know that ± 1, 2, 3, and 6 are possible zeroes of the quadratic. (And, from the factoring above, you know that the zeroes are, in fact, –3 and –2.) How would you use synthetic division to check the potential zeroes? Well, think about how long polynomial divison works. If we guess that x = 1 is a zero, then this means that x – 1 is a factor of the quadratic. And if it's a factor, then it will divide out evenly; that is, if we divide x2 + 5x + 6 by x – 1, we would get a zero remainder. Let's check:

As expected (since we know that x – 1 is not a factor), we got a non-zero remainder. What does this look like in synthetic division? Copyright © Elizabeth Stapel 2002-2011 All Rights Reserved

First, write the coefficients ONLY inside an upside-down division symbol:

 

 

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dmitriy555 [2]

The points in the brackets are on the graph are (0, 2) and (6, 10)

<h3>How to determine if the points in the brackets are on the graph?</h3>

The equation of the line is given as

4x - 3y = -6

The points are given as:

{(0,2),(1,3),(4,7),(6,10)}

Rewrite as

(x, y) = {(0,2),(1,3),(4,7),(6,10)}

Next, we substitute the x and y values in the equation 4x - 3y = -6

So, we have

(0, 2):

4(0) - 3(2) = -6

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4(1) - 3(3) = -6

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(4, 7):

4(4) - 3(7) = -6

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(6, 10):

4(6) - 3(10) = -6

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Step-by-step explanation:

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2. Solve substitution: -1

3. Compare -1 to calculated answers when c=1

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Solve the following differential equation using using characteristic equation using Laplace Transform i. ii y" +y sin 2t, y(0) 2
kifflom [539]

Answer:

The solution of the differential equation is y(t)= - \frac{1}{3} Sin(2t)+2 Cos(t)+\frac{5}{3} Sin(t)

Step-by-step explanation:

The differential equation is given by: y" + y = Sin(2t)

<u>i) Using characteristic equation:</u>

The characteristic equation method assumes that y(t)=e^{rt}, where "r" is a constant.

We find the solution of the homogeneus differential equation:

y" + y = 0

y'=re^{rt}

y"=r^{2}e^{rt}

r^{2}e^{rt}+e^{rt}=0

(r^{2}+1)e^{rt}=0

As e^{rt} could never be zero, the term (r²+1) must be zero:

(r²+1)=0

r=±i

The solution of the homogeneus differential equation is:

y(t)_{h}=c_{1}e^{it}+c_{2}e^{-it}

Using Euler's formula:

y(t)_{h}=c_{1}[Sin(t)+iCos(t)]+c_{2}[Sin(t)-iCos(t)]

y(t)_{h}=(c_{1}+c_{2})Sin(t)+(c_{1}-c_{2})iCos(t)

y(t)_{h}=C_{1}Sin(t)+C_{2}Cos(t)

The particular solution of the differential equation is given by:

y(t)_{p}=ASin(2t)+BCos(2t)

y'(t)_{p}=2ACos(2t)-2BSin(2t)

y''(t)_{p}=-4ASin(2t)-4BCos(2t)

So we use these derivatives in the differential equation:

-4ASin(2t)-4BCos(2t)+ASin(2t)+BCos(2t)=Sin(2t)

-3ASin(2t)-3BCos(2t)=Sin(2t)

As there is not a term for Cos(2t), B is equal to 0.

So the value A=-1/3

The solution is the sum of the particular function and the homogeneous function:

y(t)= - \frac{1}{3} Sin(2t) + C_{1} Sin(t) + C_{2} Cos(t)

Using the initial conditions we can check that C1=5/3 and C2=2

<u>ii) Using Laplace Transform:</u>

To solve the differential equation we use the Laplace transformation in both members:

ℒ[y" + y]=ℒ[Sin(2t)]

ℒ[y"]+ℒ[y]=ℒ[Sin(2t)]  

By using the Table of Laplace Transform we get:

ℒ[y"]=s²·ℒ[y]-s·y(0)-y'(0)=s²·Y(s) -2s-1

ℒ[y]=Y(s)

ℒ[Sin(2t)]=\frac{2}{(s^{2}+4)}

We replace the previous data in the equation:

s²·Y(s) -2s-1+Y(s) =\frac{2}{(s^{2}+4)}

(s²+1)·Y(s)-2s-1=\frac{2}{(s^{2}+4)}

(s²+1)·Y(s)=\frac{2}{(s^{2}+4)}+2s+1=\frac{2+2s(s^{2}+4)+s^{2}+4}{(s^{2}+4)}

Y(s)=\frac{2+2s(s^{2}+4)+s^{2}+4}{(s^{2}+4)(s^{2}+1)}

Y(s)=\frac{2s^{3}+s^{2}+8s+6}{(s^{2}+4)(s^{2}+1)}

Using partial franction method:

\frac{2s^{3}+s^{2}+8s+6}{(s^{2}+4)(s^{2}+1)}=\frac{As+B}{s^{2}+4} +\frac{Cs+D}{s^{2}+1}

2s^{3}+s^{2}+8s+6=(As+B)(s²+1)+(Cs+D)(s²+4)

2s^{3}+s^{2}+8s+6=s³(A+C)+s²(B+D)+s(A+4C)+(B+4D)

We solve the equation system:

A+C=2

B+D=1

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The solutions are:

A=0 ; B= -2/3 ; C=2 ; D=5/3

So,

Y(s)=\frac{-\frac{2}{3} }{s^{2}+4} +\frac{2s+\frac{5}{3} }{s^{2}+1}

Y(s)=-\frac{1}{3} \frac{2}{s^{2}+4} +2\frac{s }{s^{2}+1}+\frac{5}{3}\frac{1}{s^{2}+1}

By using the inverse of the Laplace transform:

ℒ⁻¹[Y(s)]=ℒ⁻¹[-\frac{1}{3} \frac{2}{s^{2}+4}]-ℒ⁻¹[2\frac{s }{s^{2}+1}]+ℒ⁻¹[\frac{5}{3}\frac{1}{s^{2}+1}]

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3 0
3 years ago
Let A = {1, 2, 3, 4, 5} and B = {2, 4}. What is A ∩ B?
tatyana61 [14]

Answer:

A ∩ B = {2, 4}

Step-by-step explanation:

Definition: The intersection A ∩ B of two sets A and B is the set that contains all elements of A that also belong to B.

For example:

If x∈A and x∈B

then;

x∈ A ∩ B

Given the sets:

A = {1, 2, 3, 4, 5} and B = {2, 4}

We have to find  A ∩ B.

By definition we have;

A ∩ B = {2, 4}

Therefore, the value of  A ∩ B is, {2, 4}

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