Finding Polynomials with Specific Properties: How Do I Do That?

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In summary, Fred provides a summary of the content and provides equations for p1 and p1'. He also provides equations for p2 and p2'. He states that if the polynomials are of highest degree 3, they may be written as p_{1}=a_{1}x^{3}+b_{1}x^{2}+c_{1}x+d_{1} and p_{2}=a_{2}x^{3}+b_{2}x^{2}+c_{2}x+d_{2}. He uses a matrix system to solve these equations and finds that d_1=d_2 and c_1=c_2. He
  • #1
Mathman23
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Hi

I got a Linear Algebra question.

I'm suppose to find two polynomials p1 and p2 both of highest degree 3, and which satisfies the following:

p1(-1) = 1

p1'(-1) = 0

p2(1) = 3

p2'(1) = 0

p1(0) = p2(0)

p1'(0) = p2'(0)


I hope that there is somebody out there who can explain to me how I do that ?

Thanks in advance.

Fred
 
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  • #2
Hint:
If your polynomials are of highest degree 3, they may be written as:
[tex]p_{1}=a_{1}x^{3}+b_{1}x^{2}+c_{1}x+d_{1}[/tex]
[tex]p_{2}=a_{2}x^{3}+b_{2}x^{2}+c_{2}x+d_{2}[/tex]

The letters are the constants you must determine with the aid of the given relations (you will gain a matrix system you''ll need to solve)
 
  • #3
Thanks for Your answer.

If I understand You correctly.

I take these two polynomials p1 and p1'.

I then insert the terms into the polynomial equations? So I get the following

p1 = a1*(-1)^3 + b2*(-1)^2 + c1*(-1) +d = 1

p1' = 3*a1*(-1)^2 + 2*b1*(-1) +c1 = 0

I use a matrix system to solve these two equations so they satisfy the given conditions?

And then find the variables for p2 ?

Sincerely
Fred
 
  • #4
you have far more than just those two equations. for instance the last two tell you that d_1=d_2 and c_1=c_2
 
  • #5
You mean I have these equations:

p1 = a1*(-1)^3 + b1*(-1)^2 + c1*(-1) +d1 = 1

p1' = 3*a2*(-1)^2 + 2*b2*(-1) +c2 = 0

p2 = a1*(1) ^3 +b1 * (1) ^2 + c1*(1) + d1 = 3

p2' = 3*a2*(1)^2 + 2*b2*(-1) +c2 = 0

a1*(0)^3 + b1*(0)^2 + c1*(0) +d1 = a1*(0) ^3 +b1 * (0) ^2 + c1*(0) + d1

3*a2*(0)^2 + 2*b2*(0) +c2 = 3*a2*(0)^2 + 2*b2*(0) +c2

and solve these as one system of equations ?

/Fred
 
  • #6
not quite since you appear to have several mistakes in there, let us use arildno's notation properly

1=-a_1+b_1-c_1+d_1
from (p_1(-1)=1)

0=3a_1-2b_1+c_1
from (p_1'(-1)=0)

3=a_2+b_2+c_2+d_2
from (p_2(1)=3)

0=3a_2+2b_2+c_2
from (p_2'(1)=0)

d_1=d_2
from (p_1(0)=p_2(0))

c_1=d_1
from (p_1'(0)=p_2'(0))

note you have 8 unkowns and 6 equations
 
  • #7
okay Thank You.

Fred
 
  • #8
I get the following solution values:

a1, a2 = -1/2

b1, b2 = 3/2

c1,c2 = 0

d1, d2 = 2

Can anybody tell me if they are correct ?

Thanks again.

Sincerely

Fred
 
  • #9
But you've all these identities that allow you to check your answer by yourself, you don't need us to verify it.
 

What is a polynomial?

A polynomial is a mathematical expression that consists of variables and coefficients combined using arithmetic operations such as addition, subtraction, multiplication, and non-negative integer exponents.

How do you determine the degree of a polynomial?

The degree of a polynomial is determined by looking at the highest exponent of the variable in the expression. For example, in the polynomial 3x^2 + 5x + 1, the degree is 2 because the highest exponent is 2.

What is the process for factoring a polynomial?

The process for factoring a polynomial involves finding the common factors of each term and using the distributive property to rewrite the expression in a simplified form. This process may also involve using methods such as grouping or the difference of squares.

How do you solve a polynomial equation?

To solve a polynomial equation, you can use methods such as factoring, the quadratic formula, or completing the square. These methods allow you to find the values of the variable that make the equation true.

What is the difference between a monomial, binomial, and trinomial?

A monomial is a polynomial with only one term, a binomial is a polynomial with two terms, and a trinomial is a polynomial with three terms. The number of terms in a polynomial is important because it helps determine the degree of the polynomial.

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