Building A Polynomial Function

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mathdad

mathdad

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Use f(x) = a(x - r_1)(x - r_2)(x - r_3), where r represents the zero values to make a polynomial function.

Zeros: -1, multiplicity 1; 3, multiplicity 2; degree 3

Solution:

f(x) = a(x + 1)(x - 3)^2

f(x) = a(x + 1)(x^2 - 6x + 9)

f(x) = a(x^3 - 6x^2 + 9x + x^2 - 6x + 9)

f(x) = ax^3 - 4x^2 + 3x + 9)

I see that a is 1 for this function.

f(x) = x^3 - 4x^2 + 3x + 9

Yes?
 
I would go back to:

[MATH]f(x)=a(x+1)(x-3)^2[/MATH]
Now you can choose any real non-zero value for \(a\), and you will have a member of the family of polynomials meeting the requirements of the problem.

You have a two minor errors that I see in your work expanding the function...can you spot them?
 
Check your arithmetic; one number is wrong. You also left out a parenthesis.

Also, it is not really a matter of "seeing that a is 1", but of arbitrarily choosing to make a=1 for simplicity. Nothing in the problem dictates what a should be, so you could multiply your answer by any number except 0, and the result would be a valid answer.
 
MarkFL said:
I would go back to:

[MATH]f(x)=a(x+1)(x-3)^2[/MATH]
Now you can choose any real non-zero value for \(a\), and you will have a member of the family of polynomials meeting the requirements of the problem.

You have a two minor errors that I see in your work expanding the function...can you spot them?
Click to expand...

What minor errors?
 
Dr.Peterson said:
Check your arithmetic; one number is wrong. You also left out a parenthesis.

Also, it is not really a matter of "seeing that a is 1", but of arbitrarily choosing to make a=1 for simplicity. Nothing in the problem dictates what a should be, so you could multiply your answer by any number except 0, and the result would be a valid answer.
Click to expand...

Understood. At least, I get the gist of it.
 
mathdad said:
The value of a could be any number as long as it is not 0.
Click to expand...

No, when you go from here:

f(x) = a(x^3 - 6x^2 + 9x + x^2 - 6x + 9)

to here:

f(x) = ax^3 - 4x^2 + 3x + 9)

You are missing the opening bracket that should go after the constant a, and -4x^2 should be -5x^2.
 
MarkFL said:
No, when you go from here:

f(x) = a(x^3 - 6x^2 + 9x + x^2 - 6x + 9)

to here:

f(x) = ax^3 - 4x^2 + 3x + 9)

You are missing the opening bracket that should go after the constant a, and -4x^2 should be -5x^2.
Click to expand...

You are right.
 
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