Find the polynomial f(x) that has the roots of -3, 5 of multiplicity 2. Explain how you would verify the zeros of f(x)
It is easy! ¿What are the roots? They are the values of X that makes f ( x ) = 0. Then, you will take advantage of the fact that if one factor of a product becomes cero, then all the products become cero. For example (a - x)(b + x) If the first (a - x) or the second (b + x) factor becomes cero, all the product is cero, is it ok?
So you wouldhave (x+3) (x-5)
Then...you could use that knowledge to build your function. You want a root of x = -3 ? Ok, then we need a factor that turns into cero when x = -3. It is (x + 3) , so x = -3 , then (-3+3) = 0. You do the same with the other root x = 5, so...(x+3)(x-5)
Yes you do
Ok what about the multiplicty partof it
Is redundant...multiplicity 2 is to have a x^2 as the "max" exponent. If you distribute that you will arrive at some expression of the form ax^2 + bx + c and that is the form of all the polinomics of multiplicity two, is just a quadratic.
So what would be the polynomial im so confused lol
The polinomial is what you have, (x+3)(x-5) but you have to do the distribute. x^2 -5x + 3x -15 = x^2 -2x - 15 Then you have a quadratic equatin. And a quadratic equation is a polinomial of multiplicity 2.
A quadratic equation is a KIND of polinomial of multiplicity 2.
x^3 + x +1 And this is a polinomial of multiplicity 3. (The higher exponent is what matters)
Ok thanks and its says how would you verify the zeros is that through synthetic division?
Read the first three paragraphs; http://www.purplemath.com/modules/synthdiv.htm
I don´t know exactly what are you meaning with "synthetic division" because I learned maths in spanish and some words or "titles of the concepts" escape me :P. I guess the first three paragraphs will help us.
how would you check if the zeros are right
Give X the values x = 5 and x = -3
would you subsitute it into what the equation was that you got
But is has no sense...because you have just build the expression so that 5 and -3 are roots!
f(x) = (x-5)(x+3) or which is the same f(x) = x^2 -2x - 15 Let x be -3 and compute, then do the same with 5.
Did I convince you? Is it ok?
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