contestada

Form a polynomial whose zeros and degree are given. ​Zeros: −3​, 3​, 2​; ​ degree: 3 Type a polynomial with integer coefficients and a leading coefficient of 1

Respuesta :

If a polynomial P(x) has a zero equal to a, then (x-a) is a factor of this polynomial. So if a polynomial has zeros a, b and c then it has we could write:

         P(x)=(x-a)(x-b)(x-c).

Here we can clearly see that a, making the left hand side 0 because of the factor (x-a), makes the left hand side 0 as well. This means that P(a)=0. This illustrates the discussion above.


Thus, substituting a, b, c with −3​, 3​, 2 we can write P(x)=(x+3)(x-3)(x-2).

We can expand the right hand side to have the polynomial in standard form:

[tex]P(x)=(x+3)(x-3)(x-2)=P(x)=[(x+3)(x-3)](x-2)[/tex] [tex]=(x^2-9)(x-2)=x^3-2x^2-9x+18[/tex]

We see that all conditions are satisfied.



Answer: [tex]P(x)=x^3-2x^2-9x+18[/tex]
facundo3141592

We want to write a polynomial given that we know the zeros and the leading coefficient of the polynomial.

The solution is: p(x) = x^3 - 2*x^2 - 9*x + 18

Let's see how to get that.

For a polynomial of degree N, with N zeros given by: {x₁, x₂, ..., xₙ} and a leading coefficient A, the polynomial can be written as:

p(x) = A*(x - x₁)*(x - x₂)*...*(x - xₙ)

Now, in this case, we know that the zeros are: {-3, 3, 2} and the leading coefficient is 1.

Then the polynomial will be:

p(x) = 1*(x - (-3))*(x - 3)*(x - 2)

Now we can expand this, so we get:

p(x) = (x + 3)*(x^2 - 5*x + 6)

p(x) = x^3 - 2*x^2 - 9*x + 18

So we just found the polynomial of degree 3, with the zeros {-3, 3, 2} and a leading coefficient of 1.

If you want to learn more, you can read:

https://brainly.com/question/15522547

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