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

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:

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80° is the measure of gh

step-by-step explanation:

circle f is congruent to circle j

we have two circle with center f and j.

both circles are congruent. the radius of the circle would be equal.

see the attachment for the circle.

we are given ∠efd ≅ ∠gjh

therefore, arc (ed) = arc (gh)     (∴ circle f is congruent to circle j)

but ∠dfe=80°=arc (ed)

therefore, arc (gh) = 80°

thus, 80° is the measure of gh

look at photo