Find a polynomial of degree 3 with real coefficients and zeros of -3,-1, and 4, for which f(-2)=-18.

We want to find a polynomial of degree 3 with the given roots and that meets the given boundary condition.

We will find the polynomial:

f(x) = -3*(x + 3)*(x + 1)*(x - 4)

We know that for a polynomial of degree N with roots {x₁, ..., xₙ} and a leading coefficient a, it can be written as:

p(x) = a*(x - x₁)*...*(x - xₙ)

Now we know that we have a polynomial of degree 3 and that the roots are: {-3, -1, 4}

Then we can write:

f(x) = a*(x - (-3))*(x - (-1))*(x - 4)

f(x) = a*(x + 3)*(x + 1)*(x - 4)

Now we also know that:

f(-2) = -18

We can use this to find the leading coefficient.

Then we can write:

f(-2) = -18 = a*(-2 + 3)*(-2 + 1)*(-2 - 4) = a*6

         -18/6 = a = -3

Then our polynomial is:

f(x) = -3*(x + 3)*(x + 1)*(x - 4)

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Step-by-step explanation:

hello :

f(x) = a(x+3)(x+1)(x-4)

now calculate a  if f(-2)=18   means when x=-2   y=18

so : a(-2+3)(-2+1)(-2-4)=18

6a=18

a=18/6=3

conclusion : f(x) = 3(x+3)(x+1)(x-4)

8(-10x+3.5y-7)

8×-10x + 8×3.5y + 8×-7

-80x + 28y - 56

therefore

-80x+28y-56 is the correct answer

the equation cos(35) =a/25 can be used to find the length of bc what is the length of bc? round to the nearest tenth

step-by-step explanation: