Question

Write a polynomial f (x) that satisfies the given conditions. Polynomial of lowest degree with zeros of -3 (multiplicity 3) 2(multiplicity 1)and with f(0)=270 .

Answer 1

Answer:

f(x) = -5(x+3)(x+3)(x+3)(x-2)

Step-by-step explanation:

If the polynomial has three zeros in -3 and one zero in 2, the lowest degree we need is four, as the polynomial has these four zeros, so we can use a generic form of a fourth degree polynomial:

y = a(x-x1)(x-x2)(x-x3)(x-x4)

Where x1, x2, x3 and x4 are the zeros, so we have that:

y = a(x+3)(x+3)(x+3)(x-2)

Now, to find the value of the constant 'a', we need to use the information that f(0) = 270:

270 = a*3*3*3*(-2)

-54a = 270

a = -5

So the polynomial is:

f(x) = -5(x+3)(x+3)(x+3)(x-2)