If f(x) is a polynomial with real coefficients and zeros of 4(multiplicity 2 ), -1(multiplicity2 ),-1+8i , and 2-2i, what is the minimum degree of f(x) ?

The minimum degree of polynomial f(x) = 6

Step-by-step explanation:

Multiplicity for a polynomial equation is the number of times the polynomial has that particular root.

The roots of this polynomial is given as

4(multiplicity 2 ), -1(multiplicity2 ), (-1+8i), and (2-2i)

Hence, the true roots of this particular polynomial are 4, 4, -1, -1, (-1+8i), and (2-2i)

Making a total of 6 roots.

The number of roots of a polynomial is the minimum degree of such a polynomial.

Hence, the minimum degree of this polynomial is 6.

Hope this Helps!!!

step-by-step explanation:

equation

you are asked to write the equation

  expression = 0

using the given expression. when you do that, you have

  2x^2-2x-12 = 0

factoring

you can see that each of the coefficients is even, so a factor of 2 can be removed immediately.

  2(x^2 -x -6) = 0

if you multiply the binomials (x+a)(x+b), you get

  (x +a)(x +b) = x^2 +(a+b)x +ab

this tells you the coefficient of x is the sum of two factors of the constant term. here, that means -1 is the sum of two factors of -6. we want to know those factors so we can factor the trinomial.

  -6 = -1×6 = -2×3 = -3×2 = -6×1

the factor pair that sums to -1 is {-3, 2}, so our factorization is

  2(x -3)(x +2) = 0

solutions

the solutions to this equation are the values of x that make the product zero. a product is zero only when only of the factors of the product is zero. here, that means

  x - 3 = 0

  x = 3 . . . add 3 to both sides of the equation

or

  x +2 = 0

  x = -2 . . . add -2 to both sides of the equation

the solutions to the equation are x=3 or x=-2. both of these values of x satisfy the equation—they make it be true.

answer:

sometimes

*i think