1. Find a polynomial with a zero at 3 and a zero of multiplicity 2 at -1.
My answer: x^3 - x^2 - 5x - 3 = 0
2. Find the zeros and multiplicities of f(x) = x^5 - 16x^4 + 64x^3.
My answer: x = 0 of multiplicity 3 and x = 8 of multiplicity 2.
right on both.
To find a polynomial with a zero at 3 and a zero of multiplicity 2 at -1, you can use the fact that if a polynomial has a zero at a certain value, then the polynomial can be factored by that value.
First, let's consider the zero at 3. This means that (x - 3) is a factor of the polynomial.
Next, let's consider the zero of multiplicity 2 at -1. This means that (x + 1)^2 is a factor of the polynomial because it appears twice.
To find the complete polynomial, multiply these factors together:
(x - 3)(x + 1)^2 = (x - 3)(x + 1)(x + 1) = (x - 3)(x^2 + 2x + 1)
Expanding this expression gives:
(x - 3)(x^2 + 2x + 1) = x^3 - x^2 + 2x^2 - 2x + x - 3 = x^3 + x^2 - x - 3
Thus, a polynomial with a zero at 3 and a zero of multiplicity 2 at -1 is x^3 + x^2 - x - 3.
Now, let's move on to the second question.
To find the zeros and multiplicities of f(x) = x^5 - 16x^4 + 64x^3, we need to set the polynomial equal to zero and factor it.
x^5 - 16x^4 + 64x^3 = 0
First, we notice that there is a common factor of x^3 in each term:
x^3(x^2 - 16x + 64) = 0
Next, we can solve the quadratic equation x^2 - 16x + 64 = 0 by factoring or using the quadratic formula. We get:
(x - 8)^2 = 0
This gives us a zero of multiplicity 2 at x = 8.
And since we have an additional factor of x^3, we have a zero at x = 0 with multiplicity 3.
So, the zeros and multiplicities of f(x) = x^5 - 16x^4 + 64x^3 are:
x = 0 (multiplicity 3) and x = 8 (multiplicity 2).