Question

Polynomial of lowest degree with zeros of −4/3(multiplicity 2) and 1/2(multiplicity 1) and with =f016.

Answer 1

To construct a polynomial with specified zeros and multiplicity, we'll use the factored form of a polynomial. The zeros given are −4/3 with multiplicity 2 and 1/2 with multiplicity 1. The factorization would look like:

`(x+ (4)/(3)) ^(2) (x- (1)/(2))`

This represents a polynomial with roots at −4/3,−4/3,1/2. The multiplicity indicates how many times a particular root is repeated. In this case, −4/3 has a multiplicity of 2, meaning it's a double root, and 1/2 has a multiplicity of 1.

Now, to find the polynomial of the lowest degree, we multiply the factors:

`f(x) = (x+ (4)/(3)) ^(2) (x- (1)/(2))`​

To ensure that the constant term is 016, we can introduce a constant multiplier. Let a be this constant:

`f(x) = a (x+ (4)/(3)) ^(2) (x- (1)/(2))`

Now, plug in x=0 to find a:

`f(0) = a ((4)/(3)) ^(2) (-(1)/(2)) = 16`

Solving for a, we get a =
`-(3)/(32)`

Therefore, the polynomial of the lowest degree with the specified zeros and multiplicity, and with a constant term of 016, is:

`f(x) = -(3)/(32) (x+ (4)/(3)) ^(2) (x- (1)/(2))`