Question

pre calculus Write the factored form of a polynomial function with a degree of five that passes through (0,8) with zeros of 2i, -3, and 4 (multiplicity of 2).

Answer 1

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

Explanation:

The zeroes are

2i,-3, and 4. So in factored form,

we have

`(x - 2i)(x + 3)(x - 4)`

4 has a multiplicity of 2, so we have

`(x - 2i)(x + 3)(x - 4) {}^(2)`

Rember that if 2i, is a zero, -2i, it's conjugate, is as well.

So we have

`(x - 2i)(x + 2i)(x + 3)(x - 4) {}^(2)`

So let expand this out

`( {x}^(2) + 4)(x - 4) {}^(2) (x + 3)`

`( {x}^(2) + 4)(x + 3)(x - 4) {}^(2)`

`( {x}^(3) + 3 {x}^(2) + 4x + 12)( {x}^(2) - 8x + 16)`

To avoid confusion, just look at the leading terms and the constant terms.

`{x}^(3) * {x}^(2) = {x}^(5)`

So this is a fifth degree polynomial.

However, if we plug in 0, we get

`12 * 16 = 192`

So how do we get this term to 8.

we divide the whole polynomial by 24.

or multiply it by 24

`(1)/(24) (x + 2i)(x - 2i)(x + 3)(x - 4) {}^(2)`

Here is the graph above:

The zeroes at 4, bounces off.

Their is a zero at -3,

and we have a y intercept of 8.