88.1k views
9 votes
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).

User Prcu
by
5.0k points

1 Answer

4 votes

Answer:


(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.

pre calculus Write the factored form of a polynomial function with a degree of five-example-1
User Fsimon
by
5.7k points