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

Answer:

$\frac{1}{24} (x {}^{2} + 4)(x + 3)(x - 4) {}^{2}$

Step-by-step 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.

${x}^{3} \times {x}^{2} = {x}^{5}$

So this is a fifth degree polynomial.

However, if we plug in 0, we get

$12 \times 16 = 192$

So how do we get this term to 8.

we divide the whole polynomial by 24.

or multiply it by 24

$\frac{1}{24} (x + 2i)(x - 2i)(x + 3)(x - 4) {}^{2}$

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.