Answer:
![(1)/(24) (x {}^(2) + 4)(x + 3)(x - 4) {}^(2)](https://img.qammunity.org/2023/formulas/mathematics/high-school/drt0necnh3s29h03d0ml4j4v3g04fx631m.png)
Explanation:
The zeroes are
2i,-3, and 4. So in factored form,
we have
![(x - 2i)(x + 3)(x - 4)](https://img.qammunity.org/2023/formulas/mathematics/high-school/2gco98ijxv5icvma9y9nhomnu3u9hg5faq.png)
4 has a multiplicity of 2, so we have
![(x - 2i)(x + 3)(x - 4) {}^(2)](https://img.qammunity.org/2023/formulas/mathematics/high-school/utvojevsljvef93i16hf33n5ephnv29tpf.png)
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)](https://img.qammunity.org/2023/formulas/mathematics/high-school/nv7llrjwx1pi7qihp0gvrtac01etxl4b1s.png)
So let expand this out
![( {x}^(2) + 4)(x - 4) {}^(2) (x + 3)](https://img.qammunity.org/2023/formulas/mathematics/high-school/fyvg1m4j9v3jicud875kvrehmiodtiddpd.png)
![( {x}^(2) + 4)(x + 3)(x - 4) {}^(2)](https://img.qammunity.org/2023/formulas/mathematics/high-school/8h5gihomb5ekj482dlna2jhdzse76prk17.png)
![( {x}^(3) + 3 {x}^(2) + 4x + 12)( {x}^(2) - 8x + 16)](https://img.qammunity.org/2023/formulas/mathematics/high-school/9vsungx188jrngyy02sao5osvstysc0st1.png)
To avoid confusion, just look at the leading terms and the constant terms.
![{x}^(3) * {x}^(2) = {x}^(5)](https://img.qammunity.org/2023/formulas/mathematics/high-school/40ji24lkju7d4igpucxtun2xuebcsdgr4x.png)
So this is a fifth degree polynomial.
However, if we plug in 0, we get
![12 * 16 = 192](https://img.qammunity.org/2023/formulas/mathematics/high-school/rf5ix5aurrqy8svyp5uxagqac99rumymja.png)
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)](https://img.qammunity.org/2023/formulas/mathematics/high-school/xu5ihethlvif2yi3amncz38lxpwweoylhv.png)
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.