The zeros of the equation is given below as
![-2,4,5](https://img.qammunity.org/2023/formulas/mathematics/college/zfbj7ksm8gisdi6oceyc7oy87vtc2xi9yk.png)
Step 1:
Express the zeros as factors of the polynomial P(x), we will have
![\begin{gathered} x=-2 \\ (x+2) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/p4qn4exvwnt8rrnz0tayzje8ab7gmzui8b.png)
![\begin{gathered} x=4 \\ (x-4) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/cmity9nee29pb0ju69goj8huzlu1rhr3yg.png)
![\begin{gathered} x=5 \\ (x-5) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/omy4z0ax08birftr6o02ld2d8k6z5wduoj.png)
Step 2:
Multiply the first two factors by expanding them
![\begin{gathered} (x+2)(x-4) \\ x(x-4)+2(x-4) \\ x^2-4x+2x-8 \\ (x^2-2x-8) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/5zoqoymujc8m5hr33lbdj0gm3rxamk8pb7.png)
Step 3 :
Multiply the expression gotten in step 2 by (x-5) from step 1
![\begin{gathered} (x^2-2x-8)(x-5) \\ P(x)=x(x^2-2x-8)-5(x^2-2x-8) \\ P(x)=x^3-2x^2-8x-5x^2+10x+40 \\ P(x)=x^3-2x^2-5x^2-8x+10x+40 \\ P(x)=x^3-7x^2+2x+40 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/x55eauvzgpkxtygq226y34vcq33e5p9qtw.png)
Therefore,
The final answer is
![P(x)=x^3-7x^2+2x+40](https://img.qammunity.org/2023/formulas/mathematics/college/vxsasremrnzux6xn9o6sijat56ciif5qyu.png)