Answer:
• 1st zero: x=-2/3 with a multiplicity of 5.
,
• 2nd zero: x=5 with a multiplicity of 2.
Explanation:
Given the function:
![F\mleft(x\mright)=\mleft(3x+2\mright)^5\mleft(x^2-10x+25\mright)](https://img.qammunity.org/2023/formulas/mathematics/college/dlx3gvw3n1lnxxfwng3jdp4hzsfp3eq6sr.png)
First, factorize the quadratic expression: x²-10x+25
![\begin{gathered} x^2-10x+25=x^2-5x-5x+25 \\ =x(x-5)-5(x-5) \\ =(x-5)(x-5) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/ucwziuiw6qg39vq59cizru2h397ej5sa7u.png)
Therefore, we can rewrite F(x) as:
![F(x)=(3x+2)^5(x-5)^2](https://img.qammunity.org/2023/formulas/mathematics/college/mvzcbimxe27hthq56vk4ky6f3mjtvhubsh.png)
Solving for the zeroes:
![\begin{gathered} (3x+2)^5=0 \\ \text{Subtract 2 from both sides} \\ 3x=-2 \\ \text{Divide both sides by 3} \\ x=-(2)/(3)\text{ (5 times)} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/beqz1z30rtkue1lo0u53b7vkptg43wazkx.png)
• 1st zero: x=-2/3 with a multiplicity of 5.
![\begin{gathered} (x-5)^2=0 \\ x=5\text{ (twice)} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/qd3kwvwhkxcjrtltqnybvghdrdezknqztp.png)
• 2nd zero: x=5 with a multiplicity of 2.