Answer:
• 1st zero: x=-2/3 with a multiplicity of 5.
,
• 2nd zero: x=5 with a multiplicity of 2.
Explanation:
Given the function:
[tex]F\mleft(x\mright)=\mleft(3x+2\mright)^5\mleft(x^2-10x+25\mright)[/tex]
First, factorize the quadratic expression: x²-10x+25
[tex]\begin{gathered} x^2-10x+25=x^2-5x-5x+25 \\ =x(x-5)-5(x-5) \\ =(x-5)(x-5) \end{gathered}[/tex]
Therefore, we can rewrite F(x) as:
[tex]F(x)=(3x+2)^5(x-5)^2[/tex]
Solving for the zeroes:
[tex]\begin{gathered} (3x+2)^5=0 \\ \text{Subtract 2 from both sides} \\ 3x=-2 \\ \text{Divide both sides by 3} \\ x=-\frac{2}{3}\text{ (5 times)} \end{gathered}[/tex]
• 1st zero: x=-2/3 with a multiplicity of 5.
[tex]\begin{gathered} (x-5)^2=0 \\ x=5\text{ (twice)} \end{gathered}[/tex]
• 2nd zero: x=5 with a multiplicity of 2.
