Answer :
Given the zeros:
• -1 with a multiplicity of 2
,• -3
,• -5
Let's find the equation of the polynomial.
Using the given zeros, we have the following:
x = -1, x = -1, x = -3, x = -5
Now, rewrite each zero as a factor:
[tex]\begin{gathered} x=-1 \\ Add\text{ 1 to both sides:} \\ x+1=-1+1 \\ x+1=0 \\ \\ x=-1 \\ x+1=0 \\ \\ x=-3 \\ \text{ Add 3 to both sides:} \\ x+3=-3+3 \\ x+3=0 \\ \\ \\ x=-5 \\ \text{ Add 5 to both sides:} \\ x-5+5=-5+5 \\ x+5=0 \end{gathered}[/tex]Now, we have the equation:
[tex]f(x)=(x+1)(x+1)(x+3)(x+5)[/tex]Let's expand the equation usng the FOIL method and distributive property:
[tex]\begin{gathered} f(x)=((x+1)(x+1))(x+3)(x+5) \\ \\ f(x)=(x(x+1)+1(x+1))(x+3)(x+5) \\ \\ f(x)=(x^2+x+x+1)(x+3)(x+5) \end{gathered}[/tex]Solving further:
[tex]\begin{gathered} f(x)=(x^2+2x+1)(x+3)(x+5) \\ \\ f(x)=(x(x^2+2x+1)+3(x^2+2x+1))(x+5) \\ \\ f(x)=(x^3+2x^2+x+3x^2+6x+3)(x+5) \\ \\ \end{gathered}[/tex]Combine like terms:
[tex]\begin{gathered} f(x)=(x^3+2x^2+3x^2+x+6x+3)(x+5) \\ \\ f(x)=(x^3+5x^2+7x+3)(x+5) \end{gathered}[/tex]Now, apply FOIL method and distributive property once more:
[tex]\begin{gathered} f(x)=x(x^3+5x^2+7x+3)+5(x^3+5x^2+7x+3) \\ \\ f(x)=x^4+5x^3+7x^2+3x+5x^3+25x^2+35x+15 \\ \\ \text{ Combine like terms:} \\ f(x)=x^4+5x^3+5x^3+7x^2+25x^2+3x+35x+15 \\ \\ f(x)=x^4+10x^3+32x^2+38x+15 \end{gathered}[/tex]Therefore, the equation of a polynomial with the given zeros is:
[tex]f(x)=x^4+10x^3+32x^2+38x+15[/tex]• ANSWER: C
[tex]f(x)=x^{4}+10x^{3}+32x^{2}+38x+15[/tex]