ANSWER:
A.
[tex]x=-1,-3,11[/tex][tex]f(x)=(x+3)(x-11)(x+1)[/tex]
STEP-BY-STEP EXPLANATION:
We have the following function:
[tex]f(x)=x^3-7x^2-41x-33[/tex]
To find the zeros of the function we must set the function equal to 0 in the following way:
[tex]x^3-7x^2-41x-33=0[/tex]
We reorganize the equation in order to be able to factor and calculate the zeros of the function, like this:
[tex]\begin{gathered} x^3-7x^2-41x-33=0 \\ -7x^2=-8x^2+x^2 \\ -41x=-33x-8x \\ \text{ Therefore:} \\ x^3-8x^2+x^2-33x-8x-33=0 \\ x^3-8x^2-33x=-x^2+8x+33 \\ x(x^2-8x-33)=-(x^2-8x-33) \\ x^2-8x-33 \\ -8x=3x-11x \\ x^2+3x-11x-33 \\ x(x+3)-11(x+3) \\ (x+3)(x-11) \\ \text{ we replacing} \\ x(x+3)(x-11)=-1 \\ x(x+3)(x-11)+(x+3)(x-11)=0 \\ (x+3)(x-11)(x+1)=0 \\ x+3=0\rightarrow x=-3 \\ x-11=0\rightarrow x=11 \\ x+1=0\rightarrow x=-1 \end{gathered}[/tex]
Therefore, the zeros are:
[tex]x=-1,-3,11[/tex]
And in its factored form the expression would be:
[tex]f(x)=(x+3)(x-11)(x+1)[/tex]
