To find the zeros in a function you have to find those values that make that:
[tex]f(x)=0[/tex]In this case you have an cubic equation so first we are going to factorize it:
[tex]f(x)=x^3-x^2-5x+125^{}[/tex]Knowing that 125 is equal to:
[tex]125=5^3[/tex]this function can be expressed like:
[tex]f(x)=(x^3+5^3)-x^2-5x[/tex]The sum of cubes is:
[tex]a^3+b^3=(a+b)(a^2-ab+b^2)[/tex]We can use this to the first part of the equation:
[tex](x^3+5^3)=(x+5)(x^2-5x+25)[/tex]Now we can factorize the other part as follow:
[tex]-x^2-5x=-x(x+5)[/tex]So the equation now is:
[tex]f(x)=\text{ }(x+5)(x^2-5x+25)-x(x+5)[/tex]Now we can factorize the (x+5) as a common term
[tex](x+5)(x^2-5x+25-x)[/tex]And if we organice this one we get:
[tex]f(x)=(x+5)(x^2-6x+25)[/tex]using he first part ( x + 5) we can find one zero, as follow:[tex]x+5=0[/tex][tex]x=-5[/tex]Finally we can use the quadratic equation in the second part we can find the other zeros, as follow:
[tex]x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex][tex]x=\frac{-(-6)\pm\sqrt[]{(-6)^2-4(1)(25)}}{2(1)}[/tex][tex]x=\frac{6\pm\sqrt[]{36-100}}{2}=\frac{6\pm\sqrt[]{-64}}{2}[/tex]As we get a root for a negative munber we can use the imaginary number:
[tex]\sqrt[]{-64}=8i[/tex]So:
[tex]x=\frac{6\pm8i}{2}[/tex]The zeros are now calculated with the two solutions ( + and -)
[tex]x_1=\frac{6+8i}{2}=3+4i[/tex][tex]x_2=\frac{6-8i}{2}=3-4i[/tex]Then so, the zeros of the function
[tex]f(x)=x^3-x^2-5x+125^{}[/tex]are:
- 5
3 + 4i
3 - 4i
option B
