From the problem, we have :
[tex]f(x)=x^2+8x-32[/tex]Set f(x) = 0 :
[tex]0=x^2+8x-32[/tex]Add 32 to both sides of the equation :
[tex]\begin{gathered} 32=x^2+8x-32+32 \\ 32=x^2+8x \end{gathered}[/tex]Next is to add a number to both sides of the equation using the formula (b/2a)^2
That will be :
[tex](\frac{b}{2a})^2=(\frac{8}{2})^2=16[/tex]The equation will be :
[tex]\begin{gathered} 32+16=x^2+8x+16 \\ 48=x^2+8x+16 \end{gathered}[/tex]Take note that the right side of the equation is now a perfect square trinomial and we can factor it by :
[tex]\begin{gathered} 48=x^2+8x+16 \\ 48=(x+4)^2 \\ (x+4)^2=48 \end{gathered}[/tex]Take the square root of both sides :
[tex]\begin{gathered} \sqrt[]{(x+4)^2}=\sqrt[]{48} \\ x+4=\pm4\sqrt[]{3} \end{gathered}[/tex]Subtract 4 to both sides of the equation :
[tex]\begin{gathered} x+4-4=-4\pm4\sqrt[]{3} \\ x=-4\pm4\sqrt[]{3} \end{gathered}[/tex]Take note that zeros and x-intercepts are the same.
Therefore, the answer is :
A. The zeros and the x-intercepts are the same. They are -4 + 4√3 and -4 - 4√3