Answer :
Step 1:
The equation given can be written as:
[tex]9x^2-64=0[/tex]Using the general formula:
[tex]x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]we have that:
[tex]\begin{gathered} x=\frac{-0\pm\sqrt[]{0^2-4(9)(-64)}}{2(9)} \\ x=\frac{\pm\sqrt[]{2304}}{18} \\ x=\frac{\pm48}{18} \\ x=\frac{\pm8}{3} \end{gathered}[/tex]This means that we can factor the equation as:
[tex]\begin{gathered} (x-\frac{8}{3})(x+\frac{8}{3})=0 \\ (3x-8)(3x+8)=0 \end{gathered}[/tex]Therefore the factorization is:
[tex](3x-8)(3x+8)=0[/tex]Another way to do this is by noticing that the equation:
[tex]9x^2-64=0[/tex]is a difference of squares, using the general formula for difference of squares:
[tex]x^2-y^2=(x-y)(x+y)[/tex]Using this we get the same result as before:
[tex](3x-8)(3x+8)=0[/tex]Step 2:
From the factorization and using the fact that the multiplication between two numbers can only be zero if one of them is zero we have that:
[tex]\begin{gathered} (3x-8)(3x+8)=0 \\ \text{if and only if} \\ 3x-8=0 \\ or \\ 3x+8=0 \end{gathered}[/tex]Solving each linear equation we have that:
[tex]\begin{gathered} 3x-8=0 \\ 3x=8 \\ x=\frac{8}{3} \end{gathered}[/tex]or
[tex]\begin{gathered} 3x+8=0 \\ 3x=-8 \\ x=-\frac{8}{3} \end{gathered}[/tex]Therefore the solutions of the equation are x=8/3 and x=-8/3