Answer:
The step used (x-5)(x-3)=0
C is correct
Step-by-step explanation:
Given: [tex]8x=x^2+15[/tex]
We need to factor it.
First we move all term same side and then we factor is.
[tex]x^2-8x+15=0[/tex]
Now we split middle term
[tex]x^2-5x-3x+15=0[/tex]
Make group and factor each group
[tex](x^2-5x)+(-3x+15)=0[/tex]
[tex]x(x-5)-3(x-5)=0[/tex]
[tex](x-5)(x-3)=0[/tex]
Hence, The step used (x-5)(x-3)=0
