Question 999760


a).  

{{{x^2+25= (x+5)(x+5)}}} => this is not equal,
{{{x^2+5^2<> (x+5)(x+5)}}} because {{{x^2+5^2}}} is the sum of squares and the rule for factoring  the sum of squares:{{{a^2+b^2}}} can be factorized as {{{(a+ib)(a-ib)}}}
if your factors were {{{(x-5i)(x+5i)}}}, it would work because {{{x^2+25=(x-5i)(x+5i)}}}
 
   b). 

{{{10x^3+15x^2+5x = 5x(2x^2+3x) }}}....missing one more term in parentheses

{{{10x^3+15x^2+5x = 5x(2x^2+3x+1) }}}
 
 
 
   c). 

{{{x^2-2x+24= (x-6)(x+4) }}}=> multiply {{{(x-6)(x+4) }}}=>{{{x^2+4x-6x-24=x^2-2x-24 }}}; as you can see {{{x^2-2x+24<> x^2-2x-24 }}}

{{{x^2-2x+24=(x^2-2x+b^2)-b^2+24=(x^2-2x+1^2)-1^2+24=(x-1)^2+23}}}; so, here we will have complex solutions again