Question 405860
{{{x^2-10x+24=0}}}
Here's how I do it:
{{{x^2 - 10x = -24}}}
Take 1/2 of the coefficient of {{{x}}}, square it, and add it
to both sides.
{{{x^2 - 10x + (-10/2)^2 = (-10/2)^2 - 24}}}
{{{x^2 - 10x + 25 = 25 - 24}}}
{{{(x - 5)^2 = 1}}}
Take the square root of both sides
{{{x - 5 = 1}}}
{{{x = 6}}}
and, also
{{{x - 5 = -1}}} (the other square root of 1)
{{{x = 4}}}
from these, I get {{{x-6 = 0}}} and {{{x-4 = 0}}}, so
The factors are
{{{(x - 6)*(x - 4)}}}